CFJ's The One I Cannot Kill [Game Ended]

[or]

Vote count 1:

Ircher:
Elements:
TemporalLich :

Takes 2 to lynch.

[Elements]

yea, primes just break it

[Elements]

it's like the 21 game

[Ircher]

Whoever proves the sqrt of two is irrational to my satisfaction gains immunity from me.

[Elements]

although that works slightly better with 3 players

[Ircher]

Whoever proves the sqrt of two is irrational first to my satisfaction gains immunity from me.

EBWOP

[Elements]

let a^2/B^2 = 2
=> 2B^2 = a^2
=>root(2)B=a
therefore B is a multiple of a such the a/b can be simplified indefinitely

[Elements]

something like that

[TemporalLich]

I might need a lead rist

[Ircher]

That's not rigorous or clear enough.

[Ircher]

You can use Google.

[TemporalLich]

the easy way out is for me to just put a wikipedia link inthread

[Elements]

https://www.homeschoolmath.net/teaching ... tional.php

[Elements]

beat you to it

[Ircher]

that would be too lazy.
At least copy/paste the proof

[TemporalLich]

congrats enjoy your immunity... I hope to earth you're town

[Elements]

First one to prove Fermat's last theorem doesn't get lynched

[Ircher]

that would be too lazy.
At least copy/paste the proof

[TemporalLich]

copypasta from wikipedia:

Proof by unique factorization

An alternative proof uses the same approach with the fundamental theorem of arithmetic which says every integer greater than 1 has a unique factorization into powers of primes.

Assume that √2 is a rational number. Then there are integers a and b such that a is coprime to b and √2 = a/b. In other words, √2 can be written as an irreducible fraction.
The value of b cannot be 1 as there is no integer a the square of which is 2.
There must be a prime p which divides b and which does not divide a, otherwise the fraction would not be irreducible.
The square of a can be factored as the product of the primes into which a is factored but with each power doubled.
Therefore, by unique factorization the prime p which divides b, and also its square, cannot divide the square of a.
Therefore, the square of an irreducible fraction cannot be reduced to an integer.
Therefore, √2 cannot be a rational number.

This proof can be generalized to show that if an integer is not an exact kth power of another integer then its kth root is irrational. For a proof of the same result which does not rely on the fundamental theorem of arithmetic, see: quadratic irrational.

[Elements]

why earth?

[Ircher]

Not what I would go with (there's a slightly simpler proof that employs the same idea), but I'll accept that.
VOTE: Elements

[TemporalLich]

because if town loses the earth loses one iota

[Elements]

I prefer the divisible by 2 proof

[TemporalLich]

elements, claim your favorite color now

[Elements]

one iota of what?