NationStates Jolt Archive

Some people, and some mathmaticians would tell you that 0.9reccuring EQUALS 1

Infact some mathematical concepts rely on this

For example 3 thirds 'added together' on a calculator will equal 1

However, if a number (such as 1 devided by 3 eg 0.3r) is reccuring how can we define it?

A contradiction to this beliefe in mathematics is an asymptote, when we have a curve 'approaching' an x or y value but never reaching it, we say this line gets infinately close to the line but never touches it.

Another opinion shared by some is that 0.9r gets so close to 1 that it might aswell be one.


This debate raises many issues, the concept of infinity, the question of nature relating to mathematics (can infinity exist in the natural world? this raises many sub issues)

There is some 'proof' for and against 0.9r =1, but this is a question of concept and even philosphy.

Feel free to use mathematical proof in your argument for and against however, as this allways raises new issues.

I will start by stating my opinion that 0.9r does not = 1, by definition there is allways a slight difference between 0.9r and 1 however infinately small, since this cannot exist in reality but only in concept you must take it as concept and the concept 'states' that there is an infinate number of '9's on the end and therefore an infinately small difference between 1 and 0.9r

Furthermore if a number is undefinable then it does not exist, 0.9r is an impossible concept.

Ill leave it there for now, interested to hear other opinions

0.99999r * 10 = 9 + 0.999999r

1 * 10 = 9 + 1

so therefor 0.99999r = 1 because it's not possible for 2 distinct numbers to have the exact same properties

Actually, a friend of mine and I got together and tried to figure out the Rules of Dimentional Existence. Being that everyone knows the 1st to 3rd Dimention, we continued until 9 Dimentions. I won't describe them all now (you can email me if you want to discuss it) but the interesting thing is:


Furthermore if a number is undefinable then it does not exist, 0.9r is an impossible concept.


We concluded that all objects exist, wheter a True Object (TO+/-CO) or Conceived Object (CO). The fact that there is something less than 1 but greater than 0 is a reality, and the "fact" that .9r = 1 is "reality" because someone thinks so. However, .9r is a CO, not necessarily a TO. Likewise, simply because it is concieved does not mean it is true, although the fact that you do not "know" it is true does not rule out its possibility. If you ran thru the 9 dimensions (or as we call it, "The Nine Rules For Existence") you would find yourself at the paradox of the 9th Dimension, being- "Does the Object Exist in Reality, or in Conception, or in Both?"

Don't worry, it's a paradoxal proposition, but I'd be glad to run it thru off-line.

Interestingly the proof usually used for this is the thirds argument. However

A third is not actually 0.3333333r. As you have pointed out if this is multiplied by three then it doesn't quite make one.
As by definition a third multiplied by three HAS to make one, then this is merely a representation of a third which can be used to help us understand what the number is.
The only true definition of a third can be 1/3.

Although this doesn't specifically address the argument, it does help to disprove one of the vital points in the argument.

Therefore while 0.99999r may or may not be 1, it doesn't have any properties in relation to fractions.

there are many proofs for this, for example sum to infinity also proves this

As i said its more a case of philosophy, 0.9r is non definable

for example, devide 1 by 3 and you are left with 0.3r... however get ten stones, arrange them in a circle and try to split that circle into 3 equal parts. Now on the assumption that there is some component in the physical world that makes up all other components, particles atoms etc... this is impossible in the natural world... the concept of 0.9r in itself suggests that either mathematics is flawed with respect to applying it to physics, or that physical things that exist in the universe can potentially be infinately small.

If a man travels half the distance between himself and a wall he is walking towards each step, does he ever arrive at the wall?


Also, can you have 0.9r apples? Does that even make sense? lol

Edit> wow you ppl post too fast lol! this was in response to the second post

Which also proves a point many people will be well aware of. That the majority of maths bears no resembelance what so ever to real life. Although various branches of maths coincide with real life problems, the most pure mathmatical ideas deal only with concepts. For example imaginary numbers, although the name gives it away somewhat in that case.

If a man travels half the distance between himself and a wall he is walking towards each step, does he ever arrive at the wall?
Can you travel half a distance? Or do you just travel a distance with half it's lenght?

Which also proves a point many people will be well aware of. That the majority of maths bears no resembelance what so ever to real life. Although various branches of maths coincide with real life problems, the most pure mathmatical ideas deal only with concepts. For example imaginary numbers, although the name gives it away somewhat in that case.
Imaginary numbers were used by the greeks (I think) to solve third power equations.

x = 0.9999r
10x = 9.9999r
10x - x = 9.9999r - 0.9999r
9x = 9
x = 1

Indeed, they have a use, but they are just a concept.

Indeed, they have a use, but they are just a concept.
Civilization is built on concepts.

Indeed, they have a use, but they are just a concept.

imaginary numbers are used in position vectors, because the properties of imaginary numbers are accurate representations of 2D geometry. this is used in all branches of physics and sciences to calculate position, velocity, acceleration, etc.

if you want to get technical about this, all words in language are concepts too. the words "love," "anger," even "red" or "table" are concepts and ideas just like the number sqrt(-1).

he travels half the distance between the wall and the point he is at after taking the previous step...

Again this is hypothetical, the question is is it possible to take infinately small steps?

But if
x = 0.999999r
9x = 0.999999r1

So which is right?

imaginary numbers are used in position vectors, because the properties of imaginary numbers are accurate representations of 2D geometry. this is used in all branches of physics and sciences to calculate position, velocity, acceleration, etc.

if you want to get technical about this, all words in language are concepts too. the words "love," "anger," even "red" or "table" are concepts and ideas just like the number sqrt(-1).

Yep, but mathmatically most things have a more substantial base in reality than imaginary numbers and 0.999999r

But if
x = 0.999999r
9x = 0.999999r1

So which is right?
Not true..... there is no 1 at the end, because the number has no end


oh and it's 8.9999999r

So what happens to the small difference between 0.9999999r and 1

Yep, but mathmatically most things have a more substantial base in reality than imaginary numbers and 0.999999r
maybe certain things are more easily comprehended through experiences in physical existence, but ideas exist and are just as real as a tangible physical object like a rock.

he travels half the distance between the wall and the point he is at after taking the previous step...

Again this is hypothetical, the question is is it possible to take infinately small steps?
We don't know yet, distance may either be naturally discrete or continuous... if distance is discrete then we can't take infinitely small steps, but if it is continuous then we can. However no research has determined whether distance is discrete or not.

oh and it's 8.9999999r

Yeah, sorry typo

So what happens to the small difference between 0.9999999r and 1
By definition .999~ = 1. There is no small difference between those two representations. They are one and the same. 0.999~ is just the best decimal representation of 1/3. the apparent unwieldiness of .9999999~ reveals problems and uncertainties with the way we write out numbers, not with the numbers themselves.

Put it this way... If you have 0.9999999r cents in your bank account, then no matter how many of those little nines you have, you're never going to be able to buy anything worth a cent.

Half distances can't be travelled. You travel a distance with half the length

for example, devide 1 by 3 and you are left with 0.3r... however get ten stones, arrange them in a circle and try to split that circle into 3 equal parts. Now on the assumption that there is some component in the physical world that makes up all other components, particles atoms etc...
There is no scientific proof that you can base such an assumption on. We don't know yet what the smallest possible distances or divisions of matter are.

this is impossible in the natural world... the concept of 0.9r in itself suggests that either mathematics is flawed with respect to applying it to physics, or that physical things that exist in the universe can potentially be infinately small.

Current mathematics and physics are both flawed and incomplete. Researchers simply have not figured out all the workings of the universe and all mathematicaal relationships. Yes, physical things potentially can be infinitely small. But we don't know for sure, just like we don't know for sure what is in the middle of the earth. We have good guesses, but until definitive experiments are carried out, you can't say for sure either way.

This debate is very interesting, especialy to someone who is studying A-Level maths(me). However the way i learnt maths, and indeed still learning is that when working something out i always stick to fraction as apposed to decimal places, and then convert if possible(or use that damm handy F-D button on your scientific calculator)

Some people, and some mathmaticians would tell you that 0.9reccuring EQUALS 1

Infact some mathematical concepts rely on this

For example 3 thirds 'added together' on a calculator will equal 1

However, if a number (such as 1 devided by 3 eg 0.3r) is reccuring how can we define it?

A contradiction to this beliefe in mathematics is an asymptote, when we have a curve 'approaching' an x or y value but never reaching it, we say this line gets infinately close to the line but never touches it.

Another opinion shared by some is that 0.9r gets so close to 1 that it might aswell be one.


This debate raises many issues, the concept of infinity, the question of nature relating to mathematics (can infinity exist in the natural world? this raises many sub issues)

There is some 'proof' for and against 0.9r =1, but this is a question of concept and even philosphy.

Feel free to use mathematical proof in your argument for and against however, as this allways raises new issues.

I will start by stating my opinion that 0.9r does not = 1, by definition there is allways a slight difference between 0.9r and 1 however infinately small, since this cannot exist in reality but only in concept you must take it as concept and the concept 'states' that there is an infinate number of '9's on the end and therefore an infinately small difference between 1 and 0.9r

Furthermore if a number is undefinable then it does not exist, 0.9r is an impossible concept.

Ill leave it there for now, interested to hear other opinions
Mathematicians are anal. In engineering, it's close enough.

By definition .999~ = 1. There is no small difference between those two representations. They are one and the same. 0.999~ is just the best decimal representation of 1/3. the apparent unwieldiness of .9999999~ reveals problems and uncertainties with the way we write out numbers, not with the numbers themselves.

I know that by definition 0.9999999r is one. But that doesn't mean that it works perfectly.

The fact that we can actually have this discussion reveals that 0.999999rm isn't 1. If it was then there would be no debate.

well

0.000r1 = 0 as its infinate zeros with a one on the end. ( so theoreticly you never get round to putting the 1 on the end.)

For example 3 thirds 'added together' on a calculator will equal 1


i agree with you on the rest, but a third is 1/3 of 1, that's why 3 thirds is 1, not .999999, and it's only shown as .3333 cause that's the closest the calculator can get to 1/3 without fractions.

I know that by definition 0.9999999r is one. But that doesn't mean that it works perfectly.
yes it does. In no mathematical situation will .99~ not equal 1. The concept is airtight and perfect. If you want to get into the nitty gritty details of philosophy, then lots of things break down, and the assumptions of mathematics are not self-evident, and even human language becomes illegitimate because the argument goes that people percieve things differently. In philosophy I can plausibly argue that I'm hooked into the Matrix and you are all computer simulations to delude me so the robots can harvest my energy, and you can't prove me wrong, but I can't prove you wrong either. But math isn't philosophy, and math has certain universally agreed upon definitions that can't be altered. For example, .9~ will always be 1.

The fact that we can actually have this discussion reveals that 0.999999rm isn't 1. If it was then there would be no debate.
in the 1600s there was debate about whether the earth went around the sun. the presence or absence of discussions prove nothing. in this case the presence of a discuision only suggests surprisingly large numbers of people aren't aware of the fact .9~ = 1.

There isn't always a direct connection between a fraction and the decimal representation.

If 5/9 = .55555r

and

If 8/9 = .88888r

then

What is 9/9 = ? .9999999? 1?

For all practical purposes, .9999r = 1. In reality, .9999r is the closest any number can get to 1 without going over (kinda like if this was the price is right).

.333r * 3 = .999r =/= 1

1/3 * 3 = 1

.333r can never fully represent 1/3.

There isn't always a direct connection between a fraction and the decimal representation.
THAT WHY THEY'RE FRACTIONS!

