User:Paxinum/Proof styles

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The square rooth of 2 is irrational theorem[edit]

This result uses the following: [hide]
Definition of rational number.
Definition of prime and coprime.
Definition of square rooth.
Gödels incompleteness theorem =)

The square rooth of 2 is irrational,


This is a proof by contradiction, so we assumes that and hence for some a, b that are coprime.

This implies that . Rewriting this gives .

Since the left-hand side of the equation is divisible by 2, then so must the right-hand side, i.e., . Since 2 is prime, we must have that .

So we may substitute a with , and we have that .

Dividing both sides with 2 yields , and using similar arguments as above, we conclude that .

Here we have a contradiction; we assumed that a and b were coprime, but we have that and .

Hence, the assumption were false, and cannot be written as a rational number. Hence, it is irrational.


Some nice history about the one that first proved this theorem.