2

What is wrong with the given proof? $$ \sqrt 2 + \sqrt 6 < \sqrt 15 \\ (\sqrt 2 + \sqrt 6)^2 < 15 \\ 2 + 6 + 2\sqrt2\sqrt6 < 15 \\ 2\sqrt2\sqrt6 < 7\\ 2^2 \cdot 2 \cdot 6 < 49 \\ 48 < 49 $$

Viper
  • 49
  • There is nothing wrong, as $\sqrt 2 + \sqrt 6 < \sqrt 15$ is true. – Yash Jain Feb 01 '18 at 00:41
  • 6
    Well, for starters, there are no verbs connecting one line to the next. Lacking such direction, the usual supposition is that one line implies the next. In this case, you have assumed the thing that you are trying to show, then shown that it implies that $48 < 49$. However, if you mean that line $n$ is true if and only if line $n+1$, then your proof is fine. – Xander Henderson Feb 01 '18 at 00:42
  • 3
    It's correct if you write it in reverse order. – CY Aries Feb 01 '18 at 00:52
  • 4
    No, reversing the lines does not make that correct. Words are missing — proofs are explanations of why something is true, and explanations use words, verbs, sentences, connectors and so on. A proof is a piece of text. – Mariano Suárez-Álvarez Feb 01 '18 at 01:04

1 Answers1

5

As it stands, the proof is flawed, but not for any error of reasoning (really), but because it is argued in a slightly nonsensical manner. Typically, when you write a sequence statements, you want to connect them to each other. For example, "$A$ implies $B$" or "$A$ if and only if $B$". If no such connecting verbs are stated, the usual assumption is that early statements imply later statements. Thus, eliding all of the intermediate steps, one might reasonably read your proof as:

If $\sqrt{2} + \sqrt{6} < \sqrt{15}$, then $48 < 49$.

This is, in fact, the converse of what you want to show. There are two ways to make this more clear:

  1. You could use language or notation to show that one line is true if and only if the previous line is true. If you attacked it this way, you might write \begin{align} \sqrt{2} + \sqrt{6} < \sqrt{15} &\iff \left( \sqrt{2} + \sqrt{6} \right)^2 < \sqrt{15}^2 && \text{(since both sides are positive)} \\ &\iff 2 + 2\sqrt{2}\sqrt{6} + 6 < 15 && \text{(distribution)} \\ &\iff 2\sqrt{2}\sqrt{6} < 7 && \text{(law of additive cancelation)} \\ &\iff 2^2\cdot 2 \cdot 6 < 7^2 && \text{(since both sides are positive)} \\ &\iff 48 < 49. && \text{(simplify)} \end{align} Since $48<49$, the proposition is proved.
  2. Alternatively, if you want to look clever, you write the proof as a series of implications: we know that $48 < 49$. Therefore \begin{align} 48 < 49 &\implies \sqrt{48} < \sqrt{49} \\ &\implies 2 \sqrt{2}\sqrt{6} < 7 \\ &\implies 2 + 2\sqrt{2}\sqrt{6} + 6 < 15 \\ &\implies \left( \sqrt{2} + \sqrt{6} \right)^2 < 15 \\ &\implies \sqrt{2} + \sqrt{6} < \sqrt{15}, \end{align} where $\sqrt{x}$ denotes the principle (or positive) square root.

Another rather pathological approach might be to prove by contradiction. Suppose that $\sqrt{2} + \sqrt{6} \ge \sqrt{15}$. Using the basic computations that were originally given, but replacing $<$ with $\ge$, we obtain $$ 48 \ge 49, $$ which is obvious nonsense. Therefore it cannot be that $\sqrt{2} + \sqrt{6} \ge \sqrt{15}$, which proves the proposition.

Xander Henderson
  • 29,772
  • 1
    Except steps $1$ and $4$ in method 1. are not equivalences unless you explicltly state that you are using the positiveness of the two sides. – dxiv Feb 01 '18 at 00:58
  • 2
    @dxiv Hence the caution that each step should really be checked carefully. Though I'll admit that I was rather sloppy with that argument, as I think that it is kind of the Wrong Way(tm) to do things. EDIT: And fixed. – Xander Henderson Feb 01 '18 at 01:01
  • Could it be written as?

    $$ \begin{align} \sqrt{2} + \sqrt{6} < \sqrt{15} &\implies \left( \sqrt{2} + \sqrt{6} \right)^2 < \sqrt{15}^2 \ &\implies 2 + 2\sqrt{2}\sqrt{6} + 6 < 15 \ &\implies 2\sqrt{2}\sqrt{6} < 7 \ &\implies 2^2\cdot 2 \cdot 6 < 7^2 \ &\implies 48 < 49. && \end{align} $$

     – Viper Feb 01 '18 at 12:54
  • @Viper No. That is the way that one would have interpreted the original writeup, and that proves the converse of the desired statement, i.e. if $\sqrt{2} + \sqrt{6} < \sqrt{15}$, then $48 < 49$. All you have done is shown that if the thing you want to prove is true, then you also know that $48 < 49$. The fact that you know this for other reasons is utterly irrelevant. – Xander Henderson Feb 01 '18 at 14:17