Jeez :rolleyes:

If a man travels half the distance between himself and a wall he is walking towards each step, does he ever arrive at the wall?
No, but that's because you're not properly stating this well known problem as you include the concept of 'stepping' and so as defined your man is decelerating.


Also, can you have 0.9r apples? Does that even make sense? lol


Yes because 0.9r=1! The proof that 0.9r = 1 is basic but the misconceptions and misunderstandings that arise from the use of infinite processes are fairly common.

The proof can be found on Wikipedia (http://en.wikipedia.org/wiki/Recurring_decimal).

Some people, and some mathmaticians would tell you that 0.9reccuring EQUALS 1

Infact some mathematical concepts rely on this

For example 3 thirds 'added together' on a calculator will equal 1

However, if a number (such as 1 devided by 3 eg 0.3r) is reccuring how can we define it?

A contradiction to this beliefe in mathematics is an asymptote, when we have a curve 'approaching' an x or y value but never reaching it, we say this line gets infinately close to the line but never touches it.

Another opinion shared by some is that 0.9r gets so close to 1 that it might aswell be one.


This debate raises many issues, the concept of infinity, the question of nature relating to mathematics (can infinity exist in the natural world? this raises many sub issues)

There is some 'proof' for and against 0.9r =1, but this is a question of concept and even philosphy.

Feel free to use mathematical proof in your argument for and against however, as this allways raises new issues.

I will start by stating my opinion that 0.9r does not = 1, by definition there is allways a slight difference between 0.9r and 1 however infinately small, since this cannot exist in reality but only in concept you must take it as concept and the concept 'states' that there is an infinate number of '9's on the end and therefore an infinately small difference between 1 and 0.9r

Furthermore if a number is undefinable then it does not exist, 0.9r is an impossible concept.

Ill leave it there for now, interested to hear other opinions


Simple ... truncation roundoff error (Ie .9999 does not = .9999... )
That is what happens when you are not exact (see this all the time with storage variables in programming

A real number can be defined to be a "Cauchy sequence" of rational numbers, and two such sequences a_1, a_2... and b_1, b_2... define the same real number if the distance |a_1 - b_1| between corresponding terms converges to zero.

The infinite decimal 0.999. . . means the real number defined by the sequence of finite decimals 0, 0.9, 0.99, 0.999, . . . , while 1.000. . . means the sequence 1, 1.0, 1.00, 1.000, . . . (which is the constant sequence 1, 1, 1, . . . ). These two sequences describe the same real number because the differences between corresponding terms are 1-0 = 1, 1-0.9 = 0.1, 1-0.99 = 0.01, 1-0.999 = 0.001, . . . and this sequence is converging to zero.

Which also proves a point many people will be well aware of. That the majority of maths bears no resembelance what so ever to real life. Although various branches of maths coincide with real life problems, the most pure mathmatical ideas deal only with concepts. For example imaginary numbers, although the name gives it away somewhat in that case.
Imaginary numbers have applications to quantum physics.

Oh, wait... you said applications in real life. Nevermind. :D

A real number can be defined to be a "Cauchy sequence" of rational numbers, and two such sequences a_1, a_2... and b_1, b_2... define the same real number if the distance |a_1 - b_1| between corresponding terms converges to zero.

The infinite decimal 0.999. . . means the real number defined by the sequence of finite decimals 0, 0.9, 0.99, 0.999, . . . , while 1.000. . . means the sequence 1, 1.0, 1.00, 1.000, . . . (which is the constant sequence 1, 1, 1, . . . ). These two sequences describe the same real number because the differences between corresponding terms are 1-0 = 1, 1-0.9 = 0.1, 1-0.99 = 0.01, 1-0.999 = 0.001, . . . and this sequence is converging to zero.
Convergence does not mean attainment of the asymptote

Imaginary numbers have applications to quantum physics.

Oh, wait... you said applications in real life. Nevermind. :D
We call them complex number and they represent very real quantities in engineering.

Convergence does not mean attainment of the asymptote
What is this? Do I have to prove Cauchy now?

Yes because 0.9r=1! The proof that 0.9r = 1 is basic but the misconceptions and misunderstandings that arise from the use of infinite processes are fairly common.

The proof can be found on Wikipedia (http://en.wikipedia.org/wiki/Recurring_decimal).
I concur with Wdig. I'd try to explain this by talking about infinite sums, but Wikipedia seems to include that in its explanation, already.

--The Democratic States of Cogitation

What is this? Do I have to prove Cauchy now?
No you have to prove that infinitely close = same

It is all a matter of precision and where you decide to truncate (or round)

Similar behavior does not mean identically

x = 0.9999r
10x = 9.9999r
10x - x = 9.9999r - 0.9999r
9x = 9
x = 1


i think your reasoning is faulty here. Try this equation with another number, lets say .3333r

x=.3333r
10x=3.3333r
10x-x=3.3333r-.3333r
9x=3
x=3 .3333r=/=3


also, try it with .1111r, shows an interesting relationship, but i think invalidates your reasoning

If you have too many people with not enough to do, do you get more done than with one person that has too much to do?

Any idiot knows that 0.9r does not equal 1.
It takes a mathematician to "prove" that it does...
It takes an engineer to prove that it does not matter...

What's the next topic?

No you have to prove that infinitely close = same

It is all a matter of precision and where you decide to truncate (or round)

Similar behavior does not mean identically

That's what Cauchy already did. Look up his theorems of convergence. There's no sense in even arguing about it.

No you have to prove that infinitely close = same

It is all a matter of precision and where you decide to truncate (or round)

Similar behavior does not mean identically
So what is the difference between 1 and 0.9r??? What is 1-0.9r???
Myrmidonisia tells You no lie :)

9x=3
x=3 .3333r=/=3

I think that it should be rather:
9x=3
x=3/9=1/3 and .3r=1/3 :rolleyes:

Just pulling something out of my ass, in all fairness I just woke up.

What I think is that all of this is spawned from our base-10 number system. We consider 10 to be a more or less round number, probably coming from the number of fingers we have. However, different number bases has been used. In ancient china, for example, a base 6 system was used. So, one third in the base six number system would be .20, because the numbers seven, eight, and nine did not exist. So, counting would go 0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 14, 15, 16, 20 etc. etc. (I think).

I could be wrong, but it's just an idea.

You know, this problem goes away if the calculator you plug this into is in any way modern. 1/3 =0.33r. 2/3=0.67. 0.33 + 0.67 = 1.

There are any number of mathematical concepts that have only a passing interaction with reality, yet are perfectly valid expressions of the way numbers interact. I challenge you to look out your window and show me i, for instance. (i = the square root of -1.)


Edit: hexadecimal number systems aren't nearly the most consfusing I've ever seen. The ancient Sumerians had a number system that operated off alternating sixes and thirties, and it was only used when counting certain things. They had other numeral systems depending on what you were counting (You'd count cattle on a different base system than ships). I'd give examples, but it's headache-inducing.

i think your reasoning is faulty here. Try this equation with another number, lets say .3333r

x=.3333r
10x=3.3333r
10x-x=3.3333r-.3333r
9x=3
x=3 .3333r=/=3


also, try it with .1111r, shows an interesting relationship, but i think invalidates your reasoning

please note, if 9x=3, then x is not = 3

x = 0.3333r
10x = 3.3333r
10x-x = 3.3333r - 0.3333r
9x = 3
x = 3/9
3/9 = 1/3 = 0.3333r


Also, 2/3 = 1/3*2 = 0.3333r*2 = 0.6666r

please note, if 9x=3, then x is not = 3

x = 0.3333r
10x = 3.3333r
10x-x = 3.3333r - 0.3333r
9x = 3
x = 3/9
3/9 = 1/3 = 0.3333r


Also, 2/3 = 1/3*2 = 0.3333r*2 = 0.6666r

Been a long time since I've done proofs, but is that third move valid in a proof? Subtracting a variable on one side of the equation and a defined value on the other? Where does that come from?

please note, if 9x=3, then x is not = 3

x = 0.3333r
10x = 3.3333r
10x-x = 3.3333r - 0.3333r
9x = 3
x = 3/9
3/9 = 1/3 = 0.3333r


Also, 2/3 = 1/3*2 = 0.3333r*2 = 0.6666r


Actually, the math is wrong in that proof. You got the right answer, but you don't subtract an x, or 0.3333r either.

x = 0.3333r
10x = 3.3333r
10x/10 = 3.3333r/10
x = 0.3333r

x = 0.3333r

I AM subtracting an x, look: 10x - x

if I wanted to say that x = x, I wouldn't have any trouble proving that.

I'm proving that 0.3333r = 3/9 or 1/3


The variable is defined in the beginning as x = 0.3333r
IF x = 0.3333r then the proof can prove that 0.3333r = 1/3
IF x is not = 0.3333r then I can't use this equation to prove that 0.3333r = 1/3

Also, 3x = 1 when x = 0.3333r, seeing as how 9x = 3 and 3 / 3 = 1, 9x / 3 = 3x

x = 0.3333r

I AM subtracting an x, look: 10x - x

if I wanted to say that x = x, I wouldn't have any trouble proving that.

I'm proving that 0.3333r = 3/9 or 1/3


The variable is defined in the beginning as x = 0.3333r
IF x = 0.3333r then the proof can prove that 0.3333r = 1/3
IF x is not = 0.3333r then I can't use this equation to prove that 0.3333r = 1/3

But you don't subtract an x from the equation. It's an unnecessary step. Why subtract an x from both sides, then divide by 9 when all you have to do is divide by 10 to begin with?

x is indeed 0.3333r in that proof, but the way that proof gets there isn't correct.

let's try your way then...

x = 0.3r <= true, because we define x as being such
10*x = 3.3r <= true, because 0.3r*10=3.3r
10*x/10 = 0.3r <= true, because 10x/10=x
x = 0.3r

that doesn't get us anywhere

However:

x = 0.3r
10*x = 3.3r
10*x-x = 3.3r-0.3r <= this doesn't need to be proved, we're just subtracting after all, and we're allowed to do that.

In normal arithmetic then: 3.3r-0.3r = 3 and 10*x-x=9*x
so 9x = 3 <= true because 10*x-x = 3.3r-0.3r
so x = 1/3 <= true because 9x/9 = 3/9

Therefore 1/3 = 0.3r QED


If you can show me how dividing by ten gets 1/3 = 0.3r, please do
If you can point out a mistake somewhere, please do
Please note: we're using normal arithmetic.

The maths is perfectly fine and is a useful way of finding how to express recurring decimals as fractions as shown

let's try your way then...

x = 0.3r <= true, because we define x as being such
10*x = 3.3r <= true, because 0.3r*10=3.3r
10*x/10 = 0.3r <= true, because 10x/10=x
x = 0.3r

that doesn't get us anywhere

However:

x = 0.3r
10*x = 3.3r
10*x-x = 3.3r-0.3r <= this doesn't need to be proved, we're just subtracting after all, and we're allowed to do that.

In normal arithmetic then: 3.3r-0.3r = 3 and 10*x-x=9*x
so 9x = 3 <= true because 10*x-x = 3.3r-0.3r
so x = 1/3 <= true because 9x/9 = 3/9

Therefore 1/3 = 0.3r QED


If you can show me how dividing by ten gets 1/3 = 0.3r, please do
If you can point out a mistake somewhere, please do
Please note: we're using normal arithmetic.

