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.


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.

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    $\begingroup$ Your argument is not valid. You are assuming that a certain function of $p$ is an increasing function without actually proving it. $\endgroup$ – Kavi Rama Murthy Nov 28 '18 at 11:49
  • $\begingroup$ When the demominator is increased, the fraction does not increase. $\endgroup$ – William Elliot Nov 28 '18 at 12:00
  • $\begingroup$ @KaviRamaMurthy Please explain your comment; I don't fully understand it. I did not make any mention of a function in $p$ $\endgroup$ – E.Nole Nov 28 '18 at 12:11
  • $\begingroup$ @WilliamElliot What are you driving at? Well, both the denominator and numerator have a term in $p$ and $p^2$ so when one increases then so does the other. $\endgroup$ – E.Nole Nov 28 '18 at 12:18
  • $\begingroup$ Both the denominator and numerator increase in $p$, that's true. But you have to show that the fraction increases as well. In general, increasing the numerator increases the fraction, but increasing the denominator decreases the fraction. So you have two opposing effects at play here. $\endgroup$ – MaoWao Nov 28 '18 at 12:37

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.

  • $\begingroup$ Please see my edit to question $\endgroup$ – E.Nole Nov 28 '18 at 13:29
  • $\begingroup$ The edit explains why you are interested in the particular claim you are supposed to prove. Your "proof" of the claim is still not a proof. $\endgroup$ – David K Nov 28 '18 at 13:54
  • $\begingroup$ Correct me if I' wrong but what I'm getting from this is that if the fraction was such that the numerator is an upper bound of ${4p^2+8p+4}$ and the denominator was a lower bound of ${p^2+4p+4}$, and this happened to equal 2, then only then could the conclusion that $q^2<2$ hold? $\endgroup$ – E.Nole Nov 29 '18 at 9:52
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    $\begingroup$ If you could set up the fraction that way, yes, that would work. $\endgroup$ – David K Nov 29 '18 at 12:46

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!)

  • $\begingroup$ Could you please explain further? Isn't $2$ an upper bound for all the $p^2$? $\endgroup$ – E.Nole Nov 28 '18 at 12:20
  • $\begingroup$ Please see my edit to question $\endgroup$ – E.Nole Nov 28 '18 at 13:29
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    $\begingroup$ Regarding your edit. Let us take the positive rational numbers in any open interval $S$, $(a/b,c/d)$ on the real line (positive semiaxis). Assume that you have found a least element $m/n$ of this set. Then, the fraction $(a+m)/(b+n)$ is lower than $m/n$ and larger than $a/b$ and is on $S$ as well. Therefore, $m/n$ is not the least element of $S$. One can, similarly prove that there is no largest member in $S$. $\endgroup$ – Frobenius Nov 28 '18 at 13:41
  • $\begingroup$ I understand your last comment. However, I am more interested in how I could modify my approach to make it right because I really want to understand where I went wrong. It's still unclear to me when (and how), then can one use the given data that $p>0$ and $p^2<2$ in proving the claim? I tried to solve your last inequality without substitution. I got $p^2>1- \frac{1}{2} = \frac{1}{2}$. Given $p^2<1$, it's possible that $1>p^2>\frac{1}{2}$. So with or without substitution, I'd conclude it's possible. You say the inequality does not hold which probably means I'm missing something. $\endgroup$ – E.Nole Nov 29 '18 at 9:38
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    $\begingroup$ In my inequality, obviously, if $p^2<1$ holds, then $1-p^2<1/2$ does not hold for every value of $p$. For example, for $p=0$ ($p^2<1$), $1-p^2=1<1/2$, which is absurd. Do you see? $1-p^2<1/2$ does not hold generally for all $p^2<1$. You replace specific numerical values on nonlinear functions and draw wrong conclusions from it. You should look globally and carefully. While $p^2<1$ (the domain) is just satisfied only by the interval $-1<p<1$, the function $1-p^2<1/2$ is a parabola with a maximum equal to 1 at $p=0$ and you cannot say that $1-p^2<1/2$ holds for every point in the domain. $\endgroup$ – Frobenius Nov 29 '18 at 10:14

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