# 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

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

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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