Because you're not just dividing by ten. You're treating X differently just because it's on a different side of the equation. This is what happens when you divide by 10.

x=0.3333r

10x=3.3333r
10*0.3333r=3.3333r
3.3333r=3.3333r
3.3333r/10=3.3333r/10
0.3333r=.3333r
x=0.3333r

And here's what happens when you subtract

10x=3.3333r
10*.3333r=3.3333r
3.3333r=3.3333r
-.3333r = -.3333r
3 = 3
3/9 = 3/9
.3333r=.3333r

I don't see how the unessecary steps prove anything, since you defined x at the top of the equation anyway, and I'm not sure what relationship this equation has with .99r being equal to 1.

Because you're not just dividing by ten. You're treating X differently just because it's on a different side of the equation. This is what happens when you divide by 10.

x=0.3333r

10x=3.3333r
10*0.3333r=3.3333r
3.3333r=3.3333r
3.3333r/10=3.3333r/10
0.3333r=.3333r
x=0.3333r

And here's what happens when you subtract

10x=3.3333r
10*.3333r=3.3333r
3.3333r=3.3333r
-.3333r = -.3333r
3 = 3
3/9 = 3/9
.3333r=.3333r

I don't see how the unessecary steps prove anything, since you defined x at the top of the equation anyway, and I'm not sure what relationship this equation has with .99r being equal to 1.

I am trying to prove that 9x = 3

1x = 0.3r
10x = 3.3r

so far, we agree

10x-1x=3.3r-0.3r

do you disagree?

therefore: 9x = 3 OR x = 1/3
x = 1/3 AND x = 0.3r
This is the proof that 1/3=0.3r
1/3*3 = 3/3 = 1
0.3r*3 = 0.9r

but we know that 1/3 = 0.3r
Therefore: 1/3*3 = 0.3r*3
Therefore: 1 = 0.9r

Fallanour is right.

However, it is easier to see the relationship when you understand that .9r is actualy the sum of k = 0 to infinity of 9 9(1/10^k).
When you take the sum you get the

Limit as k-> infinity of

9/10(1-1/10^k) / (1-1/10) = (9/10)/(9/10) = 90/90 = 1


It isn't a matter of .9r being infinitly close to one, it is a matter of if you keep adding 9s infinitly you will end up with one.

Just pulling something out of my ass, in all fairness I just woke up.

What I think is that all of this is spawned from our base-10 number system. We consider 10 to be a more or less round number, probably coming from the number of fingers we have. However, different number bases has been used. In ancient china, for example, a base 6 system was used. So, one third in the base six number system would be .20, because the numbers seven, eight, and nine did not exist. So, counting would go 0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 14, 15, 16, 20 etc. etc. (I think).

I could be wrong, but it's just an idea.
China has always had a base 10 number system.

(NOTE -- If you say, express, or imply 'I don't understand this math, therefore it's incorrect', then go away. There are enough people trying to prove me wrong.)

Let .999... = .999~.

.999... = sigma(.9*[.1]^[n-1]).

.999... = .9 + .09 + .009 + .0009 + .00009 + .000009 ... (ad infinitum)
Let a, 1st term = .9
Let r, common ratio = 10-¹
Infinite geometric progression: a + ar + ar² + ar³ ... (ad infinitum)
Let the above be expression 'P'.
Omega P when | r | < 1 = a / (1 - r) = .9 / (1 - .1) = .9 / .9 = 1
.999... = 1

Let S = .999...
S = .9 + .09 + .009 + .0009 + .00009 + .000009 ... (ad infinitum)
S = 0.9 + (1/10)S
(9/10)S = .9
1 = S
.999... = 1

Thanks to Blizzard for this proof:
lim(m --> inf.) sum(n = 1)^m (9)/(10^n) = 1
0.999... = 1
Thus, x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999... (NOTE: You may substitute one for another)
9x = 9
x = 1
.999... = 1

.4242424242... = 42/99
.402402402402402... = 402/999
-.888... = -8/9
-.777... = -7/9
-.666... = -6/9
-.555... = -5/9
-.444... = -4/9
-.333... = -3/9
-.222... = -2/9
-.111... = -1/9
.000... = 0/9
.111... = 1/9
.222... = 2/9
.333... = 3/9
.444... = 4/9
.555... = 5/9
.666... = 6/9
.777... = 7/9
.888... = 8/9
.999... = 9/9
9/9 = 1
.999... = 1

.333... = sigma(n = 1, n -> inf.) 3/(10^n) = Definition of geometric series.
3 * sigma(n = 1, n -> inf.) (1/10)^n = Property of a series.
3 * 1/9 = (common ratio, r/[1 - r], r = .1)
3/9 = 1/3
1/3 = .333...
1/3 * 2 = 2/3
.333... * 2 = .666...
2/3 = .666...
.333... + .666... = .999...
1/3 + 2/3 = 3/3
3/3 = .999...
3/3 = 1
.999... = 1

--- Corollary Proof ---
.999.../3 = .333...
1/3 = .333...
.333... = .333...
.999.../3 = 1/3
.999... = 1

For two numbers to be different, there must be a number between them.
.999... < N < 1
Find N. Because .999... has infinite nines, there are no numbers between them, thus N cannot exist. .999... = 1. You may say that .000...1 is between them, but then the difference between .999... and 1 is an infinite number of zeroes. A number of this form would imply that infinity is finite, it cannot exist; if, after any quantity of 0's, you add a 1, you don't really have an infinite number of zeroes. Therefore, for the 1 to exist, the 0's must be finite, which they are not; therefore, .000...1 = 0 and the difference between .999... and 1 is zero. When the difference between two numbers is 0, they are the same.



ARGUMENTS THAT PISS ME OFF
But .999... can't exist! sigma(.9*[.1]^[n-1]).
.999... = .999... and 1 = 1. And .999... = 1. What's your point?
You can only get .999... = 1 by rounding! Show me ONE proof where ANYONE rounds and is still logically correct.
They are not equal because .999 =/= 1. You're right, .999 =/= 1, but .999... = 1.
There is always one nine of a difference. There are INFINITE nines. If you add another one, then nothing will happen, since there is already an infinite amount.
STFU, nobody cares. You do, obviously, since you posted in this topic. I do, obviously, since I'm copy + pasting this. The other posters care, obviously, because they posted in the topic. The topic creator cares, obviously, because he/she made the topic. *Note: I stopped making topics these topics. I only reply in them now.

DON'T TAKE MY WORD FOR IT?
http://mathforum.org/library/drmath/view/55748.html
http://mathforum.org/dr.math/faq/faq.0.9999.html
http://www.newton.dep.anl.gov/newton/askasci/1995/math/MATH070.HTM

Viola.

So what is the difference between 1 and 0.9r??? What is 1-0.9r???
Myrmidonisia tells You no lie :)
Thanks. I'm glad someone sees the advantage in simplicity.

Think of it this way: what can you add to 0.999999999r to make it 1?

Nothing. You would need a final decimal place to stick a 1 into, but it repeats infinitely so there's no such decimal place. You can add nothing to it to make it equal 1. And if A + 0 = B then it stands to reason that A = B.

As for defining it, Calculus takes care of that. First rule of limits is that lim(a->1)[a]=1. The limit of A as it approaches 1 is equal to 1.

One of my professors mentioned a field of mathematics called Analytical mathematics, i dont know a lot about this but apparantly they disected a lot of concepts (for example sum to infinity) and found problems, or exceptions in many of them... especially when dealing with undefinable numbers.

The point is, 0.9r is not a concrete concept agreed upon by the mathematical community, it is, and allways has been debated.

Apart from the idea that it doesnt exist in nature, there is the view that it doesnt (or shouldnt) exist in maths.

Yet again this is all about philosophy and the impact this concept ALONE has on the universe, i was hoping to discuss more the possible implications of making this comparison.

For example it 'RAISES' the question of 'infinately small ''objects'' ' or limits upon that, on infinately small spaces, or again limits on that, and again infinately small speeds... and again limits upon that.

It also raises questions when approaching probability, i was told by one of my tutours that sometimes in stats the probability of 1 is viewed as 0.9r (or sometimes this difference is called dx) because there is a certain philosophy which states nothing unless given as a condition (or is allready true) can ever have a probability of 1, however these conditions are used in special cases, and not very often.

Also can probabilities be infinately small??? If the difference between 1 and 0.9r is given as the difference between 1 and a probability is this a case of misuse of concepts or can probabilities actually be 'infinately small'....


Take all of the variables in the universe which may have an effect on any one event, potentially a star billions of lightyears away can have a miniscule effect on which newspaper you buy, whether directly or indirectly... however, is there a limit on these variables? Supposedly there is a limited amount of matter and energy in the universe... does this mean probability has an 'upper limit' (if you catch my drift)

I raise these points because they are interesting to me, and i would like to hear the opinions of people who are obviously more intelligent than me

Isn't this the concept of limits?

x = 0.9999r
10x = 9.9999r
10x - x = 9.9999r - 0.9999r
9x = 9
x = 1
Hey, that's great.

Also, .3r IS 1/3, and is NOT a mathematical representation. All a repeating decimal IS is a simplified fraction. So, .3r*3=.9r, and 1/3*3=1, so .9r=1.

it has been my experience that in the mathematics community is that

(1/infinity) = 0

therefore, if you have 0.999..., you are (1/infinity) or "one infinity-th" away from the next integer, 1.0..., therefore

1 - (1/infinity) = 1 - 0 = 0.999...

or simply,

1 = 0.999...

some sort of logic behind (1/infinity) = 0, would be that if the integer 1 was divided infinitely, each division would approach 0 in width...

yeah, its all about the limits

therefore,

0.999... DOES EQUAL 1.000...

This thread is bad for my hangover

Viola.
= stringed instrument

Think of it this way: what can you add to 0.999999999r to make it 1?

Nothing. You would need a final decimal place to stick a 1 into, but it repeats infinitely so there's no such decimal place. You can add nothing to it to make it equal 1. And if A + 0 = B then it stands to reason that A = B.

As for defining it, Calculus takes care of that. First rule of limits is that lim(a->1)[a]=1. The limit of A as it approaches 1 is equal to 1.

1 - 0.0r1 = 0.9r

To debate the existance of 0.0r1 is to debate the existance of 0.9r. You cant pick and choose.

1 - 0.0r1 = 0.9r

To debate the existance of 0.0r1 is to debate the existance of 0.9r. You cant pick and choose.

What exactly do you mean by existence of? They are both theoretical representations of real fractions. It would be pointless to debate the existence of theoretical things, which is why that's not what I'm doing.

What I am indicating is that, in the world of limits, there is nothing you can add 0.9r to make it 1. And if you can add nothing to it to make it 1, then it is the same as 1. Not all that complicated a deduction.

what about adding 0.2r and 0.8r do they not equal 1? (correct me if im wrong) on the other hand there is nothing you can add to 0.8r (apart from the aformentioned) to make it one and vice versa.... does this not indicate some difference between the concept of a recurring 0.9 and 1?

sorry if this is idiotic just my current train of thought

0.8r + 0.1r = 0.9r

Have you reached 1? Since you can't add anything else to the recurring fraction to arrive at 1, and since anything you do add will result in a number greater than 1, yes you have.

We don't know yet, distance may either be naturally discrete or continuous... if distance is discrete then we can't take infinitely small steps, but if it is continuous then we can. However no research has determined whether distance is discrete or not.
What about the Planck length? Wasn't that supposed to be the definitively smallest possible unit of distance? Or is it the smallest possible unit that has meaning before space and time break down?
I don't remember entirely, and haven't ever had that really explained.

