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This was a test problem that I solved and got 4/10 points on, but I don't see where I went wrong.

Prove that $\sqrt {28}$ is irrational.

Suppose to the contrary that $\sqrt {28}$ is rational. Then $\sqrt {28} =\frac{a}{b}$ where $a$ and $b$ are integers and $\gcd(a,b) = 1$.

$28=\frac{a^2}{b^2}$

$28b^2 = a^2$

$2(14b^2) = a^2$

So $a^2$ is even. Clearly then $a$ is also even.

Let $a =2k$ where k is some integer.

Then $28b^2 = 4k^2$

And $14b^2 = 2(k^2)$

$b^2$ is even, so $b$ is also even.

Contradiction: $a$ and $b$ share a factor of 2, so $\gcd(a,b)\neq 1$

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    $\begingroup$ Saying that $14b^2=2k^2$ does not imply that $b$ is even, since $14$ is even. Indeed, $\sqrt {28}=2\sqrt 7$ so it's not a great idea to work with the prime divisor $2$. Far better to work with $7$. $\endgroup$
    – lulu
    Commented Oct 26, 2020 at 21:41
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    $\begingroup$ $14b^2$ even does not imply $b^2$ is even $\endgroup$
    – QC_QAOA
    Commented Oct 26, 2020 at 21:41
  • $\begingroup$ You just plugged $28$ in the standard proof for $2$, I guess. This does not work. $\endgroup$
    – user65203
    Commented Oct 26, 2020 at 21:42
  • $\begingroup$ Do the same thing, but with the $7$. $\endgroup$
    – Darsen
    Commented Oct 26, 2020 at 21:43
  • $\begingroup$ Your grader was kind enough to give you 4/10 for a completely wrong proof... As Brian said, you merely mimicked the proof for irrationality of $\sqrt{2}$ without actually understanding what you were doing. Next time, always ask yourself "WHY?" for every single claim you make. Why is $a^2$ even? Why is $a$ even? Why is $b^2$ even? Aha, that was a completely unjustifiable claim... $\endgroup$
    – user21820
    Commented Oct 27, 2020 at 6:16

2 Answers 2

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The fact that $14b^2=2k^2$ does not imply that $b^2$ is even; it appears that you were paraphrasing the proof that $\sqrt2$ is irrational without thinking very hard about what you were saying.

The fact that $14b^2=2k^2$ does, however, tell you that $7b^2=k^2$, so $k^2$ is a multiple of $7$, and from that you can infer that $k$ is a multiple of $7$, say $k=7\ell$. Then $7b^2=(7\ell)^2=49\ell^2$, so $b^2=7\ell^2$, $b^2$ is a multiple of $7$, and finally $b$ is a multiple of $7$. Thus, $a=2k=14\ell$ and $b$ are both multiples of $7$, contradicting the assumption that the fraction was in lowest terms.

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You assumed that the 2 factor came from the b squared but it could have also come from the 14 since 14 = 2 * 7

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