# How is it that there are 'gaps' in rational numbers and yet between any two rational numbers, there exists another rational number?

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?

• gaps are because of the existence of irrational numbers! – OmG Jan 7 '19 at 11:10
• How do you define a "gap"? – 5xum Jan 7 '19 at 11:20
• 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"? – Ned Jan 7 '19 at 11:21
• 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. – gandalf61 Jan 7 '19 at 11:21
• 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). – Adam Higgins Jan 7 '19 at 11:23

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 $$x such that no rational is between $$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: 02\}$$ and let $$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.