.8 + .2 = 1
.88 + .22 = 1.1
as you can see, .888r + .222r would be 1.111r, so that is obviously not the case.

however, if you consider:
1-.8 = .2
1-.88 = .12
1-.888 = .112
etc...
then 1-.888r = .111r + an infinitely small amount, which we can just drop
hence you could approximate and say that 1-.888r = .111r
therefore .888r + .111r = 1
therefore .999r = 1

http://www.jwgh.org/miscwritings/math/division.html it can be proved in goemetry.
x = .9 reapeating both sides times 10
10x = 9.9 reapting
- x = -.9 subtration
9x = 9 identity propery and division
x = 9/9 = 1/1 = 1 nuff said

it works out. ANd alegebra has been proven so it works

yeah wurd my brain is on strike at the moment, i was not sure about that one :D

What exactly do you mean by existence of? They are both theoretical representations of real fractions. It would be pointless to debate the existence of theoretical things, which is why that's not what I'm doing.

What I am indicating is that, in the world of limits, there is nothing you can add 0.9r to make it 1. And if you can add nothing to it to make it 1, then it is the same as 1. Not all that complicated a deduction.

But there is something you can add to 0.9r to achieve 1.

0.9r + 0.0r1 = 1

though im on the 0.9r =/= 1 bandwagon, i must point out that that issue has been adressed

0.999999999r is exactly equal to one.

take for example the formula for the sum of an infinite geometric series:

Sum=first term/(1-common ratio)

in this instance the first term is 9/10 and the common ratio is 1/10 1-1/10 = 9/10

(9/10)/(9/10)=(9/10)(10/9)=90/90=1

call me an idiot but i dont understand the relation to 0.9r... forgive me i am tired

I don't expect everyone to get it, here let me put it another way (trust me, you're not an idiot)

.9999999999r = .9+.09+.009+.0009+...+9/(10^n)+... infinitely where 9/(10^n)= the nth term.

well 1/10 is a common ratio between terms, that is to say each new term is 1/10th of the previous term.

There is a geometric theorem that states when the common ratio between the terms of a geometric series is both greater than -1 and less than 1 then the sum of this infinitely long number is:

(a_1)/(1-r)

a_1 is the first term and r is the common ratio.

also note that the nth term is a_n therefore the first term is 9/(10^1) or .9 is that better?

By the way, you cannot add .0r1 to anything. I tried that way back in geometry, I found--and confirmed with a very qualified teacher--that point anything repeating followed by something else does NOT exist in a word it is a false number, one cannot use such an argument. It's like saying "what's five plus rabbit?" which is a pointless statement.

It's like saying "what's five plus rabbit?" which is a pointless statement lol one of my mathematics tutours allways speaks like that, your name wouldnt happen to be colin would it? :D

No, but I have a friend that calls himself that.

And while I'm here on these boards, are there any other controversial math things I can lay to rest? I've had enough experience with it to probably settle any disputes. Then again this is coming from someone who does math for fun and who was ready for more even after his BC Calculus AP-Exam.

Why don't you just switch from base 10 to base 9?

base nine? so we can have 0.1111111r? what would that do?

And while I'm here on these boards, are there any other controversial math things I can lay to rest?

well there is a discussion about probability, and whether probability can = 1 among other things in another thread...

(by the way is he irish with funny hair?! :D)

no, he's italian.

also, what do you mean by probablility = 1? Of course it can.

read the entire thread through, i have the distinct feeling that while you may not completely change your opinion you will at least question it...

Its not a discussion about eg 'i have an apple... probability that i have an apple = 1) it questions whether probability even exists... anyway dont take my word for it check it

what's the link.

base nine? so we can have 0.1111111r? what would that do?
Please forgive me, I know little about math. But what I was thinking was 1,2,3,4,5,6,7,8,10. 1/3 of 10 is 3. 3X3=10.

here

http://forums.jolt.co.uk/showthread.php?t=413987

reccomend reading from start to finish, otherwise it will get extremely confusing

the base indicates what each digit stands for.

11010 in base two indicates 1 times 2^4+1 times 2^3+0 times 2^2 + 1 times 2^1 + 0 times 2^0 = 16+8+2=26. likewise each position right of the decimal represents a negative power of the base. so 0.11 base 2 = 1*2^-1+1*2^-2=1/2+1/4=3/4

If you have too many people with not enough to do, do you get more done than with one person that has too much to do?

Any idiot knows that 0.9r does not equal 1.
It takes a mathematician to "prove" that it does...
It takes an engineer to prove that it does not matter...

What's the next topic?
I thought this needed repeating. It's always funny when a few highschool students on the internet try to disprove mathematics that has been accepted for a hundred years with semantics.

dealt with.

as for the quote above my last post, I agree some people think they can just state something and say it's true, but that's not necessarily so.

the base indicates what each digit stands for.

11010 in base two indicates 1 times 2^4+1 times 2^3+0 times 2^2 + 1 times 2^1 + 0 times 2^0 = 16+8+2=26. likewise each position right of the decimal represents a negative power of the base. so 0.11 base 2 = 1*2^-1+1*2^-2=1/2+1/4=3/4

Never the less dear, my point was, why go on about 0.9999r when you can simply change the rules?

I thought this needed repeating. It's always funny when a few highschool students on the internet try to disprove mathematics that has been accepted for a hundred years with semantics.

Is there anything wrong with questioning? Simply because something is accepted does nopt mean that everyone must automatically assymilate it without understanding it first...

For example, it was widely accepted that the earth was flat just imagine if no one questioned that!

Is there anything wrong with questioning? Simply because something is accepted does nopt mean that everyone must automatically assymilate it without understanding it first...

For example, it was widely accepted that the earth was flat just imagine if no one questioned that!
My point was not that you shouldn't question but that to question mathematical concepts you have to use mathematical tools. Everyone who has done this agrees (indeed have proved) that .9r=1. Everyone who disagrees has used words. ie but they is some infinitesimal difference blah blah blah I don't know what I'm talking about.

Which also proves a point many people will be well aware of. That the majority of maths bears no resembelance what so ever to real life. Although various branches of maths coincide with real life problems, the most pure mathmatical ideas deal only with concepts. For example imaginary numbers, although the name gives it away somewhat in that case.
amen

no worries, but remember, the stupidest questions are the ones that are not asked

ohh ooh wait, it takes a wise man to ask stupid questions

i dunno, both of thoes might be actual quotes but i might have made them both up, its 3:44 am

Never the less dear, my point was, why go on about 0.9999r when you can simply change the rules?

change the rules? You cannot change the laws of mathematics simply because you have a new base. Sure things will display differently, but that changes nothing. And besides, the ones digit is outside of base rules because it is the base to the 0 power times whatever the digit is which means that the digit is the number of ones, regardless of base.

change the rules? You cannot change the laws of mathematics simply because you have a new base. Sure things will display differently, but that changes nothing. And besides, the ones digit is outside of base rules because it is the base to the 0 power times whatever the digit is which means that the digit is the number of ones, regardless of base.

.3r and, by extension, .9r are all about display. .3r = 1/3 = 3^-1

Displayed in decimal 3^-1 =
sum of k = 1 to infinity 3*(1/10^k)

There is no way to properly display 3^-1 in decimal so we use .3r .

In a base three system, it is much easier.

3^-1 in base 3 = .1 or 1*3^-1


In decimal, 3(3^-1) = 3(.3r) = .9r

In base 3, 3(3^-1) = 3(.1) = 1

Base-10? Base-9? Hah, that's nothing. Let's go back to the Babylonian system, base-60.

.3r and, by extension, .9r are all about display. .3r = 1/3 = 3^-1

Displayed in decimal 3^-1 =
sum of k = 1 to infinity 3*(1/10^k)

There is no way to properly display 3^-1 in decimal so we use .3r .

In a base three system, it is much easier.

3^-1 in base 3 = .1 or 1*3^-1


In decimal, 3(3^-1) = 3(.3r) = .9r

In base 3, 3(3^-1) = 3(.1) = 1

I understand thatbut that still doesn't change the rules, it's just showing it in a different way.

I understand thatbut that still doesn't change the rules, it's just showing it in a different way.

It depends on what rules you are talking about. There are two sets of rules in mathmatics, the immutable rules that are grounded in real geometry and the arbitrary rules of notation.

There is such a thing as incorrect notation
The rules of notation are mutable and arbitrary, but they are necessary because without them no one would be able to understand any one else.

.3r is a probuct the arbitrary rules of notation. Using a different set of rules, those of base-3, gets rid of it.

It depends on what rules you are talking about. There are two sets of rules in mathmatics, the immutable rules that are grounded in real geometry and the arbitrary rules of notation.

There is such a thing as incorrect notation
The rules of notation are mutable and arbitrary, but they are necessary because without them no one would be able to understand any one else.

.3r is a probuct the arbitrary rules of notation. Using a different set of rules, those of base-3, gets rid of it.

With that I shall concur.

.999... = .9 + .09 + .009 + .0009 + .00009 + .000009 ... (ad infinitum)
Let a, 1st term = .9
Let r, common ratio = 10-¹
Infinite geometric progression: a + ar + ar² + ar³ ... (ad infinitum)
Let the above be expression 'P'.
Omega P when | r | < 1 = a / (1 - r) = .9 / (1 - .1) = .9 / .9 = 1
.999... = 1

Heh, I remember doing this in primary school. All you people who still claim that .9r=/=1 must have done it as well. So you try to tell me that they make all the poor children live in mistake???

Search me, all I know is it's true.

For me, they are indeed 2 seperate values. 0.9r =/= 1, because no matter how close you get, there is still an infinitely small value between 0.9r and 1. For terms of limits, yes, it would be 0, but it never truely hits that, making it different.

Next, the value 1/3 does NOT have an appropriate decimal value. Yes, it can be represented as 0.3r, which results in the 3 being trunkcated at infinity. For 1/3, this is fine. However, even at infinity, 2/3 would not trunkate. The next value after the trunkation at infinity would also be a 6, resulting in 2/3 actually being 0.6r7. Similarly, 3/3 would be 0.9r[10], which equals 1. So while the theory of a repeating number is to not trunkate, for a fraction, it must force an improper rounding based on infinty + 1, which is how you can get the supposed 1/3 as a decimal to equal 1 when multiplied by 3. A mere 0.9r does NOT however, invoke the fraction part unless it was explicitly stated that it were to be 1/3, making 0.9r =/= 1.

(Now to have everyone say that it doesn't make sense...)

this is what's traditionally called an asymtote. Approaching, but never actually reaching. Now, can there really be big infinities and small infinities?
is x>0 bigger than x>1?

I like the most common proof listed so far because it sidesteps both infinity and the quantity .000r1. It hasn't been stated that way, but that is the basis of the algebraic proof.

Let x = 0.99r be a non-terminating decimal. (This is the only part anyone should have any trouble with, once you realize that there is no end, all the complexities disappear.)

Then, we have x = 0.99r
Multiplying both sides by ten yields
10 x = 9.99r (Note that shifting the decimal place does not stop the repetition. It doesn't end, that is the definition)

Subtracting x from both sides yields
9x = 9.99r - x = 9 + .99r - .99r

Dividing by nine: 9x / 9 = 9 / 9
Simplifying: x = 1

No infinity, no dealing with the "end."


(Granted, limits work great, but you have to accept the precepts of limits. Since the precepts of limits were rejected by some, I resort to this proof.)

1) x = 0.9r
=> 10*x - x = 9.9r - 0.9r = 9
=> 9*x = 9
=> x = 1

=> no discussion possible

2) You can't just see 0.9r as any other (finite number) and say that it's not equal to one because there is a difference. by definition:

0.9r = limit(sum(9/10^i, i=1..n), n->infinity) = 1

And to be correct it should be treated like a limit, always.

3) Now where does the 'problematic concept' of 0.9r = 1 come from? Why is it a problem for some?

