0
$\begingroup$

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.

$\endgroup$
  • 6
    $\begingroup$ Hint: use the hint to show $\, 3\mid n^2\,\Rightarrow\,3\mid n,\,$ then proceed as in the proof for $\,\sqrt{2}\ $ $\endgroup$ – Bill Dubuque Apr 3 '15 at 21:19
  • 1
    $\begingroup$ 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 $n-3 = 3k+r-3 = 3(k-1)+r = 3k'+r$ $\endgroup$ – JMoravitz Apr 3 '15 at 21:33
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.