# Proof attempt that $\sqrt{28}$ is irrational

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 $$\gcd(a,b)\neq 1$$

• 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 $7$. – lulu Oct 26 '20 at 21:41
• $14b^2$ even does not imply $b^2$ is even – QC_QAOA Oct 26 '20 at 21:41
• You just plugged $28$ in the standard proof for $2$, I guess. This does not work. – Yves Daoust Oct 26 '20 at 21:42
• Do the same thing, but with the $7$. – Darsen Oct 26 '20 at 21:43
• 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... – user21820 Oct 27 '20 at 6:16

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.