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.

Prove that $\sqrt{2}$ is an Irrational Number

We want to prove that $\sqrt{2}$ is irrational. What assumption should we begin with?

Suppose $\sqrt{2}$ is rational, so it can be written as $\dfrac{p}{q}$ .

What can we say about the integers $p$ and $q$ ?

Where $p, q \in \mathbb{Z}$ and $q \ne 0$ .

What assumption do we make about the form of $\dfrac{p}{q}$ ?

Assume that $\dfrac{p}{q}$ is in its lowest terms.

How do we epxress $\sqrt{2}$ as a rational number?

We can write $\sqrt{2}$ as $\sqrt{2} = \dfrac{p}{q}$

Simplify

$\Rightarrow 2 = \dfrac{p^2}{q^2}$

Simplify

$\Rightarrow p^2 = 2q^2 \quad \quad [1]$

What can we say about $p^2 = 2q^2$ ?

The R.H.S is even because it is a multiple of 2.

If R.H.S is even, then;

$\Rightarrow$ L.H.S is also even.

If L.H.S is even, then;

$\Rightarrow p^2$ is even

What can we say about squaring integers?

The square of an even number is always even and the square of an odd number is always odd.

If $p^2$ is even, then;

$\Rightarrow p$ is even $\quad \quad [2]$

How do we represent $p$ to reflect that it’s even?

If $p$ is even, we can say $p = 2p'$ where $p'$ is also an integer.

How do we write $p^2$ in term of $p'$

$\Rightarrow p^2 = 4p'^2$

How do we use the relation $p^2 = 4p'^2$ ?

Substitute $p^2$ with $4p'^2$ in $[1]$

Substitution results in?

$\Rightarrow 4p'^2 = 2q^2$

Simplify

$\Rightarrow 2p'^2 = q^2 \quad \quad [3]$

What can we say about $2p^2 = q^2$ ?

The L.H.S. is even because it is a multiple of 2.

If L.H.S. is even, then;

$\Rightarrow$ R.H.S is also even.

If R.H.S is even, then;

$\Rightarrow q^2$ is even $\quad \quad [4]$

If $q^2$ is even, then;

$\Rightarrow q$ is even

From $[2]$ and $[4]$

Since both $p$ and $q$ are even, they share a common factor of 2.

What conclusion do we draw from this contradiction?

This contradicts the assumption that $\dfrac{p}{q}$ is in lowest terms.

Final conclusion?

$\therefore$ Our initial assumption must be false. We cannot write $\sqrt{2}$ in the form of $\dfrac{p}{q}$ . Thus, $\sqrt{2}$ is irrational.

End of Lesson

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

Algebraic Method

Prove that 2\sqrt{2}2​ is an Irrational Number

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

Prove that $\sqrt{2}$ is an Irrational Number