Prove by contradiction (not using a calculator) that $\sqrt6 + \sqrt2 < \sqrt{15}$.

How do you approach such a problem? I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. I tried to square it as I read online but I'm stuck with having $\sqrt3$ and not knowing how to get rid of it to have the answer as clear as the sun. squaring them yielded: $(\sqrt2 + \sqrt6)^2 = 8 + 4\sqrt3 < 15$ which indeed holds but the problem is that I am not allowed to use calculator to check the little difference.

My final answer was $4\sqrt3 < 7$.

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    $\begingroup$ Why not try squaring again? $\endgroup$ – Mark Bennet Nov 14 '17 at 15:15
  • $\begingroup$ It seems that this problem appeared as an assignment here, where you can also find a solution. $\endgroup$ – Martin Sleziak Nov 14 '17 at 15:18
  • $\begingroup$ As on MO, I mention that your first step when proving anything by contradiction is to contradict it. Of course, moving on from there might be might or less difficult, but at least it gets you started! $\endgroup$ – LSpice Nov 14 '17 at 15:37
  • $\begingroup$ Incidentally, you can find a solution here. $\endgroup$ – Martin Sleziak Nov 14 '17 at 15:47

Let$$\sqrt6+\sqrt2\geq\sqrt{15}.$$ Thus, $$8+2\sqrt{12}\geq15$$ or $$4\sqrt3\geq7$$ or $$48\geq49,$$ which is contradiction.

Id est, our assuming was wrong, which says $$\sqrt6+\sqrt2<\sqrt{15}.$$

  • $\begingroup$ how did you get the? 48? $\endgroup$ – user481197 Nov 14 '17 at 15:19
  • $\begingroup$ @Abdul Malek Altawekji Because $\left(4\sqrt3\right)^2=48$. $\endgroup$ – Michael Rozenberg Nov 14 '17 at 15:20
  • $\begingroup$ oh of course thanks $\endgroup$ – user481197 Nov 14 '17 at 15:21
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    $\begingroup$ For what it's worth, since my students are very fond of using "let" where it doesn't belong, I should mention that you can't let $\sqrt6 + \sqrt2$ be greater than or equal to $\sqrt{15}$; it is or it isn't. What you can do instead is suppose that the former is greater than or equal to the latter, and derivative a contradiction from the supposition. (By way of contrast, although there's no reason to do so, you could begin your proof by saying "Let $x$ be $\sqrt{15}$"; since there's no $x$ yet, you can let it be whatever you want.) $\endgroup$ – LSpice Nov 14 '17 at 15:34

The claim is that $\sqrt{2} + \sqrt{6} < \sqrt{15}$, so its negation is simply the statement

$\sqrt{2} + \sqrt{6} < \sqrt{15}$ is false

or alternatively,

$\sqrt{2} + \sqrt{6} \ge \sqrt{15}$

Now in order to prove the claim, suppose that $\sqrt{2} + \sqrt{6} \ge \sqrt{15}$ and square both sides, getting

$$2 + 6 + 2\sqrt{12} \ge 15$$

Simplifying, this leads to

$$2 \sqrt{12} \ge 7$$

Do you see how to derive a contradiction now?


This is an old question but recently mentioned. I wanted to comment that there is no gain from using contradiction:

A proof by contradiction is like

claim $2*3=6$. Proof: assume otherwise. But $2*3=3+3=6. $CONTRADICTION TO WHAT WE ASSUMED!!!!

Here is the first proof reconfigured. I tried to stick to the form there.

Claim: $$\sqrt6+\sqrt2\lt \sqrt{15}.$$ Proof: Equivalently (squaring both sides) the claim is $$8+2\sqrt{12}\lt 15$$ or $$4\sqrt3\lt 7$$ or (squaring again) $$48\lt 49,$$ which is true.

Id est, our claim is true , which says $$\sqrt6+\sqrt2<\sqrt{15}.$$


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