*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

\[\begin{eqnarray} & \sqrt{2} &=& \dfrac pq \\ \implies & 2 &=& \dfrac{p^2}{q^2} \small {\color{blue} ( \text{squaring both sides} )} \\ \implies & 2q^2 &=& p^2 \end{eqnarray}\]

Thus \(2 | p^2 \implies 2 | p\).

So let \(p = 2k\) for some integer \(k\). Substituting \(p\) in \(2q^2 = p^2\) gives

\[\begin{eqnarray} & 2q^2 &=& {(2k)}^2 \\ \implies & 2q^2 &=& 4k^2 \\ \implies & q^2 &=& 2k^2 \end{eqnarray}\]

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

\[\begin{eqnarray} & \sqrt{4} &=& \dfrac pq \\ \implies & 4 &=& \dfrac{p^2}{q^2} \small {\color{blue} ( \text{squaring both sides} )} \\ \implies & 4q^2 &=& p^2 \end{eqnarray}\]

*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

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

*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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