Someone has allready mentioned cauchy, and i'm not going to do it again. But the 'problem' lies with the decimal representation of a number. Real numbers can't always be represented decimaly like integers can be represented. From our experience with integer, and finite decimal representations, we (those that make the error) assume that each number has a unique decimal representation.

A decimal representation is not a number. You can construct a decimal respresentation from each real number using the 'constructual' definition (it's a definition that tells you how to construct the decimal representation) of a decimal representation.

It basicly comes down to this:

-You start with the interval [0,1] (once you go past the .)
-You split it up in 10 (for decimal) intervals of equal length, [a(i),a(i+1)] i =0..9
-If your number lies in the interval [a(i),a(i+1)], add the value of 'i' to the end.
-Apply the same principle to [a(i),a(i+1)]

A problem arises when your number is such an a(i) -> it will be in 2 intervals during one of your choices. In this case, you are free to chose and you will end up with 2 decimal representations of the same number.

If this happens, you can get 4.999999r, if you chose the 'lowest' interval, or with 5.000000r if you chose the 'higher' interval. They will remain the same number tho ...

a sum limit is unnecessary, it's just an infinite geometric series; that's all.

a sum limit is unnecessary, it's just an infinite geometric series; that's all.

an 'infinite series' is the limit of a sum ...

an infinite series is summed from a constant to infinity, the limit does = the sum, but using limit notation is unecessary. It's like trying to find the eigenvalues for an orthogonal projection, why on earth would you do it unless you want to be really picky.

Also true for proving limits with the epsilon delta definitions.

The limits are required in formal mathematics. Without them, you are either taking a shortcut, or are confusing someone, or are confused. Pick two.

The limits are required in formal mathematics. Without them, you are either taking a shortcut, or are confusing someone, or are confused. Pick two.

Shortcut, but its so much simplier. Formula for an infinite geometric series is

a/(1-r)

which =

limit n->infinity a(1-r^n)/(1-r)

However, the limit does get in the way when you're doing Maclurin power series of 1/(1-x) and its varients or tan^-1 (x)

1/(1-x) = 1 + x + x^2 -x^4 +x^5 .... ect.

which = f(0)/0! + f'(0)x/1! + f''(0)(x^2)/2! ect.

which does eventuall = sum k = 0 to infinity of x^k but since microprocessors can't understand limits they get confused. Aproximations are better that having a computer do a single opperation an infinite number of times.

The remark about the computer and approximation is more something for applied math, numerical algorithmes.

Maclurin power series? The last time i saw that (and the first time) was somewhere in 11th grade (for the americans out there). Taylor is our god.

The remark about the computer and approximation is more something for applied math, numerical algorithmes.

Maclurin power series? The last time i saw that (and the first time) was somewhere in 11th grade (for the americans out there). Taylor is our god.

Same thing, different x value.

Same thing, different x value.

Ah, now i remember. Maclurin series as a special case of Taylor.

My point was not that you shouldn't question but that to question mathematical concepts you have to use mathematical tools. Everyone who has done this agrees (indeed have proved) that .9r=1. Everyone who disagrees has used words. ie but they is some infinitesimal difference blah blah blah I don't know what I'm talking about.

if x=0.999r
10x-x<>9.999r-0.999r
9.999r0-0.999r<>9.999r-0.999r
8.999r1<>9

It has been said earlier in this post that the above fomula yields 9=9, however, this is not the case. When you multiply the x by ten, it shifts the decimal over one place, thus leading the infinite remainder a zero where x's remainder is 9. If this is not the case, then:

infinity + infinity <> 2 (infinity)

Now I could be wrong on this, it has happend before. I believe that:

infinity + 1 > infinity; infinity + 2 > infinity +1;

and don't believe:

infinity + infinity = infinity

if x=0.999r
10x-x<>9.999r-0.999r
9.999r0-0.999r<>9.999r-0.999r
8.999r1<>9

It has been said earlier in this post that the above fomula yields 9=9, however, this is not the case. When you multiply the x by ten, it shifts the decimal over one place, thus leading the infinite remainder a zero where x's remainder is 9. If this is not the case, then:

infinity + infinity <> 2 (infinity)

Now I could be wrong on this, it has happend before. I believe that:

infinity + 1 > infinity; infinity + 2 > infinity +1;

and don't believe:

infinity + infinity = infinity


infinity + 1 = infinity +2 = infinity


The way we prove this is simple.

(infinity + 1)/ infinity = 1+(1/infinity) = 1+0 = 1

(infinity + 2)/ infinity = 1+(2/infinity) = 1+0 = 1

Of course, that is incorrect notation since you techincally can't perform any operations on infinity directly. It requires taking a limit. Howveer, Newton did such things and got away with them.

Correct notation
The limit as b-> infinity of (b+1)/b = 1+(1/b) = 1
The limit as b-> infinity of (b+2)/b = 1+(2/b) = 1

When dealing with infinity all real numbers go to zero.

Incidently, infinity + infinity can be considered to be twice as large as infinity but still equal to infinity.

Limit as b-> infinity of b+b = 2b = infinity

Limit as b-> infinity of (b+b)/b = 1+1 = 2

(infinity + infinity)/infinity = 1+1 = 2

Infinity^2 is also larger than infinity even though both are equal to infinity.

Limit as b-> infinity of b^2/b = b = infinity
Limit as b-> infinity of b/b = 1 = 1

Limit as b-> infinity of (b^2 + b)/b = b+1 = infinity

So you are saying, if I have an infinite number of apples, and you have an infinite number of apples, we do have more apples, but yet we still have an infinite number of apples?

I guess my only question would be, how in the hell did we get all these apples?



I do still think:

x=0.999r
10x-x<>9.999r-0.999r

.9r is 1 by definition. The infinately small difference between the two numbers is also known as 0.

.9r is 1 by definition. The infinately small difference between the two numbers is also known as 0.

Since infinity cannot be defined in anything other than theory, the actual difference between 0.999r and 1 would be 0.000r1. However, in mathematical theory, I suppose the definition could be 0.

So you are saying, if I have an infinite number of apples, and you have an infinite number of apples, we do have more apples, but yet we still have an infinite number of apples?

I guess my only question would be, how in the hell did we get all these apples?



I do still think:

x=0.999r
10x-x<>9.999r-0.999r

That's pretty much correct. Although I have no idea what <> means in your notation. About equal to?

Actually, if you have an infinite pile and apples and I have an infinite pile of apples then we have 2 infinite piles of apples.

2 infinite piles of apples is more than 1 infinite piles of apples.

However, together they can never produce more than infinity apples because it is impossible to reach infinity.

If we put them together then redivided them we were still have 2 infinite piles of apples.

If we had one infinite pile of apples and tried to divide it amongst outselves, we would be unable to do it because there would always be only 1 infinite pile.

If we had an infinite number of apples and tried to divide it among an infinite number of people then everyone would have 1 apple.

If we had 2 infinite piles of apples and divided them amongst an infinite number of people then everyone would have 2 apples.



On the other hands, if we have an infinite number of infinite piles of apples and tried to divide it amongst an infinite number of people then everyone would have an infinite pile of apples.

Since infinity cannot be defined in anything other than theory, the "actual" difference between 0.999r and 1 would be 0.000r1. However, in mathematically theory, I suppose the definition could be 0.Numbers can only be defined in theory.

By the algebraic proof, the issue of infinity is sidestepped by thinking of it from a different perspective (non-terminating decimal). Thus, 0.000r1 is nonsensical as you have an "end" to something which does not end.

That's pretty much correct. Although I have no idea what <> means in your notation. About equal to?


The <> means is not equal to when you cannot make the equal sign with the slash.

If we had one infinite pile of apples and tried to divide it amongst outselves, we would be unable to do it because there would always be only 1 infinite pile.

If we had an infinite number of apples and tried to divide it among an infinite number of people then everyone would have 1 apple.

This isn't acually quite right. You can divide an infinite pile of apples between a finite number of people. To take the simple example of two people, you give all the odd numbered apples to the first person and all the even numbered apples to the second. Each person gets an infinite number of apples.

If we had an infinite number of people and an infinite number of apples, we could give each person as many apples as we want. Number the apples, and give the first n apples to the first person, the apples numbered n+1 to 2n to the second person, and so on. So the mth person gets the apples numbered from (m-1)n+1 to mn. By this scheme each person clearly has exactly n apples, and no one is left out since we can clearly identify which apples any prson got.

More generally, mathematicians think of two sets of being equal in size if we can systematically pair up elements from one set with elements from the other. So there are the same number of natural numbers (numbers like 1, 2, 3.. etc) as there are even natural numbers since we can match 1 up with 2, 2 up with 4, 3 up with 6, and n up with 2n.

A set is called countable if you can match it up with the natural numbers this way, and believe it or not there are uncountable sets :D.

Ok, that's enough babbling for me. I hope people find this useful or interesting :)

Base-10? Base-9? Hah, that's nothing. Let's go back to the Babylonian system, base-60.
Is that where the 60 minutes per hour came from?

The <> means is not equal to when you cannot make the equal sign with the slash.

Ahh.. that makes more since. I use =! because that is what every software caculator I have used recognizes.

9.999r does = 10(0.99r)

This is proven by taking the sum

The sum of k = 1 to infinity of 9/(10^k) =
(9/10) + (9/100) + (9/1000) +....... =
.9 +.09 +.009 +..... = .9r

10 X The sum of k = 1 to infinity of (9/10^k) =
The sum of k= 1 to infinity of 90/(10^k) =
The sum of k= 1 to infinity of 9/(10^[k-1])=
(9/1) + (9/10) + (9/100) +..... =
9 +.9 +.09 +...... = 9.9r

For me, they are indeed 2 seperate values. 0.9r =/= 1, because no matter how close you get, there is still an infinitely small value between 0.9r and 1. For terms of limits, yes, it would be 0, but it never truely hits that, making it different.Why are you animating numbers? The difference between 0.9r and 1 never reaches 0 because it's already there. This is not a 2-dimensional problem. You're confusing concepts.

Ok, to make a post that's slightly more on topic, the fact that .9999r=1 is essentially a matter of definition. We choose this definition because otherwise we get nonsensical results. For instance, if we don't require that .999r=1, then we know that 1/3=0.333r so 3*(1/3)=.999r just by the standard rules of multiplication (I'm glossing some subtle points over here, like why 1/3=.333r exactly), but 3*(1/3)=1 by the definition of 1/3, since 1/3 is defined as the multiplicative inverse of 3. So if we don't require that 0.999r=1 then we end up with a logically inconsistent numbering system.

Needless to say, that's a bad thing.

This isn't acually quite right. You can divide an infinite pile of apples between a finite number of people. To take the simple example of two people, you give all the odd numbered apples to the first person and all the even numbered apples to the second. Each person gets an infinite number of apples.

If we had an infinite number of people and an infinite number of apples, we could give each person as many apples as we want. Number the apples, and give the first n apples to the first person, the apples numbered n+1 to 2n to the second person, and so on. So the mth person gets the apples numbered from (m-1)n+1 to mn. By this scheme each person clearly has exactly n apples, and no one is left out since we can clearly identify which apples any prson got.


But I wasn't talking about dividing up the apples I was talking about dividing up the pile. The pile itself is infinite.

If you give a all the odd apples to person A and all the even apples to person B and keep going till the end of time persons A and B have a finite countable number of apples while the original pile is still infinite. The pile will never get smaller therefore it cannot be evenly divided this way.

This is, of course, a logical distinction. Mathmatically you are correct. Infinity/2 = Infinity.

In the second example you are making a logical distinction instead of a mathmatical one. Infinity/Infinity = 1 mathamticly. Although, logically, you are correct.

