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The rational numbers form a countably infinite set, just as the integers do. For the integers, there is no integer between 4 and 5 (I.e., there is such a thing as two consecutive integers) So, I am thinking that there must be such a thing as two consecutive rational numbers, such that there is no other rational number that is between them. The difference between two consecutive integers is 1. If there is such a thing as two consecutive rational numbers, is the difference between them 0 (even though there are irrationals between them)?

Thanks!

David
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    If a < b, a < (a+b)/2 < b. – William Elliot Apr 08 '18 at 08:30
  • The easiest way to see that this is not the case is taking $$\frac{a+b}{2}$$ if $a<b$ are given rational numbers. Then, we have $$a=\frac{2a}{2}<\frac{a+b}{2}<\frac{2b}{2}=b$$ and $\frac{a+b}{2}$ is rational again. – Peter Apr 08 '18 at 08:34
  • I did not notice that this was already mentioned before by William – Peter Apr 08 '18 at 08:35

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