Is it acceptable to prove the following inequality by "approximation"?

Question

I am currently self-studying Rudin's principles of analysis and I like to try and write my own proofs before looking at the books'.

The task is to verify, given that $p^2<2$, that $q^2 < 2$ where $q = \frac{2p+2}{p+2}$. I attempted to do it this way using the given information that $p^2<2$ :

$$q^2 = \left (\frac{2p+2}{p+2}\right)^2 = \frac{4p^2+8p+4}{p^2+4p+4}< \frac{4(2)+8(2)+4}{2+4(2)+4}=\frac{28}{14}=2 $$

My concern is that most proofs of this sort that I've come across so far rarely use numerical substitutions. I've seen some that create upper bounds in terms of, say, a formula of the nth term of a sequence rather than a number in the manner I have done. I'm still new to proofs so I'd like to know if this is a sufficient or strong enough argument.

EDIT

I think it is necessary to add a few more details. The main task is to prove that the set of rational numbers has holes. More specifically, the theorem is that there exist subsets of the rational numbers that have no smallest number, and there exist others that have no largest number. To prove this, the book proves by example that $\sqrt 2 \notin Q $. This is done by letting $A = \{ p \in \mathbb Q:p^2<2 \}$ and $B = \{ p \in \mathbb Q:p^2>2 \}$. Also note that $p>0$.

To show that A has no single largest element, we have to find a $q \in A: q>p$ and $q^2<2$. Thus, the conjecture is that if $q=\frac{2p+2}{p+2}$ then it satisfies the above criteria. It is easy to show that $q>p$ since $q=p- \frac{p^2-2}{p+2}=\frac{2p+2}{p+2}$. What's left is to show that $q^2 < 2$ which is what the question is about.

Answer 1

Let's try another proof. Let $r = \frac{p+3}{p+2}.$ Using your method, $$ r^2 = \left (\frac{p+3}{p+2}\right)^2 = \frac{p^2+6p+9}{p^2+4p+4}\stackrel?< \frac{2+6(2)+9}{2+4(2)+4}=\frac{23}{14}<2, $$ from which (according to your method) we should conclude that $p^2 < 2$ implies $r^2 < 2.$

But if $p = -1$ then $p^2 < 2$ and $ r = \frac{-1+3}{-1+2} = 2, $ and therefore $r^2 = 4.$ So we have a counterexample to the claim that $p^2 < 2$ implies $r^2 < 2.$

What went wrong? Your error is that you assume it is safe to substitute an equal or greater value in the denominator of a ratio. But a ratio of two positive numbers actually gets larger as the denominator gets smaller, so you cannot legitimately establish an upper bound on the ratio by assuming the denominator is larger than it actually might be.

Answer 2

The problem with using numerical substitutions as you do is that in this way you are not showing that for every possible $p^2<2$ the inequality is satisfied, but just only for the number that you are replacing. You have to show that $\left(\frac{2p+2}{p+2}\right)^2$ is bounded above for the values of $p$ under consideration, and that $2$ is an upper bound of $q^2$. Note that $q^2$ is non-monotonic for $-\sqrt 2 < p < \sqrt 2$. It happens to have a minimum at $p=-1$, but... what would happen if you had another function with a maximum instead of a minimum? You had to show that the maximum is lower than 2 in that case and replacing a particular numerical value could lead easily to wrong conclusions.

In brief, you have to show that for every $p$ such that $-\sqrt 2 < p < \sqrt 2$ the inequality for $q^2$ holds and you have to be careful bounding $q^2$ because it is non-monotonic in the domain considered.

Imagine the following problem. Show whether for $p^2<1$ the inequality $1-p^2<1/2$ is correct. With your method, replacing $p=1$, you would have $1-1=0<1/2$ and you would conclude that the inequality is correct. It is not, however. (Show it!)