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If there are gaps in rational numbers then lets assume we have a gap between a and b, both being rational. Then we have $\frac{a+b}{2}$ which is inside the gap which essentially makes it a non-gap. What am I getting wrong?

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    $\begingroup$ gaps are because of the existence of irrational numbers! $\endgroup$ – OmG Jan 7 '19 at 11:10
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    $\begingroup$ How do you define a "gap"? $\endgroup$ – 5xum Jan 7 '19 at 11:20
  • $\begingroup$ The rationals and its complement are both dense in the reals. That's all. The word gap is ambiguous here, e.g. If you remove one point from a line, is that a "gap"? $\endgroup$ – Ned Jan 7 '19 at 11:21
  • $\begingroup$ You have shown that there is at least one real number between $a$ and $b$ that is rational. But to "fill" the gap you would have to show that every real number between $a$ and $b$ is rational. $\endgroup$ – gandalf61 Jan 7 '19 at 11:21
  • $\begingroup$ This is because there is no well-defined notion of the “next” rational number after each rational number. (or at least the usual ordering of the rationals doesn’t allow a well-defined next rational). $\endgroup$ – Adam Higgins Jan 7 '19 at 11:23
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A gap in $\Bbb Q$ means there exist non-empty sets $A, B$ with $\Bbb Q=A\cup B,$ such that (i) every $a\in A$ is less than every $b\in B,$ and (ii) $A$ has no largest member and $B$ has no smallest member. It does NOT mean that there are rationals $x, y$ with $x<y$ such that no rational is between $x$ and $y$. No such $x,y$ exist but if they did, the pair $(x,y)$ would be called a jump.

Example: No $x\in \Bbb Q$ satisfies $x^2=2.$ Let $B=\{b\in \Bbb Q: 0<b\land b^2>2\}$ and let $A=\Bbb Q \setminus B.$ If $b\in B$ then $b>\frac {1}{2}(b+\frac {2}{b})\in B,$ so $B$ has no least member. This implies that $\{a\in A:a>0\}=\{\frac {2}{b}: b\in B\}$ has no largest member. So $A$ has no largest member.

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