Disprove that 4\sqrt{4} is irrational

You all must be familiar with the 'famous' proof by contradiction to prove the irrationality of 2\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 2\sqrt{2} is irrational.

Solution:

Let us assume that 2\sqrt{2} is rational. Then 2\sqrt{2} can be expressed in the form of pq\dfrac{p}{q} for coprime integers pp and qq with q0q \neq 0.

Now

2=pq    2=p2q2(squaring both sides)    2q2=p2\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 2p2    2p2 | p^2 \implies 2 | p.

So let p=2kp = 2k for some integer kk. Substituting pp in 2q2=p22q^2 = p^2 gives

2q2=(2k)2    2q2=4k2    q2=2k2\begin{aligned} & 2q^2 &=& {(2k)}^2 \\ \implies & 2q^2 &=& 4k^2 \\ \implies & q^2 &=& 2k^2 \end{aligned}

Thus 2q2    2q2 | q^2 \implies 2 | q.

Hence 22 is a common factor of both pp and qq. But this contradicts the fact that pp and qq are coprime integers, hence our assumption is false.

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

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


Main Question\large \text{Main Question}

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

Flawed problem:

Prove that 4\sqrt{4} is irrational.

Flawed solution:

Step 1:

Let us assume that 4\sqrt{4} is rational. Then 4\sqrt{4} can be expressed in the form of pq\dfrac{p}{q} for coprime integers pp and qq with q0q \neq 0.

Step 2:

Now

4=pq    4=p2q2(squaring both sides)    4q2=p2\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 4p2    4p4 | p^2 \implies 4 | p.

Step 4:

So let p=4kp = 4k for some integer kk. Substituting pp in 4q2=p24q^2 = p^2 gives

4q2=(4k)2    4q2=16k2    q2=4k2\begin{aligned} & 4q^2 &=& {(4k)}^2 \\ \implies & 4q^2 &=& 16k^2 \\ \implies & q^2 &=& 4k^2 \end{aligned}

Step 5:

Thus 4q2    4q4 | q^2 \implies 4 | q.

Step 6:

Hence 44 is a common factor of both pp and qq. But this contradicts the fact that pp and qq are coprime integers, hence our assumption is false.

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

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


But we know that 4=2\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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