Prove that $\sqrt3$ is irrational. Hint: To treat $\sqrt3$, for example, use the fact that every integer is of the form $3n$, $3n+1$ or $3n+2$.
Solution from the "Answers" Chapter:
Since $$ \\ (3n+1)^2 = 9n^2 +6n + 1 = 3(3n^2+2n)+1 \\ (3n+2)^2 = 9n^2 +12n + 4 = 3(3n^2+4n +1)+1 $$ He proceeds then to state that if $k^2$ is divisible by 3, then so is $k$. I have a hard time understanding what $k$ he is talking about in these equations, it hasn't been defined earlier. He then proceeds:
Supose $\sqrt3$ were rational, and let $\sqrt3 = p/q$, where $p$ and $q$ have no common factor. Then $p^2=3q^2$, so $p^2$ is divisible by 3, so $p$ must be. Thus, $p=3p'$ for some natural number $p'$ and consequently $(3p')^2=3q^2$, or $3(p')^2=q^2$. Thus, $q$ is also divisible by 3, a contradiction.
I have a hard time linking the first part where he states that "$k^2$ being divisible by 3 leads to $k$ being divisible by 3" and the second part. Where does the first conclusion come from?
NOTE: Problem statement in the book in the same manner requests the proof of $\sqrt5$ and $\sqrt6$ being irrational. However, in this question I'm interested in the proof for the case of $\sqrt3$.