Proof That The Square Root of 3 is Irrational

We recently looked at the Proof That The Square Root of 2 is Irrational. We will now proceed to prove that $\sqrt{3} \not \in \mathbb{Q}$.

Theorem 1: There exists no rational number $r = \frac{a}{b}$ ($a, b \in \mathbb{Z}$ and $b \neq 0$) such that $r^2 = 3$.

• Proof: Once again we will prove this by contradiction. Suppose that there exists a rational number $r = \frac{a}{b}$ such that $r^2 = 3$. Let $r$ be in lowest terms, that is the greatest common divisor of $a$ and $b$ is $1$, or rather $\mathrm{gcd}(a, b) = 1$. And so:

(1)

\begin{align} r^2 = 3 \\ \frac{a^2}{b^2} = 3 \\\ a^2 = 3b^2 \end{align}

• We have two cases to consider now. Suppose that $b$ is even. Then $b^2$ is even, and $3b^2$ is even which implies that $a^2$ is even and so $a$ is even, but this cannot happen. If both $a$ and $b$ are even then $\mathrm{gcd}(a,b) ≥ 2$ which is a contradiction.

• Now suppose that $b$ is odd. Then $b^2$ is odd and $3b^2$ is odd which implies that $a^2$ is odd and so $a$ is odd. Since both $a$ and $b$ are odd, we can write $a =2m - 1$ and $b = 2n - 1$ for some $m, n \in \mathbb{N}$. Therefore:

(2)

\begin{align} a^2 = 3b^2 \\ (2m - 1)^2 = 3(2n - 1)^2 \\ 4m^2 - 4m + 1 = 3(4n^2 - 4n + 1) \\ 4m^2 - 4m + 1 = 12n^2 - 12n + 3 \\ 4m^2 - 4m = 12n^2 - 12n + 2 \\ 2m^2 - 2m = 6n^2 - 6n + 1 \\ 2(m^2 - m) = 2(3n^2 - 3n) + 1 \end{align}

• We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number $r$ such that $r^2 = 3$. $\blacksquare$