NationStates Jolt Archive

Mathematical proofs can be fun, but i don't want to limit myself to proving 0.9999r = 1. So this is a thread where one person gives an 'assignment' and the first person to prove it may give the next assignment. Biscuits are handed out for giving a correct proof or poking a hole in a proof. As an example, I'll start by giving the first proof of 2 related 'problems':

Proof that sum of 1/n, n=1..infinity diverges:

1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+ ...= 1/1 + 1/2 + (1/3+1/4) + (1/5+1/6+1/7+1/8) + ...
> 1+1/2+(1/4+1/4) + (1/8+1/8+1/8+1/8) + ...
=1+1/2+1/2+...= 1+sum(1/2, n=0..infinity) which diverges.
QED

Now it's up to you to give a proof of the following:

the sum of (1/n)^a, n=1..infinity converges If And Only If a>1.

Okay.... *Steps away from the thread*

Burns thread with Kerosene Toilet rolls.

Burns thread with Kerosene Toilet rolls.

*Joins in*

Psycho…

Psycho…

I second this motion

Motion passed.

Motion passed.

*Cheers*

Very carefully plants plastic explosives around the thread. Puts on "reasonable" tone of voice....

"Now, I'm sure that you can appreciate that this action is appropriate, since I took my last maths IB paper on Wednesday...and as such I never want to have to even consider mathematical formulae again."

Flicks a switch, and detonates evil mathematics. Cackles insanely for a bit.

Motion passed.

*overruled*

*overruled*

Your outvoted!

HA HA!

*overruled*
Uses Veto Powers to counter the Overuling.

*overruled*
VETO! VETO! :D

VETO! VETO! :D
Sorry son a matter of seconds! :D :eek: :( :confused: ...............

I veto the veto.

No...more....maths!

*Steals Illich Jackals licence*

*Stamps it with "VOID"*

*Proceeds to rip up his birth certificates*

*Sits back and torches house*

Hang on...did I just veto the veto to counter the overruling of no more maths?

If so, I voluntarily veto my veto.

*Steals Illich Jackals licence*

*Stamps it with "VOID"*

*Proceeds to rip up his birth certificates*

*Sits back and torches house*

LOL.

Sadly, this action is called for!

Just looking at those numbers burns!

as much as i find this thread interesting... ...i don't.
i get enough maths at uni, i don't need to see it on a forum, if you want to discuss maths, i can send you the email of one of our tutors or lecturers.
maths is not a leisure topic

Hang on...did I just veto the veto to counter the overruling of no more maths?

If so, I voluntarily veto my veto.
I think you said you wanted Maths!

*Burns Maths books at a Nazi Rally* :headbang: :mad: :eek:
Indiana Jones is there! :cool:
:mp5: :sniper: :gundge: :mp5:

*Gets carried away and hires a steamroller*

*Proceeds to crush thread*

*Forgets about the people in*

*Stamps about in the red puddles*

Mathematical proofs can be fun,


This has to be the joke of the week?!

*Gets carried away and hires a steamroller*

*Proceeds to crush thread*

*Forgets about the people in*

*Stamps about in the red puddles*

OW! My fingernails!

This has to be the joke of the week?!

Well, I certainly fell off my chair laughing! :D

Finger nails eh?

*Ponders*

Think there is anything I can possibly do involving destroying ones livelyhood AND fingernails?

Well, I certainly fell off my chair laughing! :D

:D

Finger nails eh?

*Ponders*

Think there is anything I can possibly do involving destroying ones livelyhood AND fingernails?

sentence makes no sense

:D

:D:D

:D:D

I'm nto liking this game!

:D :D :D

*Re-reads it*

*Gives it a prod with a stick*

...I can't find the problem...

I'm nto liking this game!

:D :D :D

Me neither!

:D:D:D:D

Finger nails eh?

*Ponders*

Think there is anything I can possibly do involving destroying ones livelyhood AND fingernails?

This isn't a question.... It is a poorly worded one...

Are you saying, 'I think there is something I could do involving destroying ones livelyhood AND fingernails.'