I can show this by taking the Limit as B-> infinity of B apples/B people = 1 apple/1 person

Taking the limit as B-> infinity of the sum of K= 1 to infinity of k apples/B people produces
0+0+0+0+0+0+0+...+1 = 1

As long as the number of people and the number of apples approach infinity at the same rate the ratio will always be 1.

I'll rephrase my only point to which noone has replied yet. It simply states that 1.0r and 0.9r are simply two decimal representations of the same number that we know as '1' (the only number that has the property 1*X = X, for all X)



3) Now where does the 'problematic concept' of 0.9r = 1 come from? Why is it a problem for some?

Someone has allready mentioned cauchy, and i'm not going to do it again. But the 'problem' lies with the decimal representation of a number. Real numbers can't always be represented decimaly like integers can be represented. From our experience with integer, and finite decimal representations, we (those that make the error) assume that each number has a unique decimal representation.

A decimal representation is not a number. You can construct a decimal respresentation from each real number using the 'constructual' definition (it's a definition that tells you how to construct the decimal representation) of a decimal representation.

It basicly comes down to this:

-You start with the interval [0,1] (once you go past the .)
-You split it up in 10 (for decimal) intervals of equal length, [a(i),a(i+1)] i =0..9
-If your number lies in the interval [a(i),a(i+1)], add the value of 'i' to the end.
-Apply the same principle to [a(i),a(i+1)]

A problem arises when your number is such an a(i) -> it will be in 2 intervals during one of your choices. In this case, you are free to chose and you will end up with 2 decimal representations of the same number.

If this happens, you can get 4.999999r, if you chose the 'lowest' interval, or with 5.000000r if you chose the 'higher' interval. They will remain the same number tho ...

My AU$0.02:

A real number is defined as a number which can be put as a fraction with an integer on teh top and bottom.

0.9999r can not be put into a fraction. If you can find a faction for it, tell me.

As it is not a real number, it cannot be subtracted from a real number

I believe that it just does not exist. It will never exist. It is impossible to have a 0.999rth of an item, because the 0.0000r...1 part of it will continue getting infinitely small.

A real number is defined as a number which can be put as a fraction with an integer on teh top and bottom.

Nope, you've stumbled accross the definition of a rational number.

The set of rational numbers plus the set of irrational numbers which contains numbers such as pi, e, and the square root of 2 make up the set of real numbers...

Nope, you've stumbled accross the definition of a rational number.

The set of rational numbers plus the set of irrational numbers which contains numbers such as pi, e, and the square root of 2 make up the set of real numbers...

This one has just inspired me for another 'proof' that 0.9r = 1:

If we assume that 0.9r is not equal to 1:
-between every 2 irrational numbers, you can find a rational number
-between every 2 rational numbers, you can find a irrational number
(real number are 'archimidean?' i think)
Using these, we see that there is a number x between 0.9r and 1, that is not equal to either of these: 0.9r<x<1

delta = x-0.9r
0.9r<x => delta>0.0r
now for every delta>0.0r, we can find an epsilon(n) 0.(n)01 ('0.' then n '0', then '1') for which we have: delta>epsilon(n).

we write: x = 0.9r + delta > 0.9r + epsilon(n) = m
0.9r = 0.(n)99 + 0.(n)009r
m = 0.(n)99 + 0.(n)01 + 0.(n)009r = 1 + 0.(n)009r

and we end up with x > 1

This is impossible according to our assumptions, so these assumptions are false. => 0.9r = 1

*consults sci.math faq*

*hands out biscuits for correct answers*

Mmmmmm, recycled usenet...

I think I can explain the (2infinity) problem now that I've had some sleep.

Basicly, nothing can ever equal infinity by definition. It can only increase without bounds and thus approach infinity.

However, different sets can increase at different rates.

Take the functions

f'(t) = 2apples/second

therefore f(t) = 2t apples

and

g'(t) = 1person/second

therefore g(t) = t persons

The limit as t -> infinity of f(t)/g(t) =

limit as t->infinity of (2t apples)/(t persons) =

2 apples/person

This is because the number of apples increases twice as fast as the number of people. Both approach infinity but the apples do so at twice the rate as the people.
Because neither can ever reach infinity there wil always be a 2/1 ratio.


On the other hand, if you have
(t apples)/ 2 people

the limit as t-> infinity of (t apples)/(2 people) =
the limit as t-> infinity of (t/2)(apples/person) =
limit as t-> infinity t of (apples/person) =
infinite apples/person.

The number of apples per person increases without bounds because the number of persons is constant. However no person ever has infinite apples for any real value of t.
Assuming that t is not boundless then the number of apples that any individual can have is finite because time eventually ends. However, so long as t increases the number of aples per person will increase.

My point was not that you shouldn't question but that to question mathematical concepts you have to use mathematical tools. Everyone who has done this agrees (indeed have proved) that .9r=1. Everyone who disagrees has used words. ie but they is some infinitesimal difference blah blah blah I don't know what I'm talking about.

Not all competent mathematical theorists are incompetent linguists.

Every proof that .9r = 1 uses a bent tool (a mathematical fiction in the formulae) to get rid of that infinitesimally small and ever decreasing difference.

Intellectually honest theorists admit the use of the mathematical fiction. Folk who just want to win arguments don't admit or don't recognize that their tool is bent.

I don't mind using a bent tool. I just hope everyone who has "proven" that .9r = 1 recognizes when the fiction occurs.

0.99999r * 10 = 9 + 0.999999r

1 * 10 = 9 + 1

so therefor 0.99999r = 1 because it's not possible for 2 distinct numbers to have the exact same propertiesThis is just about the flimsiest 'proof' I have ever encountered. You have stated two expressions, that true though they may be, have not been equated.

Example:
4 * 9 = 36 = 30 + 6
2 * 9 = 18 = 12 + 6

so?

he travels half the distance between the wall and the point he is at after taking the previous step...

Again this is hypothetical, the question is is it possible to take infinately small steps?

No, because eventually you will reach the wall, won't you? It may take billions of years, but you'll get there. The initial distance between you and the wall is constant. For example, the initial distance between you and a wall is 5 meters. Take your first step, and you have 2.5m to go. Next step is 1.25 meters. So, so far you have travled a grand total of 3.75 m. Only 1.25 m to go. You can keep dividing your distances in half, but when you add them, you'll eventaully come out to 5 meters although you may not live to see yourself reach the wall. Therefore, .999r will eventually reach 1. It will just take an amount of time inconcievable to our petty human imaginations.

1 cannot be definitely divided by 3. That's the problem.
All standards have set 1 to be divisible by 3 and have the answer be infinite, not quite solving the same method reversed (meaning .33r * 3)

Simply put, 1 does NOT equal .99r

Ok.. a few things...

Mathematics isn't about realness. Yes, there are many things in which mathematics have real world applications, but objects that are of mathematical interest need not have a "real world" analogue.

That much said...

0.9999r is simply written as the LIMIT of a infinite geometric series, namely 0.9 + 0.09 + 0.009 + ...

The limit of said sequence is no doubt 1. Nobody, however, should be caught writing 0.999r, because that is just cumbersome notation. "Adding infinite times" does not make sense in the real world, but it has mathematical significance. Remember, "infinity" is not a number in and of itself.

Writing 0.9999r is simply equivalent, mathematically defined, to be this infinite series, and we have defined that to be 1. Remember, the number 0.9999r does not "exist" in the traditional sense. The statement that 0.9999r = 1 is just an abbreviation of writing limit x->inf 9/(10^x), which is clearly 1.
Anyway, this question is of a notational significance, not a real philosophical or mathematical signifiance. It is simply understood this way in the community.

Now, onto addressing the hand-waving posts and mathematical logic...

Many of you have seen the following "proof"

Thus x = 0.9999... ; 10x = 10 * 0.9999... ; 10x = 9.9999... ; 10x - x
= 9.9999... - 0.9999... ; 9x = 9 ; x = 1 ; .

An even easier argument multiplies both sides of 0.9999... = 1 by
0.3333... = 1/3 . The result is 3 .

These "proofs" are far from rigorous, as any mathematician will tell you. However, it does give a hand waving argument that 0.9999r = 1. You may also see another proof of the following nature:

Find the smallest real number, epsilon, and add it to 0.9999r. The result will be slightly more than 1, no matter how small the number epsilon is. Ergo, 0.9999r = 1. (0.0000r1 is a stupid number that doesn't exist; where do I put the 1? after a non ending string of zeros?)

Again, another interesting, but hand wavy argument.

Now, onto mathematical logic...

Remember, the premise of a mathematical proof is to show that it works. You only need one fool-proof logical argument to show that the theorem holds. It doesn't quite matter if you find another road that doesn't work - I already found a road that takes me where, so why should I try to dig up another road? If my current logical argument has no errors, then it is correct, and the result holds.

Now, remember that in order to prove that something is false, you just need a counterexample.

Anyway, remember, Mathematics is full of things that don't make sense. :)

Hope everyone have a great day.

-cyby

No, because eventually you will reach the wall, won't you? It may take billions of years, but you'll get there. The initial distance between you and the wall is constant. For example, the initial distance between you and a wall is 5 meters. Take your first step, and you have 2.5m to go. Next step is 1.25 meters. So, so far you have travled a grand total of 3.75 m. Only 1.25 m to go. You can keep dividing your distances in half, but when you add them, you'll eventaully come out to 5 meters although you may not live to see yourself reach the wall. Therefore, .999r will eventually reach 1. It will just take an amount of time inconcievable to our petty human imaginations.


That's true because the infinite sum of 1/2^k converges at 1.

However, if you only take 1/3 of a step everytime then you will only reach the half way point.

Infinite sum of 1/3^k = (1/3)/(2/3) = 1/2


Every proof that .9r = 1 uses a bent tool (a mathematical fiction in the formulae) to get rid of that infinitesimally small and ever decreasing difference.

.9r is itself a bent tool. I isn't a real number. It is a polynomial aproximation of a real number.

The only reasonable justification for its existance is the fact that microprocessors can deal with fractions directly but can deal with polynomials.

.9r should always be expressed on paper as the real number 3/3. Only using a calculator provides reason to do otherwise.

That's true because the infinite sum of 1/2^k converges at 1.

However, if you only take 1/3 of a step everytime then you will only reach the half way point.

Infinite sum of 1/3^k = (1/3)/(2/3) = 1/2


error: the first step is 1/3, but the second is (1/3)*(2/3), not (1/3)^2

Infinite sum of (1/3)*(2/3)^(k-1) = 1

how many people can be snookered into buying as "true" a bent mathematical construct?

That's true because the infinite sum of 1/2^k converges at 1.

However, if you only take 1/3 of a step everytime then you will only reach the half way point.

Infinite sum of 1/3^k = (1/3)/(2/3) = 1/2




.9r is itself a bent tool. I isn't a real number. It is a polynomial aproximation of a real number.

The only reasonable justification for its existance is the fact that microprocessors can deal with fractions directly but can deal with polynomials.

.9r should always be expressed on paper as the real number 3/3. Only using a calculator provides reason to do otherwise.
Out of curiosity, why is my proof wrong?
Look at this (http://forums2.jolt.co.uk/showpost.php?p=8722262&postcount=35) post.

Out of curiosity, why is my proof wrong?
Look at this (http://forums2.jolt.co.uk/showpost.php?p=8722262&postcount=35) post.

Its right. Roundabout, but right.

error: the first step is 1/3, but the second is (1/3)*(2/3), not (1/3)^2

Infinite sum of (1/3)*(2/3)^(k-1) = 1

Thanks for the correction.

how many people can be snookered into buying as "true" a bent mathematical construct?

