Prove that $n$ is divisible by $5$ if and only if $n^2 $is divisible by $5$ and use it to prove that $\sqrt{5}$ is irrational

Question

Not sure about what cases to consider in part 1 of the proof and how to use it to prove the next part.

Mimic the classic proof that $\sqrt{2}$ is irrational. – Bill Dubuque Oct 22 '18 at 23:49 — Bill Dubuque, Oct 22 '18 at 23:49

score 4 · Answer 1 · edited Oct 23 '18 at 00:24

First part Every natural number takes one of the following forms $(k\in\Bbb Z)$: $$5k$$ $$5k+1$$ $$5k+2$$ $$5k+3$$ $$5k+4$$

You can then use the contrapositive to verify that statement, i.e. prove that:

$$5\nmid n \to 5\nmid n^2$$ by squaring each of the forms.

Second part: take $$\sqrt5=\frac ab\to 5=\frac{a^2}{b^2}\to 5b^2=a^2$$ which implies $a^2$ is a multiple of $5$ and hence $a$ is too (because of the first part you've already proven). You can take it from there.

score 0 · Answer 2 · answered Oct 23 '18 at 00:13

0

Suppose $p$ is a prime number and $p | ab$. Then $p|a$ or $p|b$.
Use this fact. The other proposed solution also works.

answered Oct 23 '18 at 00:13

ncmathsadist

49,383

Prove that $n$ is divisible by $5$ if and only if $n^2 $is divisible by $5$ and use it to prove that $\sqrt{5}$ is irrational

2 Answers2