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November 2nd, 2018, 07:04 AM   #1
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The irrational number between irrationals

How I can prove that between every two irrational number there is irrational number?
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November 2nd, 2018, 07:12 AM   #2
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How I can prove that between every two irrational number there is irrational number?
If $p$ and $q$ are irrational numbers and (without loss of generality) $p > q$, you should be able to show there will be some other irrational number $x$ such that $x \in (q, p)$. There are a number of ways to do this. Take a crack at finding a solution and post what you come up with. If you get stuck, I'm sure someone will assist.
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November 2nd, 2018, 09:50 AM   #3
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Hint: prove first that between two rationals is an irrational, and that between any two irrationals is a rational.
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November 2nd, 2018, 10:26 AM   #4
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I know one way to show it, but it is rare and not as advanced as other methods.
Can anyone post the solution? I'm sure there are a lot of ways to show that.

Last edited by skipjack; November 3rd, 2018 at 04:21 PM.
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November 2nd, 2018, 10:27 AM   #5
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Suppose that $p > q$ with both irrational.

$p-q > 0$ and so there exists a natural number $ n$ such that $n(p-q) > 1 > 0$. Thus $p-q > \frac1n > 0$ and therefore $$p > q + \frac1n > q$$

All that you need now is to prove that $(q + \frac1n )$ is irrational. (Hint: assume that it is rational and show that $q$ must then also be rational - a contradiction - because the rationals are closed under subtraction.)
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Last edited by skipjack; November 2nd, 2018 at 10:48 AM.
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November 2nd, 2018, 11:49 AM   #6
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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 $q$.
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November 3rd, 2018, 02:52 PM   #7
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Originally Posted by idontknow View Post
I know one way to show it
What's your approach?
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November 7th, 2018, 01:11 PM   #8
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Each irrational number can equal to a sum of two irrational numbers $\displaystyle i=i_1+i_2$ and $\displaystyle i_2=i_3 + i_4 $
$\displaystyle i=i_1 + i_3 + i_4 $ such that $\displaystyle i_1 , i_3 , i_4 $ do not equal to each other , so there is a relation between them , one possible relation as $\displaystyle i_1 < i_3 < i_4 $
so $\displaystyle i_3$ is an irrational number between two irrational numbers
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November 7th, 2018, 05:08 PM   #9
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I can't say that I can see how that guarantees that two very close irrationals must have a third between them. Your irrationals also have a condition that is not met by every set of three irrationals.

I'm not saying it's definitely wrong, but I certainly don't find it very satisfying.
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November 7th, 2018, 05:31 PM   #10
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Originally Posted by idontknow View Post
Each irrational number can equal to a sum of two irrational numbers $\displaystyle i=i_1+i_2$ and $\displaystyle i_2=i_3 + i_4 $
$\displaystyle i=i_1 + i_3 + i_4 $ such that $\displaystyle i_1 , i_3 , i_4 $ do not equal to each other , so there is a relation between them , one possible relation as $\displaystyle i_1 < i_3 < i_4 $
so $\displaystyle i_3$ is an irrational number between two irrational numbers
This has several problems.

1. You haven't proved any of your claims and they are far from obvious. For example, prove that every irrational can be written as the sum of two other irrationals.

2. Even if the above claim was proved (it is true after all, just unproved), it doesn't prove what you set out to show. 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.
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