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Let $A$ be the set of all rationals $p$ such that $p^2 < 2$ and Let $B$ be the set of all rationals $p$ such that $p^2 > 2$

  1. If $p$ is in $A$, then there is some $q$ in $A$ such that $p<q$

  2. If $p$ is in $B$, then there is some $q$ in $B$ such that $p>q$

This is the final part of Baby Rudin's proof of irrationality of $\sqrt2$

I don't understand how these two statements imply the irrationality of $\sqrt2$

If someone can tell me the reasoning behind it, It'd really help.

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    $\begingroup$ I assume that it should be $B$ in $(2)$? As stated, I don't see how you deduce irrationality from that. After all, those claims are both true if you replace "$2$" by "$1$". I think you need weak inequalities instead ($≤$ instead of $<$ and so on in the definition of $A,B$). $\endgroup$
    – lulu
    Commented Oct 18, 2018 at 18:29
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    $\begingroup$ See Choice of $q$ in Baby Rudin's Example 1.1 and Proving $\sqrt{2}$ is irrational: why $ q = p - \frac{p^2 -2}{p+2}$ (both questions are about this exact proof) $\endgroup$
    – Winther
    Commented Oct 18, 2018 at 18:30
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    $\begingroup$ Those questions are about why we chose an arbitrary q. My question is that once we've done that we get the mentioned two statements are true, but why does them being true imply irrationality of root 2 $\endgroup$
    – LoneStar
    Commented Oct 18, 2018 at 18:40
  • $\begingroup$ As I said, those two statements alone (once you correct the $A,B$ confusion) do not appear to be sufficient. Suppose $A_1$ is the set of rationals $p$ with $p^2<1$ and $B_1$ is the set with $p^2>1$. Then it is perfectly true that given $p\in A_1$ we can find $q>p$ with $q\in A_1$, take $q=\frac {1+p}2$ for instance. Similarly for the other part. And, of course, $\sqrt 1$ is rational. So you need more. $\endgroup$
    – lulu
    Commented Oct 18, 2018 at 18:45
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    $\begingroup$ I pulled out Baby Rubin. You appear to be missing the point...he is not claiming that those two statements (for the third time: please correct the $A,B$ confusion) prove irrationality. Indeed, he proves irrationality the standard way (as in Euclid). The bit about the sets $A,B$ is to show that the rationals contain gaps. To be fair, I don't think the author makes this point very clearly. $\endgroup$
    – lulu
    Commented Oct 18, 2018 at 18:58

1 Answer 1

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Rudin is not attempting to prove that $\sqrt{2}$ is irrational by the two statements you have summarized in your post. He has already shown in the first two paragraphs of $\S$1.1 Example that there is no rational $p$ such that $p^2 = 2$. After this, he says, "We now examine this situation a little more closely." This examination is what you are concerned with in your question.

The bottom line of this examination is the following:

  1. It is assumed that we already know that the set of rational numbers is linearly ordered, that is, for any two rationals $p$ and $q$, we can say that either $p \leq q$ or $q \leq p$ (or both, in which case, $p = q$).
  2. Rudin has just shown that there is no rational $p$ such that $p^2 = 2$.
  3. Following this, Rudin has shown that for any rational $p > 0$ such that $p^2 < 2$, there is another rational $q > p$ such that $q^2 < 2$, too; similarly, for any rational $p$ such that $p^2 > 2$, there is another rational $q < p$ such that $q^2 > 2$, too.

Thus, the set of rational numbers appears to have "gaps" in the sense identified in 3. above. For any rational that is slightly less than $\sqrt{2}$, we can find a larger rational that is also slightly less than $\sqrt{2}$. For any rational that is slightly greater than $\sqrt{2}$, we can find a smaller rational that is also slightly greater than $\sqrt{2}$. Yet, at no point can we find a rational that is exactly equal to $\sqrt{2}$.

Hence, though the linear ordering of the set of rational numbers as in 1. suggests the mental image of a line, we must amend the naïve picture of a continuous line to note the presence of these "gaps" in this rational number line. Thus does real analysis begin.

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