Proof: $\sqrt{2}$ is an Irrational Number

The journey of theoretical mathematics usually starts with this basic proof.

$\sqrt{2}$ has an approximate value of $1.41421356237 \ldots$. And these decimals never turn repeating or terminate such that we could write an accurate fraction like $\dfrac{p}{q}$ to represent it. There are different ways to show this proof. Almost all methods involve a contradiction where first we assume that $\sqrt{2}$ is rational, and then show how this leads to a contradiction by manipulating equations or using geometry.

The method shown here uses simple algebraic manipulation to create a contradiction thus showing that $\sqrt{2}$ is not a rational number, which makes it an irrational number by derivation.

Algebraic Method

The proof first starts by assuming that $\sqrt{2}$ is rational and can be written in the form of $\dfrac{p}{q}$ where $p$ and $q$ do not have any common factors. We square the expression to remove the square root sign.

The next step is to establish that $p$ must have 2 as a factor and thus can be replaced by $2p'$. Simplifying this expression leads to the conclusion that $q$ must also have 2 as a factor. If both $p$ and $q$ have a common factor, it contradicts the early claim that we can write $\sqrt{2}$ as $\dfrac{p}{q}$ where $p$ and $q$ are integers in their lowest forms.

Best of luck in going through the step by step proof. You need to get all steps correct for the proof to count as successful.

We can use the same logic as finding both $p$ and $q$ to have the same factors for all prime numbers.