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My teacher tells me that in the vicinity of any rational number, an irrational exists. To elucidate, I presume, he further went on to say, if a function, if defined to give 1 for every rational number and 0 for every irrational, then the function would be discontinuous at each and every point.

So that means that there are two irrational numbers surrounding a rational, right? And vice versa? So if I take the A.M. of those two irrational numbers, I get a rational number? Or of those two rational numbers, an irrational? Does that make sense?

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    $\begingroup$ But the set of nonzero rational numbers is closed under addition and division, meaning that you can never average two rationals to get something irrational. Are you asking what's wrong with your reasoning, or hadn't you considered that? $\endgroup$ – pjs36 Jan 30 '16 at 5:06
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So that means that there are two irrational numbers surrounding a rational, right?

No, it is not the fact that there are only two irrationals surrounding a rational; there are actually infinitely many (for any distance of interest, say $\epsilon > 0$). This is referred to as the set being "dense".

So if I take the A.M. of those two irrational numbers, I get a rational number?

Hopefully you now see that this question doesn't make sense, because you can't uniquely identify "those two irrational numbers".

You could, if carefully selected, find two irrational numbers around any rational of equal distance, and as a consequence have their arithmetic mean be that initial rational number. However, picking irrationals at random close to a given rational would not have that as a result.

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  • $\begingroup$ One in a million chance, but the possibility exists, you're saying? $\endgroup$ – Nikhil Itty Jan 30 '16 at 5:18
  • $\begingroup$ @Anaklusmosismyfavoriteweapon Yes, the arithmetic mean of irrationals can be rational - e.g. $1+\sqrt{2}$ and $1-\sqrt{2}$. It can also be rational - e.g. $\sqrt{2}$ and $\sqrt{3}$. However, the converse can't happen - the mean of two rationals is always rational (think about why . . .). Finally, in a precise sense, it turns out that "one in a million" is wildly optimistic - two randomly-chosen irrationals will "almost never" (probability zero) have a rational mean! Making sense of this, though, requires some work . . . $\endgroup$ – Noah Schweber Jan 30 '16 at 5:27
  • $\begingroup$ @Anaklusmosismyfavoriteweapon: No, since it's a set of measure zero, it would be probability zero to pick such irrationals at random from any continuous distribution (roughly speaking: probability 1 divided by infinity [because of the infinite number of nearby irrationals], if it helps to think of that). $\endgroup$ – Daniel R. Collins Jan 30 '16 at 5:27
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    $\begingroup$ That makes more sense @Daniel... Thanks! $\endgroup$ – Nikhil Itty Jan 30 '16 at 5:32
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You have four questions. The answer to the first is yes. The answer to the second is yes. However, although given irrationals $x<y$, there is a rational $r$ with $x<r<y$, it need not be the arithmetic mean ${1\over2}(x+y)$. The arithmetic mean can easily be irrational: suppose, for example, $x=\sqrt{3}-\sqrt{2}$ and $y=\sqrt{3}+\sqrt{2}$. Similarly, given rationals $r<s$, there is an irrational $x$ with $r<x<s$, but it cannot be the arithmetic mean ${1\over2}(r+s)$ -- that will always be rational too.

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  • $\begingroup$ Your x is around 0.6, and your y around -0.6. But I meant to say, 1 is rational, right? That makes 1+h, and 1-h irrationals, right? (h-->0) Surrounding one? And the mean of those two irrationals apparently gives me one. $\endgroup$ – Nikhil Itty Jan 30 '16 at 5:14
  • $\begingroup$ My $x$ is around $0.3$, and my $y$ around $3.1$; both are positive. If $h\to 0$ then $h$ passes through both rational and irrational values -- infinitely many of each. And yes, there are many examples of two irrationals whose mean is rational, for example $1-\sqrt{2}/8$ and $1+\sqrt{2}/8$. But there are also many examples (like the one I gave) of two irrationals whose mean is irrational. $\endgroup$ – ForgotALot Jan 30 '16 at 5:30

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