# 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.

• What is your question? Jan 21, 2020 at 7:03
• Is there a question here? Your proof is not the most succinct, but seems correct. Jan 21, 2020 at 7:03
• I just want to verify my proof so that i can continue to do other proof that rely on the result of this proof,thanks. Jan 21, 2020 at 7:07
• oh my mistake sorry sir Jan 21, 2020 at 7:08
• Your question title says $\frac{\sqrt{4}}{2}$. Your quiestion body says $\frac{\sqrt2}{4}$. Which is it?
– 5xum
Jan 21, 2020 at 7:09

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