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If $x\lt y $ for arbitrary real $x$ and $y$ there exists a real r $r$ such that $x \lt r \lt y$

Prove that there is at least one r satisfying this inequality, and hence infinitly many.

I was wondering if this would be an acceptable answer to this question.

$y-x \gt 0 \to $ by use of Archimedian Principle that there exists $n$ such that $n(y-x) \gt 1$

so $n \gt \frac 1{y-x}$ so $y-x \gt \frac 1n$ for all $n \gt \frac 1 {y-x}$ so let $\frac 1n = c$ then $y-x>c$

Assume $x+c \lt x$ then $x-x-c\gt 0$ so $-c>0$ This is a contradiction due to Trichtomy so $x+c \gt x$

Assume then that $x+c \gt y$ Since $y-x \gt c$ then $x+(y-x) \gt c+ x \to y \gt x+c$ Contradiction. So $x+c \lt y$ So Let $x+c = x+ \frac 1n \forall n\gt \frac 1 {y-x} = r$, then since r is unbounded above $x \lt r \lt y$ for infinitely many r.

Would this be a correct answer? If not, what are my errors? Thank you for your help.

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If you want to show there is a rational between $x$ and $y$, you do have to go through some work with "$1/n$." However, in your case, since it only asks for a real, $(x+y)/2$ does the job. – André Nicolas Mar 14 '13 at 18:18
up vote 2 down vote accepted

Your proof is correct, but here are some comments:

1) It is an axiom of ordering that if $a < b$ then $a + c < b + c$ for all $c$. You do use this, but you can eliminate your need for trichotomy/proof by contradiction as follows:

You know $y - x > c > 0$. Adding $x$ to $c > 0$ yields $x + c > x$. Adding $x$ to $y - x > c$ gives $y > x + c$. And so, $x < x + c < y$.

2) Your last sentence is sort of awkward. I would suggest writing: "So, if $n > 1/(y - x)$, we have shown that $x + 1/n$ is a real number between $x$ and $y$. Since there are infinitely many integers greater than $1/(y - x)$, this gives infinitely many real numbers between $x$ and $y$."

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Thanks, I see what you're saying. That's a nice way of doing it. – AlexHeuman Mar 14 '13 at 18:46

You could simply take $\frac{x+y}{2}$.

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I can't, because I have to show that there are infinitely many r between x and y. Taking the average, does not show that. – AlexHeuman Mar 14 '13 at 18:16
You can repeat the argument: Start with $r$ with $x < r < y$. Now apply the argument on $x$ and $r$, giving you $x < r' < r < y$ etc. – azimut Mar 14 '13 at 18:17
I suppose that I can repeat the argument, but would there be a way to show that this works infinitely many times, or just as many times as I repeat the argument, and is there something wrong with my proof? – AlexHeuman Mar 14 '13 at 18:18
AlexHeuman: That it works as often as you like is induction. – azimut Mar 14 '13 at 18:19
I understand that, but with induction, you usually show for the case k and k+1, here you are just doing repeating processes. I could be wrong, but that just doesn't seam kosher to me. – AlexHeuman Mar 14 '13 at 18:20

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