OR

'Is there anything I can possibly do involving destroying ones livelyhood AND fingernails?'

I like how all the maths-haters see a topic called "Math", then read it and decide that they don't like it so it must be destroyed. If you don't like something, just don't enter a discussion under its name. Its simple really....

Nowt wrong with liking maths, and I wish I could remember how to solve that, becaause I have an exam o nit on tuesday and have gone really stupid.

Me neither!

:D:D:D:D

:D:D:D:D:D

Assuming intelligence is a scalar quantity and:

Because of start of this thread: intelligence(Illich Jackal)>0
Replies of other people indicate: intelligence(X)= 0 with X element of {people replying to this thread} and X != Illich Jackal AND X !=Rebecacaca .

We end up with:

sum(intelligence(X(i)), sum over {people replying to this thread}\Illich Jackal AND Rebecacaca ) = 0< intelligence(Illich Jackal)

Or in other words: I am smarter than all of you put together!

Ah, my mistake. Hmmm. Mistake eh?

...

Can't let the press find out about this...

*Flees*

I like how all the maths-haters see a topic called "Math", then read it and decide that they don't like it so it must be destroyed. If you don't like something, just don't enter a discussion under its name. Its simple really....

Nowt wrong with liking maths, and I wish I could remember how to solve that, becaause I have an exam o nit on tuesday and have gone really stupid.

I find the first sentence funny.... How can maths be fun?

Assuming intelligence is a scalar quantity and:

Because of start of this thread: intelligence(Illich Jackal)>0
Replies of other people indicate: intelligence(X)= 0 with X element of {people replying to this thread} and X != Illich Jackal AND X !=Rebecacaca .

We end up with:

sum(intelligence(X(i)), sum over {people replying to this thread}\Illich Jackal AND Rebecacaca ) = 0< intelligence(Illich Jackal)

Or in other words: I am smarter than all of you put together!

I second this motion. But I suggest an addition to this motion?

*Throws very heavy object at foundations of thread*

....

*Bails out window*

Assuming intelligence is a scalar quantity and:

Because of start of this thread: intelligence(Illich Jackal)>0
Replies of other people indicate: intelligence(X)= 0 with X element of {people replying to this thread} and X != Illich Jackal AND X !=Rebecacaca .

We end up with:

sum(intelligence(X(i)), sum over {people replying to this thread}\Illich Jackal AND Rebecacaca ) = 0< intelligence(Illich Jackal)

Or in other words: I am smarter than all of you put together!

Whatever. . . . . .



Moleland: :D:D:D:D:D:D

Ah, my mistake. Hmmm. Mistake eh?

...

Can't let the press find out about this...

*Flees*

You'd better run Egg! (http://www.tallstories.org.uk/shows/other/the-egg.jpg)

...B-bu-but I don't like eggs...

*lip trembles*

I'm nto liking this game!

:D :D :D
Wow if you add some letters and change some you get
I'm Into killing this game!
WOW!

:D :D :D

Wow if you add some letters and change some you get
I'm Into killing this game!
WOW!

:D :D :D

:eek:

@Psychopathic Warmonger]




:D :D :D :D :D :D :D :D :D :D

Beat that!

@Psychopathic Warmonger]




:D :D :D :D :D :D :D :D :D :D

Beat that!

Fine. Excuse me for a second.

*Begins marking out enormous run-up*

*Beats You*


Hows that for beating you?

...B-bu-but I don't like eggs...

*lip trembles*

DIE DIE DIE!!!! (http://reptilis.net/index4/flick.jpg)

Wow if you add some letters and change some you get
I'm Into killing this game!
WOW!

:D :D :D

WOW

*Dies*

Fine. Excuse me for a second.

*Begins marking out enormous run-up*

?

*Dies*

YES!!!!!!

?

??

??

You deleted that post... why'd it look silly?

