My question is the following:
Let $q$ be irrational. Let $a,b$ be rationals such that $a<q<b$. If there exists an element $p$ s.t. $a<p<b$, would it be valid to conclude $q=p$?
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$\begingroup$ What do you mean by continuing with this process? If you have $q$, $a$ and $b$, you have $q$, $a$ and $b$ and cannot do anything to them any more. Picking $c$ for $a$ is a different case which might yield a different $p$. $\endgroup$– JiKApr 2, 2014 at 16:56
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2 Answers
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You certainly cannot conclude that $p=q$. In fact $a \lt \frac 12(a+b) \lt b$ and (since you said $q$ is irrational) $q \neq \frac 12(a+b)$. Also, either $q+\frac 12(b-a)$ or $q-\frac 12(b-a)$ will be another irrational between $a$ and $b$.
answered Apr 2, 2014 at 16:54
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This depends crucially on the wording of the question. As your question stands, you absolutely cannot conclude that $p=q$. In fact, in any open interval there are uncountably many irrational numbers, so if you choose $a$ and $b$ before you ask whether $p=q$, you definitely can't conclude $p=q$.
However, if you're handed $p$ and told that for any $a,b\in\mathbb{Q}$ such that $a<q<b$, also $a<p<b$, then you can conclude that $p=q$.
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$\begingroup$ I think the second paragraph is what the OP was trying to get at (without the formal language for doing so); it might be worth fleshing out that this is the case (in fact, of course, this is one of the ways the real numbers are defined). $\endgroup$ Apr 2, 2014 at 17:02
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$\begingroup$ Thank you for your reply. That's what I was trying to ask, though I realize now I worded it wrong. If I wanted to prove your second remark, do you think it would it be best to use induction? $\endgroup$– JoshHApr 2, 2014 at 17:04
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$\begingroup$ @JoshH While to some extent how to prove your remark will depend on your definition of real number, one straightforward way is to argue by contradiction: assume that $p\neq q$; then $p-q\gt 0$, so there's a rational number $\epsilon\lt\frac12(p-q)$. You should be able to show that there is an integer multiple of $\epsilon$ between $p$ and $q$... $\endgroup$ Apr 2, 2014 at 21:16