## Crazy math

Post and discuss your favorite math jokes.

### Crazy math

Here's a thread from my site that annoyed quite some people, as it does stuff that mathematicians don't like. Prove things that are wrong, in ways that seem right to the common folks out there.
So enjoy (or not)

-All negative numbers are equal to their positive counterparts.

There are several ways to prove this, the most logical is using roots and powers.
Example:
Code: Select all
2² = √4 = (-2)²

You can also look for a more complex answer, literally
(i is the imaginary number)
Code: Select all
-1 = i²  (-1)² = (i²)²1 = i^4√1 = √i^41 = i²i² = -1 v i² = 1Thus proving1 = -1

- One equals two
Anthon sent me this one a few day's back,
Code: Select all
a = ba² = a*ba²-b² = a*b-b²(a+b)(a-b) = b(a-b)(a+b) = ba+a = a2a = a2 = 1

There is a problem with it tho, at a point you divide by 0, making it impossible.

- Zero equals one
Here's a fun one suggested by a friend.
Code: Select all
0 = 0 + 0 + 0 + 0 + ...0 = (1-1) + (1-1) + (1-1) + ...0 = 1 + (-1+1) + (-1+1) + (-1+1) + ....0 = 1 + 0 + 0 + 0 + 0 + ...0 = 1

Source: http://www.parascientifica.com/forums/v ... =32&t=5487
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theboss
University

Posts: 147
Joined: Thu Feb 10, 2011 4:11 pm

### Re: Crazy math

N.B:, the troll means $8$ and not $8!$.
A Holocaust Survivor wrote:I will never say anything that couldn't stand as the last thing I ever say.
capncanuck
Elementary School

Posts: 17
Joined: Thu Feb 10, 2011 3:52 pm

### Re: Crazy math

One proof I like, which may or may not beg the question is the following. You can find a similar proof on various websites and probably the xkcd forums as well.

Theorem: $\sqrt[n]{2}$ is irrational for $n \geq3$.

Proof: Let $n \geq3$ and suppose $\sqrt[n]{2}$ is rational. Then there exist relatively prime integers $p$ and $q$ so that

$\Large \frac{p}{q}=\sqrt[n]{2}.$

Then,

$\Large\frac{p^n}{q^n} = 2.$

Hence:

$\Large p^n = q^n + q^n.$

However, by Fermat's Last Theorem, no such integers $p$ and $q$ can exist. This is a contradiction. Therefore, $\sqrt[n]{2}$ is irrational. $\qed$

For $n=2$, Fermat's Last Theorem does not apply so we need a different technique for that case.
Math - It's in you to give.

SpikedMath

Posts: 133
Joined: Mon Feb 07, 2011 1:31 am

### Re: Crazy math

SpikedMath wrote:One proof I like, which may or may not beg the question is the following. You can find a similar proof on various websites and probably the xkcd forums as well.

Theorem: $\sqrt[n]{2}$ is irrational for $n \geq3$.

Proof: Let $n \geq3$ and suppose $\sqrt[n]{2}$ is rational. Then there exist relatively prime integers $p$ and $q$ so that

$\Large \frac{p}{q}=\sqrt[n]{2}.$

Then,

$\Large\frac{p^n}{q^n} = 2.$

Hence:

$\Large p^n = q^n + q^n.$

However, by Fermat's Last Theorem, no such integers $p$ and $q$ can exist. This is a contradiction. Therefore, $\sqrt[n]{2}$ is irrational. $\qed$

For $n=2$, Fermat's Last Theorem does not apply so we need a different technique for that case.

Is this a joke of the 2nd degree?
Posting a valid proof in a section for math jokes?
A Holocaust Survivor wrote:I will never say anything that couldn't stand as the last thing I ever say.
capncanuck
Elementary School

Posts: 17
Joined: Thu Feb 10, 2011 3:52 pm

### Re: Crazy math

SpikedMath wrote:For $n=2$, Fermat's Last Theorem does not apply so we need a different technique for that case.