Mathematical proofs can be fun, but i don't want to limit myself to proving 0.9999r = 1. So this is a thread where one person gives an 'assignment' and the first person to prove it may give the next assignment. Biscuits are handed out for giving a correct proof or poking a hole in a proof. As an example, I'll start by giving the first proof of 2 related 'problems':

Proof that sum of 1/n, n=1..infinity diverges:

1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+ ...= 1/1 + 1/2 + (1/3+1/4) + (1/5+1/6+1/7+1/8) + ...
> 1+1/2+(1/4+1/4) + (1/8+1/8+1/8+1/8) + ...
=1+1/2+1/2+...= 1+sum(1/2, n=0..infinity) which diverges.
QED

Now it's up to you to give a proof of the following:

the sum of (1/n)^a, n=1..infinity converges If And Only If a>1.


The Integral test.

The Limit as n-> infinity of the Integral of (1/n)^1 =
Limit as n-> infinity of ln(n)= infinity.

If a is less than 1 then the Integral will be even larger ths must also diverge.

Limit as n-> infinity of the Integral of (1/n)^(1001/1000) = Limit as n -> infinity of the integral of n^-(1001/1000) =
limit as n -> infinity of -1/1000n^(1/1000) = 0
-1<0>1 therefore it converges according to the integral test.

All larger values of a should converge as well because the series decreases.

Mine is a doozy,

Prove that sin(pi/3) = root3/2

Assuming intelligence is a scalar quantity and:

Because of start of this thread: intelligence(Illich Jackal)>0
Replies of other people indicate: intelligence(X)= 0 with X element of {people replying to this thread} and X != Illich Jackal AND X !=Rebecacaca .

We end up with:

sum(intelligence(X(i)), sum over {people replying to this thread}\Illich Jackal AND Rebecacaca ) = 0< intelligence(Illich Jackal)

Or in other words: I am smarter than all of you put together!
And look they added a Sad smilie
*Its in Bold*
I think they know that Factor Non Verba is better!

(1/1)^a = 1^a
any power of 1 equals 1, therefore n>1

The Integral test.

The Limit as n-> infinity of the Integral of (1/n)^1 =
Limit as n-> infinity of ln(n)= infinity.

If a is less than 1 then the Integral will be even larger ths must also diverge.

Limit as n-> infinity of the Integral of (1/n)^(1001/1000) = Limit as n -> infinity of the integral of n^-(1001/1000) =
limit as n -> infinity of -1/1000n^(1/1000) = 0
-1<0>1 therefore it converges according to the integral test.

All larger values of a should converge as well because the series decreases.

IT BURNS!!!!!

You deleted that post... why'd it look silly?

Because the system wouldn't let me put THAT many smileys in, so it came up with a screen of crap. Believe me it looked stupid.

Because the system wouldn't let me put THAT many smileys in, so it came up with a screen of crap. Believe me it looked stupid.

i know... I did the as many as it would let me :D

i know... I did the as many as it would let me :D

Hehehehehe. You win then :)

Hehehehehe. You win then :)

2 wins in 1 thread! I'm good!

*does victory dance*

2 wins in 1 thread! I'm good!

*does victory dance*

*Pats on back*

You deserved it.

*Plots revenge*

*Pats on back*

You deserved it.

*Plots revenge*

*Looks shifty*

*Looks shifty*

*Pulls hat over eyes*

*Pulls hat over eyes*
*Grabs Tin Foil hat of own*

'fraid I gotta go now, I have a History lecture thing to go to. Sorry.
Though maybe we can reconvene soem other day :)

*Looks sad and waves goodbye*

Prove that sin(pi/3) = root3/2

Pi rad = 180° so pi/3 rad = 60°

consider a perfectly symmetrical triangle with sides a, because the sum of the angles is 180°, one angle is 60°. draw a line that divides this angle in two angles of 30° and divides the opposing side into two parts of a/2. By symmetry this line makes an angle of 90° with the opposing side.

I end up with a triangle with one angle of 30°, one of 90° and one of 60°. A side a, a side a/2 and with pythagoras:
a^2/4 + x^2 = a^2 => x = root(3)/2*a

position yourself in the corner with the angle of 60° and you see that
sin(60°) = root(3)/2*a/a = root(3)/2

prove that: e^(ix) = cos(x) + i*sin(x)

Omg, i h8 math - lol. It is soooo boring and i h8 it!

