# Disprove that $\sqrt{4}$ is irrational

You all must be familiar with the 'famous' proof by contradiction to prove the irrationality of $\sqrt{2}$. The following is a reminder of that proof. If you already know this, you may skip to the main question.

Problem:

Prove that $\sqrt{2}$ is irrational.

Solution:

Let us assume that $\sqrt{2}$ is rational. Then $\sqrt{2}$ can be expressed in the form of $\dfrac{p}{q}$ for coprime integers $p$ and $q$ with $q \neq 0$.

Now

\begin{aligned} & \sqrt{2} &=& \dfrac pq \\ \implies & 2 &=& \dfrac{p^2}{q^2} \small {\color{#3D99F6} ( \text{squaring both sides} )} \\ \implies & 2q^2 &=& p^2 \end{aligned}

Thus $2 | p^2 \implies 2 | p$.

So let $p = 2k$ for some integer $k$. Substituting $p$ in $2q^2 = p^2$ gives

\begin{aligned} & 2q^2 &=& {(2k)}^2 \\ \implies & 2q^2 &=& 4k^2 \\ \implies & q^2 &=& 2k^2 \end{aligned}

Thus $2 | q^2 \implies 2 | q$.

Hence $2$ is a common factor of both $p$ and $q$. But this contradicts the fact that $p$ and $q$ are coprime integers, hence our assumption is false.

$\therefore \$ $\sqrt{2}$ is irrational.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left( \mathbf{Q.E.D.} \right)$

$\large \text{Main Question}$

Let us do a similar procedure with $\sqrt{4}$.

Flawed problem:

Prove that $\sqrt{4}$ is irrational.

Flawed solution:

Step 1:

Let us assume that $\sqrt{4}$ is rational. Then $\sqrt{4}$ can be expressed in the form of $\dfrac{p}{q}$ for coprime integers $p$ and $q$ with $q \neq 0$.

Step 2:

Now

\begin{aligned} & \sqrt{4} &=& \dfrac pq \\ \implies & 4 &=& \dfrac{p^2}{q^2} \small {\color{#3D99F6} ( \text{squaring both sides} )} \\ \implies & 4q^2 &=& p^2 \end{aligned}

Step 3:

Thus $4 | p^2 \implies 4 | p$.

Step 4:

So let $p = 4k$ for some integer $k$. Substituting $p$ in $4q^2 = p^2$ gives

\begin{aligned} & 4q^2 &=& {(4k)}^2 \\ \implies & 4q^2 &=& 16k^2 \\ \implies & q^2 &=& 4k^2 \end{aligned}

Step 5:

Thus $4 | q^2 \implies 4 | q$.

Step 6:

Hence $4$ is a common factor of both $p$ and $q$. But this contradicts the fact that $p$ and $q$ are coprime integers, hence our assumption is false.

$\therefore \$ $\sqrt{4}$ is irrational.

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left( \mathbf{Q.E.D.} \right)$

But we know that $\sqrt{4} = 2$ is rational.

In which step in the flawed solution shown above has the error been committed for the first time?

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