Leaving the most difficult case for last, eh?

xander

xander
University

Posts: 154
Joined: Fri Feb 11, 2011 12:14 am
Location: Sparks, NV, USA

### Re: Crazy math

capncanuck wrote:<pic>
N.B:, the troll means $8$ and not $8!$.
You can do the same thing for the diagonal of a square.
E_net4
Kindergarten

Posts: 7
Joined: Thu Feb 10, 2011 3:55 pm

### Re: Crazy math

xander wrote:
SpikedMath wrote:For $n=2$, Fermat's Last Theorem does not apply so we need a different technique for that case.

Leaving the most difficult case for last, eh?

xander

the proof for sqareroot 2 is very similar
when we had p² = q²+q² we could write it as p² = 2q² which leads to the conclusion that p² and p must be even, therefore not a prime, which contradicts the original assumption that p is a prime.

in hindsight, I think I have been trolled...
Q.E.D. , or not?
Zapp
University

Posts: 124
Joined: Thu Feb 10, 2011 3:49 pm

### Re: Crazy math

Well, I'm no expert on the proof of Fermat's Last Theorem, but I assume the joke part of the proof lies in the fact that Fermat's Last Theorem requires that the nth roots of non nth-powers are irrational. I'm not entirely sure if the proof of Fermat's Last Theorem makes some sort of implicit use of that fact somewhere (and I'm not about to try to look through the proof to figure it out ), but it seems likely something in the proof would require that.

DeathRowKitty
Mathlete

Posts: 68
Joined: Fri Feb 11, 2011 8:38 am

### Re: Crazy math

Zapp wrote:in hindsight, I think I have been trolled...

:P

I'll bet you are going to ask me to prove that there are infinitely many primes next, eh?

xander

xander
University

Posts: 154
Joined: Fri Feb 11, 2011 12:14 am
Location: Sparks, NV, USA

### Re: Crazy math

xander wrote:
Zapp wrote:in hindsight, I think I have been trolled...

I'll bet you are going to ask me to prove that there are infinitely many primes next, eh?

xander

I like the proof of infinitely many composites: Assume there are finitely many composites...
Math - It's in you to give.

SpikedMath

Posts: 133
Joined: Mon Feb 07, 2011 1:31 am

### Re: Crazy math

remember the strip with Pi = 3?
http://spikedmath.com/118.html
Q.E.D. , or not?
Zapp
University

Posts: 124
Joined: Thu Feb 10, 2011 3:49 pm

### Re: Crazy math

Consider the funciton $f(x)=x^{x^{x^{x...}}}$. Notice that we can rewrite this as $f(x)=x^{f(x)}$

Consider the equation $f(x)=2$. We can solve for x as follows:

$f(x)=2$
$x^{f(x)}=2$
$x^2=2$
$x=\sqrt{2}$

We can solve $f(x)=4$ similarly:

$f(x)=4$
$x^{f(x)}=4$
$x^4=4$
$x=\sqrt{2}$

Since each value of x corresponds to only one value of f(x), we conclude that 2=4.

DeathRowKitty
Mathlete

Posts: 68
Joined: Fri Feb 11, 2011 8:38 am

### Re: Crazy math

Zapp wrote:
xander wrote:
SpikedMath wrote:For $n=2$, Fermat's Last Theorem does not apply so we need a different technique for that case.

Leaving the most difficult case for last, eh?

xander

the proof for sqareroot 2 is very similar
when we had p² = q²+q² we could write it as p² = 2q² which leads to the conclusion that p² and p must be even, therefore not a prime, which contradicts the original assumption that p is a prime.

in hindsight, I think I have been trolled...

