📂Number Theory

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.

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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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See Also

• $\pi$ is irrational
• $e$ is irrational