Proof of the irrationality of the square root of 2 by lowest terms

Proof

We argue by contradiction. By the definition of an irrational number, it suffices to show that no fraction $a / b$ of integers with $b \neq 0$ has square $2$. The square root $\sqrt{2}$ is a real number with square $2$, so this shows that $\sqrt{2}$ is irrational.

1. Suppose that $a$ and $b$ are integers with $b \neq 0$ and $(a / b)^2 = 2$.
2. By the existence of a fraction in lowest terms, there are integers $r$ and $s$ with $s > 0$, $\gcd(r, s) = 1$ and $r / s = a / b$.
3. Then $(r / s)^2 = 2$. Multiplying by $s^2$ gives $r^2 = 2 s^2$.
4. The number $2$ is a prime number: a positive divisor of $2$ is at most $2$, so it is $1$ or $2$.
5. By step 3, $2$ divides $r \cdot r$. By Euclid's lemma, $2$ divides $r$. Write $r = 2 t$ with $t$ an integer.
6. Then $4 t^2 = 2 s^2$, so $s^2 = 2 t^2$. Thus $2$ divides $s \cdot s$, and Euclid's lemma gives $2 \mid s$.
7. So $2$ is a common divisor of $r$ and $s$. Then $\gcd(r, s) \ge 2$ by the definition of the greatest common divisor.
8. This contradicts $\gcd(r, s) = 1$. So there are no such $a$ and $b$, and $\sqrt{2}$ is irrational.