<s>[INST] isn't right to prove that $\sqrt{2}$/4 is irrational number? 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.
 [/INST] 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?
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