isn't right to prove that $\sqrt{2}$/4 is irrational number?

Question

Assume $\sqrt{2}$/4 is rational number.

rational number have p/q in the lowest term.

\begin{align}\sqrt{2}/4 = p/q\\ \sqrt{2}=4p/q\\ 2=16p^2/q^2\\ 2q^2=16p^2\\ 2q^2=2(8p^2) \\ q^2=2(4p^2)\\ \end{align}

Then we know that q is even number according to if $q^2$ is even then $q$ is even.

Also,we have $2q^2$=$16p^2$ in the above,we know that $16p^2$ is a even number.

So we can construct the form in the below: \begin{align}2(16a)=16p^2\\p^2=2a \end{align}

Then we know that p is even number according to if $p^2$ is even then $p$ is even.

Since we assumed that $\sqrt{2}$/4 is in the lowest form of rational number,but the result showed p and q have the common factor 2.It causes to contradiction and means that our assumption is wrong.

Therefore,$\sqrt{2}$/4 is irrational number.

Answer 1

Hint $:$ If $\frac {\sqrt 2} {4}$ is a rational number then so is $\sqrt 2 = \frac 1 2 \times \frac {4} {\sqrt 2}.$ Is $\sqrt 2$ a rational number?

Answer 2

Seems fine.

Alternatively, suppose that $\frac{\sqrt2}4$ is rational number, $q$. Then $\sqrt{2}=4q$ is a rational number, which contradicts that $\sqrt2$ is irrational.