Crazy math

[theboss]

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:

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2² = √4 = (-2)²

You can also look for a more complex answer, literally
(i is the imaginary number)

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-1 = i² 
(-1)² = (i²)²
1 = i^4
√1 = √i^4
1 = i²

i² = -1 v i² = 1

Thus proving
1 = -1

- One equals two
Anthon sent me this one a few day's back,

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a = b
a² = a*b
a²-b² = a*b-b²
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 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.

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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

[capncanuck]

N.B:, the troll means and not .

[SpikedMath]

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: is irrational for .

Proof: Let and suppose is rational. Then there exist relatively prime integers and so that



Then,



Hence:



However, by Fermat's Last Theorem, no such integers and can exist. This is a contradiction. Therefore, is irrational.

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

[capncanuck]

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: is irrational for .

Proof: Let and suppose is rational. Then there exist relatively prime integers and so that



Then,



Hence:



However, by Fermat's Last Theorem, no such integers and can exist. This is a contradiction. Therefore, is irrational.

For , 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?

[xander]

For , 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

[E_net4]

<pic>
N.B:, the troll means and not .

You can do the same thing for the diagonal of a square.

[Zapp]

For , 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...

[DeathRowKitty]

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.

[xander]

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

[SpikedMath]

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...

[Zapp]

remember the strip with Pi = 3?
http://spikedmath.com/118.html

[DeathRowKitty]

Consider the funciton . Notice that we can rewrite this as

Consider the equation . We can solve for x as follows:






We can solve similarly:






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

[WhoDatMath]

For , 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 where becomes the non-square number and also stands for yeah, dude, you just got rolled.

[WhoDatMath]

Consider the funciton . Notice that we can rewrite this as

.
.
.

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.

[theboss]

That graph really does look awesome

[WhoDatMath]

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

[DeathRowKitty]

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

[WhoDatMath]

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.

[theboss]

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 )

[theboss]

Damn you, your graph is correcter and posted before mine xD

Consider the funciton . Notice that we can rewrite this as

.
.
.

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.