The square root of 2 is irrational

Let $x$ be a positive number such that $x^2 = 2$. Assume for the sake of contradiction that $x$ is rational. Then there exist coprime non-negative integers $a,b$, with $b \ne 0$, such that $x = \frac{a}{b}$. Then we have $2 = x^2 = \frac{a^2}{b^2}$, so $a^2 = 2b^2$. Since $2$ is prime and $2$ divides $a^2$, $2$ must divide $a$. Say $a = 2c$ for some integer $c$. Then $4c^2 = a^2 = 2b^2$, so $2c^2 = b^2$, and therefore $2$ divides $b$. But then $a$ and $b$ are not coprime, which is a contradiction. Therefore $x$ is irrational. QED
  • Thank you for your wonderful contribution! Always love to see some real maths here.

    15°

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