Which construct are you speaking of.

The easiest way to show that .9r is 1 is to show it is an infinite sum. To do this requires limits.

Limits may be percieved as a mere mathmatical construct but they do exist. We know this because speedometers and radar guns work. Without limits it would be impossible to measure the instaneous speed of a vehicle.

OK, I'll chase this silly rabbit.

1 and 0.9r are two representations of values.

Between any two real, consecutive numbers lies an infinite number of real numbers.

So, what are the numbers between 1 and 0.9r? The difference between the two is an infinitesimally small number. The only reason 1=0.9r by mathematical proof is because the difference cannot be quantified an a real number. If we could quantify it, we'd through calculus out the window.

Infinity means two things: Unquantifiable and without end. Infinity is a mathematical inclusion because of the lack of evidence to disprove its existence. It's a disclaimer, in case we should ever find something truly indiscrete.

1/9 = .11111111r
2/9 = .22222222r
3/9 = .33333333r
4/9 = .44444444r
5/9 = .55555555r
6/9 = .66666666r
7/9 = .77777777r
8/9 = .88888888r

What comes next?

Then 0/9 = .00000000r?

try base 3

OK, I'll chase this silly rabbit.

1 and 0.9r are two representations of values.

Between any two real, consecutive numbers lies an infinite number of real numbers.

So, what are the numbers between 1 and 0.9r? The difference between the two is an infinitesimally small number. The only reason 1=0.9r by mathematical proof is because the difference cannot be quantified an a real number. If we could quantify it, we'd through calculus out the window.


Not just calculus, Newtonian physics as well.

The thing is that 0.9r isn't a real number. It is the sum of an unending series of real numbers.

1/9 = .11111111r
2/9 = .22222222r
3/9 = .33333333r
4/9 = .44444444r
5/9 = .55555555r
6/9 = .66666666r
7/9 = .77777777r
8/9 = .88888888r

What comes next?

Strangly, 9/9 = .999r works by long division.

Nope, you've stumbled accross the definition of a rational number.

The set of rational numbers plus the set of irrational numbers which contains numbers such as pi, e, and the square root of 2 make up the set of real numbers...
Ok, but the point still remains that 0.999r is a mathematically impossible number to represent. And i still believe it does not exist...

Not just calculus, Newtonian physics as well.

The thing is that 0.9r isn't a real number. It is the sum of an unending series of real numbers.
Why do you say 0.9r isn't a real number when you go on to define it as one? A real number is the limit of a Cauchy sequence of rationals. It's that simple. It means that any time we have something that looks like it should converge to a limit, we say it does. And we always say it should if it meets the requirements that Cauchy gives us. Even if the sequence does not converge to any rational number, we still say it converges to a real number.

Since every infinite decimal represents a Cauchy sequence, whether it repeats or not, every infinite decimal represents a real number. Even the finite decimals represent real numbers, since you can append an infinite string of zeros onto them to make them infinite decimals. So 0.9r has to be a real number.

Actually, a friend of mine and I got together and tried to figure out the Rules of Dimentional Existence. Being that everyone knows the 1st to 3rd Dimention, we continued until 9 Dimentions. I won't describe them all now (you can email me if you want to discuss it) but the interesting thing is:



We concluded that all objects exist, wheter a True Object (TO+/-CO) or Conceived Object (CO). The fact that there is something less than 1 but greater than 0 is a reality, and the "fact" that .9r = 1 is "reality" because someone thinks so. However, .9r is a CO, not necessarily a TO. Likewise, simply because it is concieved does not mean it is true, although the fact that you do not "know" it is true does not rule out its possibility. If you ran thru the 9 dimensions (or as we call it, "The Nine Rules For Existence") you would find yourself at the paradox of the 9th Dimension, being- "Does the Object Exist in Reality, or in Conception, or in Both?"

Don't worry, it's a paradoxal proposition, but I'd be glad to run it thru off-line.

A friend of mine, and I got together once and discussed football and girls and planes and stuff. It was good. Had a few beers, you know. We concluded that Derby were the best team, Louise from Lower Somercotes was the best girl and that all planes were equally as good as each other. (we don't really know much about planes).
Don't worry about this either. I'd be more than happy to run this thru off line too. Cheers.

Strangly, 9/9 = .999r works by long division.
Please explain this. I am running it through my head, and maybe I don't understand you concept of long divison, but I cannot figure out how

9/9=0.999r

Ok, I figured out the 9/9=0.999r thing. My whole concept of reality has now changed.

I like the most common proof listed so far because it sidesteps both infinity and the quantity .000r1. It hasn't been stated that way, but that is the basis of the algebraic proof.

Let x = 0.99r be a non-terminating decimal. (This is the only part anyone should have any trouble with, once you realize that there is no end, all the complexities disappear.)

Then, we have x = 0.99r
Multiplying both sides by ten yields
10 x = 9.99r (Note that shifting the decimal place does not stop the repetition. It doesn't end, that is the definition)

Subtracting x from both sides yields
9x = 9.99r - x = 9 + .99r - .99r

Dividing by nine: 9x / 9 = 9 / 9
Simplifying: x = 1

No infinity, no dealing with the "end."

(Granted, limits work great, but you have to accept the precepts of limits. Since the precepts of limits were rejected by some, I resort to this proof.)

A "b" postfix indicates binary, "d" decimal, "h" hexadecimal (we'll skip octal!):

x = .1rb (Let x = 1/2 + 1/4 + 1/8 + ... + 1/n)
10bx = 1.1rb (Multiply by 2, 10b in binary, shifting the decimal place by 1)
x = 1.1rb - x (Subtract x, 10b - 1b = 1b, hence just x)
x = 1 + .1rb - .1rb (Translate as to 9 + .99r - .99r as above)
x = 1 (Simplify)

(I first did the above in Base 5 - yes, Base 5, I'm a little insane - but changed it to binary as an after thought)

Less intuitive than in decimal, but it illustrates two things.

The first is that precision and presentation would make a difference in impact, though not in result - 0.999d is much closer to 1 than the binary 0.111b (0.875d) is to 1. 0.FFFh (hexadecimal) is even closer to 1 than 0.999d.

Not brought up to think in binary or hex, we think of 0.9r for purposes such as this, but in essence, it is no more correct or incorrect than 0.1r or 0.Fr, in binary and hex, respectively. Yet, all of the above repeating representations are ultimately equivalent - notwithstanding that 0.111rb translated into decimal is not 0.999d, but would be something completely different.

The second is that math, while attempting to model the Universe, is entirely conceptual, not withstanding the intuitive sense that 1 apple + 1 apple = 2 apples (but 1 apple + 1 apple also equals, say, 40 +-5 seeds, or a good snack!). Even with the above method, the concept of infinity is still involved - the assumption that 0.9r repeats indefinitely is itself an implicit invocation of the infinite!

0.9 != 1, nor does 0.99999999999999 - nor 0.9 succeeded by a googolplex of 9s. The only case in which any number of 9s in succession is taken to be equivalent to 1 is when there are an infinite number of 9s implied. Infinity is implicit and complicit in anything mathematically repeated indefinitely. Whether we represent the (infinite) binary sequence 1/2 + 1/4 + ... + 1/n as 0.1r or the (infinite) decimal sequence 1/10 + 1/100 + ... + 1/n as 0.9r, we simply use a shorthand form of representing an infinite sequence for the actual infinite sequence itself.

Numbers can only be defined in theory.

By the algebraic proof, the issue of infinity is sidestepped by thinking of it from a different perspective (non-terminating decimal). Thus, 0.000r1 is nonsensical as you have an "end" to something which does not end.

Unfortunately, a number infinitely small yet larger than 0 (akin to if not the same as a "0.0r1") is precisely what limits rest upon. Limits are built upon the concept that an infinitely small delta value can be used in progressing from one value to another, the value of which is both considered to be 0 (in order to actually "reach" the limit) and not 0 (in order to divide delta by delta, where 0/0 is undefined and hence undesired!).

Yet that's what all higher math rests upon.

Anyway, 1/aleph0 would be the appropriate representation of 0.0r1 - 1 divided by the first cardinal number (normal "infinity").

For those who haven't gotten it, 9.0/9 could be expressed as .9 with a remainder of .9.

9 will go into 9 .9 times resulting 9.0-8.1 =.9

9 will go into that .09 times resulting in .90-.81 =.09 and so on to infinity.

This also works for any other fractional reprensation of 1.

1 will go into 1.0 .9 times with a remainder of .1
1 will go into that .09 times and etcetera till the end of time.

I admit that it isn't the usual method of division but it works on 1.
It doesn't work on any other number I am aware of as it usually produces a remainder that is greater than the divisor.
It is further proof of the relationship between .9r and 1.

Is that where the 60 minutes per hour came from?

Yes. Base 60 or sexagesimal. The Babylonians in turn borrowed the system from the Sumerians.

It's thought that Base 60 came to be used as a result of a merger of sorts between two peoples, one of which used Base 12 or Base 10 (decimal), the other having used Base 5 or Base 6.

Ahh.. that makes more since. I use =! because that is what every software caculator I have used recognizes.

I use "<>" or "!="... programming habits ("=/=" looks like anything but a "not equal" sign to my coding eyes!). "<>" is found in the BASIC family of computer languages, "!=" in the C family. I avoided "!=" in this forum because my instinct would be to use "==" for assertion of equivalence ("=" is used for assignment - "x = 5" assigns the value of 5 to variable x).

Yes. Base 60 or sexagesimal. The Babylonians in turn borrowed the system from the Sumerians.

It's thought that Base 60 came to be used as a result of a merger of sorts between two peoples, one of which used Base 12 or Base 10 (decimal), the other having used Base 5 or Base 6.

Base 60 probably came into use because clocks were circular. Its easier to divide a circle into 60 units than it is to divide it into 10 or 100 units.

infinity + infinity = infinity

I can't help explain "infinity" or more generally infinite numbers. I was going to try to, but I can't do it on my own, and I can't find the book I was going to refer to.

So, I'll refer you (and anyone) to the book! (In depth reviews of its subject matter, as well.)

The Mystery Of The Aleph (http://www.amazon.com/exec/obidos/ASIN/0743422996/qid=1115375051/sr=2-1/ref=pd_bbs_b_2_1/104-7440189-2968709)

It's a wonderful read, and it does clear up infinities rather well. If you ever manage to understand it fully, let me know - I promise I won't steal the Fields Medal that would invariably follow!

Base 60 probably came into use because clocks were circular. Its easier to divide a circle into 60 units than it is to divide it into 10 or 100 units.

:)

The Babylonians invented the sundial - but marking no hours, no minutes - no clocks, though - I'm sure the Swiss might take offense. ;)

Easier to divide by a variety of different numbers is one reason behind use of Base 60, though, yes - 60 is cleanly divisible by 2, 3, 4, 5, 6, 10, 12, 15, 20, and 30 - much easier to subdivide, absolutely.

A reason it's thought that it came from a mixture of two different number systems is that the Babylonians actually used a mixed radix system - they used both 10 and 6 as bases. Where we use 10 and 100 as divisions, they used 10 and 60. That is, they counted one 1, up to one 10, then up to five 10s and nine 1s, and then one 60. 70, which we represent as seven 10s, would be represented by one 60 and one 10, while 120 (one 100 and two 10s) would be two 60s.

That's not to say that the Sumerians or their predecessors didn't sit down and decide to use 60 on top of a preexisting decimal system - it's quite possible. Base 12 and Base 5 are not as common as Base 10, but still common - Base 6 isn't very common, so it's very possible that Base 60 was a purposeful invention of the Sumerians.

