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November 8th, 2018, 02:40 AM   #11
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Quote:
 Originally Posted by SDK 1. You haven't proved ... For example, ... that every irrational can be written as the sum of two other irrationals.
Well, it is kind of trivial.$$p=\frac{p+1}{2}+\frac{p-1}{2}$$
The problem is in getting a proof that is general enough for the purpose required.

Quote:
 Originally Posted by SDK 2. ... This would show that every irrational can be inserted between two others. It does NOT show that given two irrationals, you can find one between them. The difference is you don't get to pick the two irrationals, they are given to you.
This is what was troubling me.

November 8th, 2018, 04:06 AM   #12
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Quote:
 Originally Posted by v8archie Well, it is kind of trivial.$$p=\frac{p+1}{2}+\frac{p-1}{2}$$
Yeah you are right this is obvious. Thanks for catching it.

November 10th, 2018, 08:57 PM   #13
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Quote:
 Originally Posted by v8archie Given a number $b$, there is a greatest integer $m$ such that $m \le b$ and $m+1 > b$, (because $(m+1)$ is, by definition an integer greater than $m$). Suppose that $a-b > 1$, then $$b + (a-b) = a \gt m+1 \gt b$$ With $a=np$ and $b=nq$ we then have that $np > m+1 > nq$ and so $$p > \frac{m+1}{n} > q$$ and $\frac{m+1}{n}$ is thus a rational between $p$ and $q$.
Thinking about this, I wonder whether it's better to keep the above for cases where $b > 0$ (and talk about natural numbers rather than integers). For $a < 0$, there is a greatest natural number $m$ such that $m \le -a$ and so $m+1 > -a$ and then, with $a-b > 1$ we have $$-a < m+1 < -a + (a-b) = -b \\ \implies a > -(m+1) > b$$ and thus $-\frac{m+1}{n}$ is a rational between $p$ and $q$.

In the remaining case, where $a > 0 > b$, zero is a rational between $p$ and $q$.

Last edited by v8archie; November 10th, 2018 at 09:28 PM.

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