Famous Theorems of Mathematics/Proof style

This is an example on how to design proofs. Another one is needed for definitions and axioms.

Irrationality of the square root of 2Edit

This result uses the following:

• Definition of rational number.
• Definition of prime and coprime.
• Definition of square root.

The square root of 2 is irrational, ${\sqrt {2}}\notin \mathbb {Q}$${\sqrt {2}}\notin \mathbb {Q}$

ProofEdit

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

This implies that $2={\frac {a^{2}}{b^{2}}}$$2={\frac {a^{2}}{b^{2}}}$. Rewriting this gives $2b^{2}=a^{2}\!\,$$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}$$2|a^{2}$. Since 2 is prime, we must have that $2|a$$2|a$.

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

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

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

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

NotesEdit

• As a generalization one can show that the square root of every prime number is irrational.
• Another way to prove the same result is to show that $x^{2}-2$$x^{2}-2$ is an irreducible polynomial in the field of rationals using Eisenstein's criterion.