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,  \sqrt{2} \notin \mathbb{Q}


This is a proof by contradiction, so we assumes that  \sqrt{2} \in \mathbb{Q} and hence  \sqrt{2} = a/b for some a, b that are coprime.

This implies that 2 = \frac{a^2}{b^2}. Rewriting this gives 2b^2 = a^2 \!\,.

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

So we may substitute a with 2a', and we have that 2b^2 = 4a^2 \!\,.

Dividing both sides with 2 yields b^2 = 2a^2 \!\,, and using similar arguments as above, we conclude that 2 | b .

Here we have a contradiction; we assumed that a and b were coprime, but we have that 2 | a and 2 | b .

Hence, the assumption were false, and  \sqrt{2} cannot be written as a rational number. Hence, it is irrational.


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