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Attempt:

We can use the following archimidean property or R: for all $x $ real we can find positive integer n so that $n > x$

since $b-a > 0$ another real number and so can find integers $N$ so that $N(b-a) > 1$ or in other words that $b - a > \dfrac{1}{N} $ so

$$ b > a + \dfrac{1}{N} > a $$

but doesnt seem to help. Another strategy would be to realize first that we can find $n$ so that $b < n$. Now, $an$ is a real number thus can find $m$ so that $m > na $ and thus $a < \dfrac{m}{n} < \dfrac{m}{b} $

Now, we know that $m/b$ is real number. Therefore, we have found a rational $m/n$ between two reals!

Is this a correct solution?

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James
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  • Why would $\frac{m}{b}<b$ ? – Tuvasbien Apr 17 '20 at 21:59
  • If you constructed the reals using Dedekind cuts you can find these rationals by using naive set theory. Otherwise write $b = a + p$ for some positive $p$ and consider the average of $a$ and $b$. See if you can use the Archimedean property from there. – CyclotomicField Apr 17 '20 at 22:15
  • "Therefore, we have found a rational m/n between two reals! " But not the two reals you were supposed to find it between.You had to find them between $a$ and $b$. Finding them between a different pair of reals doesn't count. Otherwise we could have juse said $1.5$ is between $1$ and $2$ and been done with it. – fleablood Apr 17 '20 at 23:29
  • you are close. Try to show that if $n(b-a) > 1$ then there is an integer $m$ so that $na < m < nb$. Hint: fi $n(b-a) >1$ then $nb > na+ 1 > na$. Is there an integer between $n+1$ and $na$. (Sadly, you will not be allowed to say "of course there is") – fleablood Apr 17 '20 at 23:42

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