# Prove that $\sqrt{3}$ is irrational *using the given hint* . Hint: Any integer has the form $3k,3k+1,3k+2.$

Please explain why the given hint is true and give me another hint to solve the problem.Previously in the text $\sqrt{2}$ was proven irrational by considering 3 cases of $\frac mn$ where 1). both m and n are odd, 2,3). either m or n is odd and the other is even. I tried the method using the given hint but could not arrive at a contradiction. Help.

• Hint: use the hint to show $\, 3\mid n^2\,\Rightarrow\,3\mid n,\,$ then proceed as in the proof for $\,\sqrt{2}\$ – Bill Dubuque Apr 3 '15 at 21:19
• The truth of the hint is usually considered a trivial observation due to the fact that $3\mathbb{Z}, 3\mathbb{Z}+1, 3\mathbb{Z}+2$ is a partition of $\mathbb{Z}$, but it seems like a circular argument to use that to prove the hint. Similarly using that dividing by 3 leaves a remainder of 0,1, or 2 also seems to be circular logic. If you wish to prove it, you could do so by a proof by induction. $k=0$ works for $n\in\{0,1,2\}$. Since $n=3k+r$ for some $k\in\mathbb{Z}$ and $r\in\{0,1,2\}$, it follows that $n+3 = 3k+r+3 = 3(k+1)+r = 3k'+r$. Similarly $n-3 = 3k+r-3 = 3(k-1)+r = 3k'+r$ – JMoravitz Apr 3 '15 at 21:33

write: $$3n^2=m^2$$
where $m,n$ are coprime then they are not both of the form $3k$ and because $m$ is of the form $3k$ you will have $n$ is either of the form $3k'+1$ or $3k'+2$. replace in the two cases and get a contradiction.
First: When you divide an integer by three, you get a remainder, right? And that remainder is 0, 1, or 2. (If it were more than 2, you'd add 1 to the quotient.) So if $n/3$ is $k$ with a remainder of $p$, that means that $$(n-p)/3 = k \\ n-p = 3k\\ n = 3k+p$$ where $p$ is 0, 1, or 2.
Second, for each of the three cases, compute what $n^2$ is. For $n = 3k + 1$, you get $$n^2 = (3k+1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$$ i.e., a multiple of three, with a remainder of 1.
When you do this for the other two cases, you'll find out that the only one for which 3 divides $n^2$ is when $n$ is a multiple of 3 already, as Bill Dubuque said. Then work onwards from there.