Why is this proof false? $\sqrt 2 + \sqrt 6 < \sqrt 15$

Question

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

Answer 1

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.