Proof that the Square Root of 2 is Irrational

We've looked at the operations of addition and multiplication over the field of real numbers. One next might question is how $\sqrt{2} \approx 1.4142...$ might arise with only these two operations since at best we can only form rational numbers, that is numbers in the set $\mathbb{Q} := \{ \frac{a}{b} : a, b \in \mathbb{Z} \: \mathrm{and} \: b \neq 0 \}$. The following proof will illustrate that $\sqrt{2} \not \in \mathbb{Q}$.

Theorem 1: There exists no rational number $r = \frac{a}{b}$ ($a, b \in \mathbb{Z}$ and $b \neq 0$) such that $r^2 = 2$.

• Proof of Theorem: We will do this proof by contradiction. Suppose that $r$ is an irrational number in lowest terms, that is the integers $a$ and $b$ have 1 as their greatest common divisors commonly abbreviated as $\mathrm{gcd}(a, b)= 1$, and $r$ is such that $r^2 = 2$. Therefore:

(1)

\begin{align} r^2 = 2 \\ \left ( \frac{a}{b} \right )^2 = 2 \\ \frac{a^2}{b^2} = 2 \\ a^2 = 2b^2 \end{align}

• Therefore it follows that $a^2$ is even which implies that $a$ is even. Since $a$ is even it can be written as $a = 2m$ for some $m \in \mathbb{Z}$. Making this substitution we get that:

(2)

\begin{align} a^2 = 2b^2 \\ (2m)^2 = 2b^2 \\ 4m^2 = 2b^2 \\ 2m^2 = b^2 \end{align}

• And we see that $b^2$ is also even which implies $b$ is even so $b$ can be written as $b = 2n$ for some $n \in \mathbb{Z}$.

• Since $a = 2m$ and $b = 2n$ for some $m, n \in \mathbb{Z}$, it follows that $\mathrm{gcd}(a, b) = \mathrm{gcd}(2m, 2n) ≥ 2$ which contradicts our original assumption that $r$ was rational. $\blacksquare$