This can be generalized to prove all roots of non-squares are irrational, it leads to $p^2 = Tq^2$ where $T$ becomes the non-square number and also stands for yeah, dude, you just got $T$rolled.
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: Crazy math

DeathRowKitty wrote:Consider the funciton $f(x)=x^{x^{x^{x...}}}$. Notice that we can rewrite this as $f(x)=x^{f(x)}$

.
.
.

Since each value of x corresponds to only one value of f(x), we conclude that 2=4.

I'd like to point out that the graph of f(x) would be sick. Undefined at zero, zero for everything else between -1 and 1, 1 at 1, -1 at -1, then infinity for everything outside of that. I'll leave the sketch of that graph as an exercise to the reader.
Attachments
Never mind, I made a sketch.
f of x.JPG (13.53 KiB) Viewed 2823 times
Last edited by WhoDatMath on Fri Feb 11, 2011 3:26 pm, edited 1 time in total.
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: Crazy math

That graph really does look awesome
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theboss
University

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### Re: Crazy math

But wait, for values between 0 and 1, f(x) would go to 1 wouldn't it?

EDIT: Yeah, it would...but I'll leave the graph up. I worked hard on it, even if it is wrong
Last edited by WhoDatMath on Fri Feb 11, 2011 3:32 pm, edited 1 time in total.
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: Crazy math

WhoDatMath wrote:I'd like to point out that the graph of f(x) would be sick. Undefined at zero, zero for everything else between -1 and 1, 1 at 1, -1 at -1, then infinity for everything outside of that. I'll leave the sketch of that graph as an exercise to the reader.

It wouldn't actually be 0 for everything else between -1 and 1. If you type out a bunch of terms on a calculator, you'll find that it will eventually converge to some non-zero number for non-zero values.

The function can actually be written in terms of a certain named function, but I won't say which one yet because I don't want to give away the answer

pre-post edit: it doesn't go to 1 either

DeathRowKitty
Mathlete

Posts: 68
Joined: Fri Feb 11, 2011 8:38 am

### Re: Crazy math

DeathRowKitty wrote:pre-post edit: it doesn't go to 1 either

Hmmm...I wonder why not. You take the root (any root) of a number less than 1, the root must be higher than the number. But it can't be one, because only one's root is one. So do it over and over, it gets bigger and bigger without reaching 1.

Logically, it seems like that would go to 1.

If it doesn't, I'll never figure out what it goes to.
WhoDatMath
Mathlete

Posts: 78
Joined: Thu Feb 10, 2011 4:53 pm

### Re: Crazy math

WhoDatMath wrote:But wait, for values between 0 and 1, f(x) would go to 1 wouldn't it?

It will converge to 0 for everything below 1 and -1, 1 and -1 themselves will stay one though.
So it will look something like the graph in my sig, where red is on and -1, blue is >1 and orange is <1
(assuming that infinity is a value that can be drawn on a graph )
Attachments
Graph of f(x) = x^(f(x))
Untitled.png (5.19 KiB) Viewed 2816 times
I have a physics/science/tech/general nerd forum, take a look if you want
Parascientifica's News Feed wrote:

Yes you can click that thing. ^

theboss
University

Posts: 147
Joined: Thu Feb 10, 2011 4:11 pm

### Re: Crazy math

Damn you, your graph is correcter and posted before mine xD
WhoDatMath wrote:
DeathRowKitty wrote:Consider the funciton $f(x)=x^{x^{x^{x...}}}$. Notice that we can rewrite this as $f(x)=x^{f(x)}$

.
.
.

Since each value of x corresponds to only one value of f(x), we conclude that 2=4.

I'd like to point out that the graph of f(x) would be sick. Undefined at zero, zero for everything else between -1 and 1, 1 at 1, -1 at -1, then infinity for everything outside of that. I'll leave the sketch of that graph as an exercise to the reader.
I have a physics/science/tech/general nerd forum, take a look if you want
Parascientifica's News Feed wrote:

Yes you can click that thing. ^

theboss
University

Posts: 147
Joined: Thu Feb 10, 2011 4:11 pm