# Prove by contradiction: For all prime 𝑝, √𝑝 is irrational

Prove by contradiction: For all prime $$p$$, $$\sqrt p$$ is irrational.

Hint: Use the following theorem: For every prime $$p$$ and all integers $$a,b$$ if $$p\mid ab$$, then $$p\mid a$$ or $$p\mid b$$.

I am currently here: Assume $$\sqrt p$$ is rational, then, $$\sqrt p = \dfrac a b$$ where a and b are integers. Then, $$p = \dfrac{a^2}{b^2}$$

Now I am stuck, I don't understand how to proceed from here nor how to use the theorem given as a hint.

• Clear of fractions. – Lubin Oct 5 '20 at 4:28
• @Atlecx Have you seen the proof that $\sqrt2$ is irrational? The approach is the same, but the divisibility conditions are different. – Toby Mak Oct 5 '20 at 8:12

We add one more point that assuming $$\sqrt{p}$$ to be rational we have that $$a,b$$ are coprime.
proceeding from your method $$b^2=\frac{a^2}{p}..(1)$$ or $$p|a^2$$ or $$p|a$$ (using the theorem given as Hint). Hence we can assume that $$a=pk$$ where $$k$$ is some positive integer.
Hint:Show that using the substituition $$a=kp$$ in (1) that $$p|b$$ and hence $$a,b$$ are not coprime.
Hint: You can assume it further by making $$\sqrt p = \dfrac a b$$ where $$\dfrac a b$$ is the reduced form fraction. Then prove by contradiction
Additional hint: Prove that $$\dfrac a b$$ is also not in reduced form, which conflicts with the original assumption