"Do you know what the man is saying? Do you? This is dialectics. It's very simple dialectics. One through nine, no maybes, no supposes, no fractions. You can't travel in space, you can't go out into space, you know, without, like, you know, with fractions. What are you going to land on, one quarter, three-eighths, what are you going to do when you go from here to Venus or something -- that's dialectic physics, OK? Dialectic logic is there's only love and hate, you either love somebody or you hate them."-Photojournalist(Dennis Hopper)- Apocalypse Now

Unfortunately, a number infinitely small yet larger than 0 (akin to if not the same as a "0.0r1") is precisely what limits rest upon. Limits are built upon the concept that an infinitely small delta value can be used in progressing from one value to another, the value of which is both considered to be 0 (in order to actually "reach" the limit) and not 0 (in order to divide delta by delta, where 0/0 is undefined and hence undesired!).

Yet that's what all higher math rests upon.

Anyway, 1/aleph0 would be the appropriate representation of 0.0r1 - 1 divided by the first cardinal number (normal "infinity").

'proving' that 0.0r1 does not exist, or if it exists, is equal to 0. Note that 0r means that the zeroes are repeated an infinite amount of times, so you cannot append a 1 in the end. This makes the notation in itself meaningless.

You said that 0.9r + 0.0r1 = 1 (0.0r1 being the difference). You also said 0.0r1>0.

if 0.0r1>0, i can find an epsilon>0 so that: 0.0r1>epsilon. I chose this epsilon to be of the form: 0.0..01, with n 0's in between.

0.9r + 0.0r1 > 0.9r + epsilon = 0.9..99 + 0.0..009r + 0.0..01 = 1 + 0.0..009r > 1.

---------------------

The proof comes down to this:
You assume that an infinitly small delta (0.0r1) can still be greater than 0. But if delta > 0, we can find an infinite amount of finite epsilons>0 for which delta>epsilon because the real numbers are not 'discrete' eg. Between every 2 non equal real numbers, you can find an infinite amount of real numbers. This in turn makes that your infinite delta is:

1) equal to 0.
OR
2) not infinite.

'proving' that 0.0r1 does not exist, or if it exists, is equal to 0. Note that 0r means that the zeroes are repeated an infinite amount of times, so you cannot append a 1 in the end. This makes the notation in itself meaningless.


Which is why I insisted on the appropriate representation of the smallest possible number greater than 0.

You said that 0.9r + 0.0r1 = 1 (0.0r1 being the difference). You also said 0.0r1>0.

Actually, I said neither. :) The first is referring, I assume, to my mention of infinitesimally small delta values used in limits. I by no means was addressing the proof (or proofs) being discussed. I simply stated that something being dismissed off hand is in fact critical to limits and calculus.

The second, sure, if you don't want to split hairs. Again, I don't care for that representation, and, anyway, I don't like repeating numbers or using them (anything repeating, after all, is the result of fractional representations or of infinite series that have yet to converge completely).

This in turn makes that your infinite delta is:

1) equal to 0.
OR
2) not infinite.

*sigh* Again, I never stated anything, delta or otherwise, was infinite. Infinitesimally small, used in calculus, sure. Not infinite.

Anyway, you've nailed the matter on the head - an infinitesimally small delta value, in calculating the limits of functions, is assumed to be equal to zero. As I said, that's necessary, as otherwise a division of zero by zero would be involved, rendering the results undefined.

It's mathematical slight of hand.

Which is why I insisted on the appropriate representation of the smallest possible number greater than 0.



Actually, I said neither. :) The first is referring, I assume, to my mention of infinitesimally small delta values used in limits. I by no means was addressing the proof (or proofs) being discussed. I simply stated that something being dismissed off hand is in fact critical to limits and calculus.

The second, sure, if you don't want to split hairs. Again, I don't care for that representation, and, anyway, I don't like repeating numbers or using them (anything repeating, after all, is the result of fractional representations or of infinite series that have yet to converge completely).



*sigh* Again, I never stated anything, delta or otherwise, was infinite. Infinitesimally small, used in calculus, sure. Not infinite.

Anyway, you've nailed the matter on the head - an infinitesimally small delta value, in calculating the limits of functions, is assumed to be equal to zero. As I said, that's necessary, as otherwise a division of zero by zero would be involved, rendering the results undefined.

It's mathematical slight of hand.

Yes, i indeed used faulty notations there; but you got the point.

But - even when considering the general case - it is not 'mathematical slight of hand' to assume an infinitesimally small delta is equal to zero.

This is because if delta>0, you can find epsilons>0 for which delta>epsilon. Because these epsilons are finite, you end up with a finite delta, not an infinitesimally small one.

I use "<>" or "!="... programming habits ("=/=" looks like anything but a "not equal" sign to my coding eyes!). "<>" is found in the BASIC family of computer languages, "!=" in the C family. I avoided "!=" in this forum because my instinct would be to use "==" for assertion of equivalence ("=" is used for assignment - "x = 5" assigns the value of 5 to variable x).Why don't you use "?" for nonequivalence? :D

My thoughts on 0.9999r and 1.000r:

The difference is very slight, but apparent. 0.99999r * 1,000,000r = 9,999,999r.99999r. For all practical purposes, .99999r = 1.

However, this does not mean that 0.000r...1 = 0, as the equation above might imply. As a mathematics book I read stated:
Technically, your chances of winning the lottery are 0.0000r...0001. This number is so small as to be virtually equivalent to zero. Therefore, should we round your chance of winning the lottery to zero? Absolutely not, say the companies. You still have a chance of winning the lottery, however small.
This is the only instance I know of where, when a = b, a - c ? b - c.

Yes, i indeed used faulty notations there; but you got the point.

But - even when considering the general case - it is not 'mathematical slight of hand' to assume an infinitesimally small delta is equal to zero.

This is because if delta>0, you can find epsilons>0 for which delta>epsilon. Because these epsilons are finite, you end up with a finite delta, not an infinitesimally small one.

The slight of hand occcurs when delta values, regarding limits, are taken to be infinitesimally small non-zero values, in order that the delta value can be reduced by dividing it by itself, in which case it must be non-zero to be defined, as zero by zero is undefined, by definition.

The slight of hand is when .0...1 is considered to be equivalent to zero, but not for purpose of division by itself.

I've been imprecise - deltas as used above are found in determining limits as far as derivatives and differentiation - sorry.

http://en.wikibooks.org/wiki/Calculus:Differentiation

It's in that sense - the whole concept rests upon taking infinitesimally small delta values, tending towards zero, considering that the number is small enough in making little difference to the result that it might as well be equivalent to zero, and then cancelling it out (by division of itself) as non-zero. Deltas are used as simultaneously both zero and non-zero values.

First, I was taught that infinity does not exist in maths

Here, I'm going to ignore that rule.

0.9r can be considered to tend towards 1 as the number of recurrances increases, as the difference between 1 and 0.9 becomes less and less. In mathematical terms, you cut off and round up to the nearest 3 digit decimal number, which rounds up to 1.

Now, as the recurrances tend towards infinity, 0.9r tends towards 1. When the recurrances reach infinity, only then can it be considered that 0.9r = 1.

However, in mathematical terms, how far you go depends on how accurate you need your calculations to be

They're just flaws in the decimal system...
We should all use base 8

They're just flaws in the decimal system...
We should all use base 8

Why not just use binary?

There are extremely few flaws in maths, and the few that exist are mostly very high logic problems (ie. the mathematical form of me saying "I am a liar" has no solution because it cannot be said whether or not I am a liar since the phrase is a paradox.
My point was going to be that there are few flaws in maths, but quite a few flaws in how we write maths, which someone else brought up.
0.9~ = 1 has been proved mathematically.

Don't you just love mathematics! :-D

It doesn't work on any other number I am aware of as it usually produces a remainder that is greater than the divisor.
It is further proof of the relationship between .9r and 1.

It will work for any terminating decimal. For example, 1/4:

4 goes into 10 2 times, 4 goes into 20 4 times, 4 goes into 40 9 times, 4 goes into 40 9 times, etc..

Thus 0.25 = 0.24999r



As far as the difference between 1 and 0.999r being 0.000r1, it is not true. Where ever you end up placing the terminating 1 will lead to a number that is greater than 1 because the 9's never terminate.

0.999r+0.000r1=1.000r999r

Still waiting for the part where everyone realizes that .9999r is exactly the same as 1. Not something infinitely close like 1-dx but in actuality the same number simply written in a less convenient form. .0000r1 cannot exist and as such does not exist. It's been proven with an infinite geometric series, a different base, patterns in numbers, and numerous other proofs.

You know what this is don't you? We all all having a maths-debate. (Say it fast and you'll get it.)

You know what this is don't you? We all all having a maths-debate. (Say it fast and you'll get it.)

A maths debate on a board full of master debators. Holy undead threads, Batman.

By the way. Infinity doesn't exist in math. Infinity doesn't exist anywhere. However, there are some variables that increase or decrease without bound. These are refered to as being infinite or negativly infinite for the sake of simplicity, since increasing without bound is as close to being infinite as one can be.

Take for example the variable x. A basic number line. When defining it we say -infinity< x<infinity. There is no limit to what x can be.

X couuld be 10^100000000. It could also be 11^0000000000000000000000000000000. As you travel along the number line x becomes so huge that mortal minds cannot fathom it. Instead of trying we simply say that x increases without bound.

Now, another way of saying this is the limit as x aproaches infinity of x is infinity.

With 0.9r we have a different situation.

The limit as n increases without bounds of the sum of k = 1 to n of 9*10^-k = 1.

As n increases 0.9r comes closer to 1.

The problem is that n will never stop increasing. Never, ever. Therefore 0.9r will never stop getting closer to 1.

0.9r isn't a single and you shouldn't treat it as such. It is a sum of a never ending succession of terms of terms. It is because the terms are never ending that the sum equals 1.

Point of clarification 1, decimal notation - The number 146 is a shorthand way of writing 1x10^(2) + 4x10^(1) + 6x10^(0)

The notation I'm using is that 10^2 means "ten to the power 2" which is 100. 10^(1) is 10 and 10^(0) is 1. This is what is meant by decimal notation, you express the number as a sum of the powers of ten.

Point of clarification 2, infinite sums - Infinite sums crop up a lot in maths. For example...

1/2 + 1/4 + 1/8 + 1/16 + ......

Define S(n) to be the sum of first n terms. In the above example then S(1)=1/2, S(2)=3/4, S(3)=7/8 etc. Then we say that the infinite sum exists and is equal to S if for an arbitrary strictly positive number (call it k), no matter how small, there is some term in the series S(N), beyond which the difference between S and S(n) is always less than k. This is simply the definition of what mathematicians mean when they write an infinite sum. Like everything else in mathematics it is true solely by definition.

For example take k= 1/8. Then beyond the third term in the series, S(3), 1-S(n) is less than 1/8. With a little thinking you can convince yourself that no matter how small you choose k to be you can always get to a point in the series beyond which you are less than distance k away from 1. Thus the infinite sum is equal to 1.

If you've wrapped your head round these 2 definitions it is easy to see how 0.99999...=1. Remembering that the definition of 10^(-1) is 1/10, 10^(-2)= 1/100 etc, the definition of the decimal notation 0.99999 is,

0x1+ 9 x 1/10 + 9 x 1/100 + 9 x 1/1000 + ......

For any k you choose, if you go far enough along this sum you will reach a point where the sum is always less than distance k away from 1, so the sum of the infinite series is 1, i.e. 0.99999...=1.

Pretty heavy going, I know, but mathematics has become so rigorous exactly because people get mixed up by ideas like 0.9999...=1.