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My textbook in real analysis proves that the rationals are dense in the reals by using first the archemidean principle and then constructing a fairly long and contrived proof. What I am wondering is, is that could you not first show that the sum of two rationals divided by a rational is a rational number (the sum can be done by contradiction, and ditto for the product of two rationals), and then just point out that the arithmetic average of any two nonequal rational numbers is gaurenteed to be in between them? Proving that step should be straightforward, assume wlog xx y+x>2x (x+y)/2>x

And then this is similar if x and y are in general real, provided that one uses the axiom that addition and division is a closed operation. Can someone point out why this is no rigorous if it isn't? Pardon the lack of latex... Still learning it! Cheers :)

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  • $\begingroup$ The midpoint of two reals may not be rational. $\endgroup$ Commented Sep 30, 2017 at 11:29
  • $\begingroup$ Very true. Thanks $\endgroup$ Commented Sep 30, 2017 at 11:43

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You can do that if you know about decimal representations (yes, you know, but it looks like you are there building reals from the axioms).

If you were to define rationals as periodic decimals and irrationals as non-peridic, then you would have, for irrational $\alpha=a_0.a_1a_2...a_n...$, the rational $a_0.a_1a_2,...a_n999....$ is $> \alpha$ but $< \alpha +1=(a_0+1).a_1a_2...a_n...$ (you should take into account here that two different decimal representations can represent same number and avoid that).

With decimals, you can easily prove that between two rationals there is a rational and irrational and that between two irrationals there is a rational and irrational.

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It is easy to prove that between two rationals there is a rational (your argument is correct) and that between two reals there is a real. But how do you deduce from that that between two reals there is always a rational number, which is the statement that you want to prove here?

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  • $\begingroup$ @SeanThrasher If my answer was helpful, perhaps that you could mark it as the accepted answer. $\endgroup$ Commented Sep 30, 2017 at 11:45
  • $\begingroup$ Did that. Fairly new to this site sorry $\endgroup$ Commented Sep 30, 2017 at 11:46
  • $\begingroup$ @SeanThrasher I know how that is. After my first technical question here got an answer, it took me weaks until I marked it as the accepted one. $\endgroup$ Commented Sep 30, 2017 at 11:48
  • $\begingroup$ I am glad you told me. Got to give credit where credit is due! $\endgroup$ Commented Sep 30, 2017 at 11:49

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