Can I prove a numerical inequality by manipulating it and ending at a true statement?

Question

Say I want to prove that $\sqrt2$+$\sqrt3$ < $\sqrt10$. The steps I would do to prove this is square both sides to get:
5+2$\sqrt6$<10
(subtract 5 from both sides): 2$\sqrt6$<5
(square both sides): 24<25
which is true.

but does this constitute as a valid proof? or am I missing something?

Answer 1

Technically, what you have is that if $\sqrt{2}+\sqrt{3}$ is less than $\sqrt{10}$, then $24$ is less than $25$. That is not a proof of what you want. The fact that the last statement is true also does not show that the first statement is true, in and of itself: “If I fall into the pool, then I will get wet.” If I am wet, that does not prove that I fell into the pool (maybe I got caught in a rainstorm?)

In this particular instance, you could try to see if you can “reverse” the steps: starting from $24\lt 25$, can you go back through your manipulations to obtain the result you want? That is, every step “reversible”? If so, then that will work. So, from $24\lt 25$ you can take square roots (since they are both positive) to get $2\sqrt{6}\lt 5$. Then add $5$ to both sides to get $5+2\sqrt{6} \lt 10$. Then note that the left hand side is the same as $(\sqrt{2}+\sqrt{3})^2$, and finally, take square roots on both sides (noting that both $\sqrt{2}+\sqrt{3}$ and $10$ are positive) to obtain the desired result. That will be a valid proof.

You may wonder why you need to check all of that. Well, imagine you want to prove that $1$ is less than $-2$. So, you start with $1\lt -2$, and square both sides to get $1\lt 4$, which is true. Voila! I guess that means $1\lt -2$, right? Well, no, because the step to go from $1\lt -2$ to $1\lt 4$ is not reversible: you can’t go backwards, so you can’t get a valid proof from that manipulation.

Answer 2

In order for this to be a proof, you need to verify that the reverse implications are true.

For $a,b>0$ you have

$$a^2 <b^2 \implies a<b \text{ or } -b <-a$$

but as you’re considering positive reals, you can’t be on the second case. Hence the reverse implication is true.

Consequently, your proof is correct. But you should mention at each step that the initial numbers are positive.