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I thought this proof was much simpler than it actually is. I used an Axiom that states, "if $p$ and $q$ are real numbers, then there is a number between them, i.e: $$\frac{(p + q)}{2}$$ My attempt at the proof was:

By the above Axiom, there exists a number, $q$, between $x$ and $y$, i.e: $$\frac{(x+y)}{2}$$ We know $q$ is rational because it is in the form $\frac ab$, where $a = x + y$ and $b = 2$. Thus, there exists a rational number, $q$ such that $q$ is between $x$ and $y$.

However, I realized that if $x=\pi$ and $y = e$, then $\frac{(x+y)}{2}$ will not be rational.

I know that to prove this, I need to use the Lemma: If $y-x>1$, ∃ n ∈ $\mathbb{Z}$ such that $x<n<y$.

I really don't know how to approach this differently because I did think it would be simpler.

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    $\begingroup$ x and y are distinct real numbers. It cannot be that both are the same. One concrete example could be pi and pi + e. Then both are real and distinct $\endgroup$ – Icycarus Feb 3 '17 at 1:35
  • $\begingroup$ We actually don't know whether $\pi+e$ is rational. It could be. $\endgroup$ – Matt Samuel Feb 3 '17 at 2:48
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  • Find the minimum integer value of $k$ for which $$|x-y| < 10^k$$

  • Round $\min(x,y)$ to the smallest multiple of $10^{k-1}$ greater than $\min(x,y)$.

  • Add $10^{k-2}$. The result is a rational number strictly between $x$ and $y$.

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Since $x$ and $y$ are different you can choose an integer $N$ large enough so that $1/N$ is less than their (positive) difference. Now imagine a ruler (number line) with marks every $1/N$th. One of those marks will have to fall strictly between $x$ and $y$. It's rational since it looks like $k/N$ for some integer $k$.

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  • $\begingroup$ So I wouldn't necessarily have to use the lemma? $\endgroup$ – Mathgirl Feb 3 '17 at 1:45
  • $\begingroup$ @Mathgirl No, you wouldn't. $\endgroup$ – Ethan Bolker Feb 3 '17 at 11:27
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If the difference between two real numbers is greater than 1, then there will be an integer between the numbers. If the difference is less than , take the difference of first decimal number from where the numbers start to differ. And that decimal number at its position has to exist between the real numbers.

Eg: $0.12345......$ and $0.123378.....$ are such real numbers then, $0.1234$ has to exist between them which is a rational number.

Similar trick Can be applied for negative numbers also.

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  • $\begingroup$ I understand the first part, but I am still a bit confused about the case when the difference is less than 1. Since $x$ and $y$ are arbitrary, is there a way to prove this would be true for all real numbers? Or would we use proof by example? $\endgroup$ – Mathgirl Feb 3 '17 at 1:55
  • $\begingroup$ Any arbitrary numbers will have decimal expansion similar to the example I gave. From there only required decimal numbers can be used to prove the existence of rational number. Try cooking up your own example and getting a decimal number between them. You will get the idea $\endgroup$ – jnyan Feb 3 '17 at 2:09

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