# A Quick Little Proof

Let’s prove that $\sqrt{3}$ is irrational. An irrational number is one that cannot be written in the form $\frac{a}{b}$, where $a$ and $b$ are both integers; in other words there is no repeating pattern in it’s decimal.

First, let’s prove that if $n^2$ is a multiple of $3$ then so is $n$. Assume that $n$ is not a multiple of $3$. Then $n$ can be written as: $n = 3m + \ell$ where $\ell \in \{1,2\}$. Then $n^2 = 9m^2 + 6m\ell + \ell^2$. We can factor out a $3$, so $n^2 = 3(3m^2 + 2m\ell) + \ell^2$. This implies $n^2$ is not a multiple of $3$ since it’s an integer times $3$ plus a number that isn’t a multiple of $3$ (that’s the $\ell$ part). We have proved the contrapositive which implies our original proposition ($P \Rightarrow Q$ is the same as $\neg Q \Rightarrow \neg P$).

Now let’s move on to the main proof. Assume, by way of contradiction that the square root of $3$ is rational. If the square root of $3$ is rational then we can write $\sqrt{3} = \frac{a}{b}$ with $a$ and $b$ not sharing any common factors. We can do some math: $3 = \frac{a^2}{b^2}$ which means $3b^2 = a^2$. But this means $a^2$ is a multiple of $3$ since it’s an integer times $3$. We just proved that if $n^2$ is a multiple of $3$ then so is $n$. This means that $a$ is a multiple of 3. But if $a$ is a multiple of $3$ it can be written as $3m$. Let’s plug this in to get $3b^2 = (3m)^2$, simplifying we get that $3b^2 = 9m^2$. Let’s divide through by $3$ to get $b^2 = 3m^2$. This means that $b^2$ is a multiple of $3$ since it’s $3$ times an integer. But again, if $b^2$ is a multiple of $3$ then so is $b$. We started off by assuming that $a$ and $b$ had no common factors and we just showed that they share a common factor of $3$. We’ve derived a contradiction to our proposition that $\sqrt{3}$ is rational, therefore it must be irrational.