Proof that the Square Root of 2 is Irrational
Theorem
$\sqrt{2}$ is irrational.
Proof
Strategy: To show $\sqrt{2}$ is irrational, we assume it can be expressed as a fraction in lowest terms and derive a contradiction. This method can be used to prove that $\sqrt{n}$ is irrational for every $n$ that is not a perfect square.
Assuming $\sqrt{2}$ is rational, it can be expressed as a fraction of two natural numbers, $a,b$ and $\displaystyle \sqrt{2} = {{ a } \over {b}}$, that are coprime. Multiplying both sides by $b$ gives $$ \sqrt{2} b= a $$ Squaring both sides yields $$ 2 b^2 = a^2 $$ Since $a^2$ is the product of $2$ and $b^2$, it is even, and thus, $a$ must also be even. This means that $a$ can be expressed as some natural number $A$ times $a = 2 A$. $$ 2 b^2 = (2A)^2 = 4 A^2 $$ Dividing both sides by $2$ gives $$ b^2 = 2 A^2 $$ Since $b^2$ is the product of $2$ and $A^2$, it is even, and thus, $b$ must also be even. This means that $b$ can be expressed as some natural number $B$ times $b = 2 B$. However, given the expression was assumed as $\displaystyle \sqrt{2} = {{ a } \over {b}}$, $$ \sqrt{2} = {{ a } \over {b}} = {{ 2A } \over {2B}} $$ This contradicts the assumption that $a$ and $b$ are coprime. Therefore, $\sqrt{2}$ is irrational.
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