Die Math Die :sniper: :mp5:


lol i'd rather bang my head than do math! :headbang:


hehehe - well, thats my opinion!


:)

exp(x)=1+x+x^2/2!+x^3/3!+... [by definition]
=> exp(ix)=1+ix-x^2/2-ix^3/3!+x^4/4!+ix^5/5!-....
=1-x^2/2!+x^4/4!-....+i(x-x^3/3!+x^5/5!-...)
=cos(x)+i sin(x) [by definition]

Maybe I can do maths then....:)

And for the next problem...show that the set of prime numbers must be infinite.

Damn, I read the thread title as, "Myrth."

Ah well... math is good too.

exp(x)=1+x+x^2/2!+x^3/3!+... [by definition]
=> exp(ix)=1+ix-x^2/2-ix^3/3!+x^4/4!+ix^5/5!-....
=1-x^2/2!+x^4/4!-....+i(x-x^3/3!+x^5/5!-...)
=cos(x)+i sin(x) [by definition]

Maybe I can do maths then....:)

And for the next problem...show that the set of prime numbers must be infinite.

Eh....Since number of intergers are infinite.
Therefore number of prime numbers must be infinite? :p

exp(x)=1+x+x^2/2!+x^3/3!+... [by definition]
=> exp(ix)=1+ix-x^2/2-ix^3/3!+x^4/4!+ix^5/5!-....
=1-x^2/2!+x^4/4!-....+i(x-x^3/3!+x^5/5!-...)
=cos(x)+i sin(x) [by definition]

Maybe I can do maths then....:)

And for the next problem...show that the set of prime numbers must be infinite.

Assume the set of prime numbers is finite, we then write all primes as p1 p2 .. pn.

we make the product(p(i),i=1..n) and add 1 to it:
now we have for all j:
product(p(i),i=1..n) + 1 = 1 modulo p(j)

and thus the product(p(i),i=1..n) + 1 is a prime that did not belong to our original set of primes. Therefore the set of primes is infinite.

next challenge: prove that Q has as many elements as N (or: the amount of natural numbers equals the amount of rational numbers)

Edit for The Alma Mater: in the proof, it is a prime according to our basic assumptions (which are then proven wrong)

And for the next problem...show that the set of prime numbers must be infinite.

According to Euclids unique prime factorization theorem every positive integer larger than one can be expressed as the product of a unique set of primes.

Assume now there is a limited number of n primes, say p1, p2, p3.. pn. Then define the number P = p1 x p2 x p2 x ... x pn +1 and let p be a prime dividing P.
If p would be from the set p1..pn it should be able to divide the difference P -p1 x p2 x p2 x ... x pn too. Clearly it cannot, since this difference is 1. Therefor p must be another prime, so it is not possible to have a non-infinite set.

Next assignment: show that the two solutions to a second order polynomial equation of the form ax^2+ bx +c = 0 are given by the quadratic formula:

x = [ -b +- sqrt(b^2 - 4ac) ] / 2a

EDIT: ack ! I should type faster :( Or wait.. no... Jackals proof is flawed :p

and thus the product(p(i),i=1..n) + 1 is a prime that did not belong to our original set of primes.

The product+1 is not necessarily a prime number. It must however be divisible by a primenumber not part of the set.

Axiom: Women = Time * Money
Axiom: Time = Money
Axiom: Money is the Root of all Evil

Then: Women = Money * Money

or: Women = Money^2

But: Money = root(Evil)

Then: Women = ((root(Evil))^2)

Thus: Women = Evil :eek:

Right? :p :D

According to Euclids unique prime factorization theorem every positive integer larger than one can be expressed as the product of a unique set of primes.

Assume now there is a limited number of n primes, say p1, p2, p3.. pn. Then define the number P = p1 x p2 x p2 x ... x pn +1 and let p be a prime dividing P.
If p would be from the set p1..pn it should be able to divide the difference P -p1 x p2 x p2 x ... x pn too. Clearly it cannot, since this difference is 1. Therefor p must be another prime, so it is not possible to have a non-infinite set.

Next assignment: show that the two solutions to a second order polynomial equation of the form ax^2+ bx +c = 0 are given by the quadratic formula:

x = [ -b +- sqrt(b^2 - 4ac) ] / 2a

EDIT: ack ! I should type faster :(

(x - [ -b + sqrt(b^2 - 4ac) ] / 2a)*(x - [ -b + sqrt(b^2 - 4ac) ] / 2a) = 0
=> (2ax+b-sqrt(b^2 - 4ac))*(2ax+b+sqrt(b^2 - 4ac)) = 0 (mult by 2a!=0)
=> (2ax+b)^2 - (b^2 - 4ac) = 0
=> 4a^2x^2 + 4abx + b^2 - b^2 + 4ac = 0
=> 4a*(ax^2+bx+c) = 0
=> (ax^2+bx+c) = 0

this shows they are the solutions :p

this shows they are the solutions :p

:D *applauds*
Please note my added comment about a small error you made in your prime proof btw ;)

Now it's up to you to give a proof of the following:

the sum of (1/n)^a, n=1..infinity converges If And Only If a>1.

The proof is as follows:

Suppose that a<1, then obviously the sum of (1/n)^a, n=1 would not converge based on Newton's non-convergence principle. The same would apply for the case where a=1.

As a result...it follows logically that the of (1/n)^a infinitely converges...and the theorem is thus proven.


Now it's my turn:

Prove that x + y = 2 when x=y=1

Good luck!

:D *applauds*
Please note my added comment about a small error you made in your prime proof btw ;)

i added a disclaimer :)

The problem at stake is:

prove that Q has as many elements as N
(or: the amount of natural numbers equals the amount of rational numbers)

And a bonus: show that R contains more numbers :p

i added a disclaimer :)

Hmm.. then your proof is apparantly different from mine... and I do not get yours :(
I see no reason to assume that P (=product +1) is a prime itself, nor that such an assumption is needed for the proof to be correct. All which I need is the observation that there must exist a primenumber that can divide P which was not part of the originally defined set. Which *could* be P itself, but doesn't have to be.

Hmm.. then your proof is apparantly different from mine... and I do not get yours :(
I see no reason to assume that P (=product +1) is a prime itself, nor that such an assumption is needed for the proof to be correct. All which I need is the observation that there must exist a primenumber that can divide P which was not part of the originally defined set. Which *could* be P itself, but doesn't have to be.

It's a proof that starts from an assumption that has to be proven false and leads to a contradiction => proving the assumption false.

I assume we have a finite (n) amount of primes, p1 to pn

i make the product and add 1 to it.
This number now has the property that it is equal to 1 modulo pj, for all j (or all primes - in the proof). It cannot be divided by any prime, so it is a prime itself. The number is not a part of our finite set of ALL primes, so this finite set does not exist.

edit: it is probably because i have the tendency to take 'bigger steps' when doing math than most people do, because they seem very clear to me but not to them. I had math teachers complaining about it in the past.
second edit: It is the same proof as yours, only written by me - almost by definition hard to read - and using other notations (I for example use product(...), modulo, you use P1*P2*...*Pn, and use the idea behind modulo, without writting it:))

prove that Q has as many elements as N
(or: the amount of natural numbers equals the amount of rational numbers)

And a bonus: show that R contains more numbers :p

*Is puzzled* I though there were more elements in Q than in N; after all, there's an infinite number of elements in the set {0<x<1:xEQ}, but none in the set {0<x<1:xEN}.

*Is puzzled* I though there were more elements in Q than in N; after all, there's an infinite number of elements in the set {0<x<1:xEQ}, but none in the set {0<x<1:xEN}.

two sets have 'the same amount of elements' when a bijection, one to one correspondance, between both sets exists. It's up to you to prove that you such a bijection exists.