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Say you have a number like $\pi$ or e. Is it possible to subtract another number from it and end up with a rational number? I mean I guess you could write an equation like $\pi-x=3$ But could there ever be a solution for x (that we could know and write out)?

Correction: Any number besides the irrational number itself. Damn math ppl are too quick..

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    $\begingroup$ You can always do things like $\sqrt{2} - (\sqrt{2} - 3) = 3$. $\endgroup$
    – levap
    Oct 22, 2015 at 7:00
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    $\begingroup$ Sure, you definitely know it and you can also write it out. $x=\pi -3$. $\endgroup$
    – cr001
    Oct 22, 2015 at 7:13
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    $\begingroup$ Note: No proof is known to determine whether $x = \pi + e$ is rational or irrational, even if we know very well that each addend is rational. $\endgroup$ Oct 22, 2015 at 13:26
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    $\begingroup$ @JeppeStigNielsen: Each addend is irrational. $\endgroup$ Oct 23, 2015 at 3:25
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    $\begingroup$ See: Is the sum and difference of two irrationals always irrational? $\endgroup$ Oct 23, 2015 at 11:09

3 Answers 3

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If you take any real number $x\in\mathbb{R}$ you can show that the set of $y\in\mathbb{R}$ such that $x-y\in\mathbb{Q}$ is exactly $x+\mathbb{Q}$.

Let $y$ be in $\mathbb{R}$ with $x-y$ being rational then $y=x-(x-y)$ so that $y\in x+\mathbb{Q}$. On the other hand, if $y=x+q$ with $q\in\mathbb{Q}$ then $x-y=-q\in\mathbb{Q}$.

So this is kind of easy. However I think that it is not known wether $\pi+e$ is rational or not... What we know is that either $\pi e$ or $\pi+e$ (maybe both) is irrational...

Assume that both $\pi e$ and $\pi+e$ are rational then :

$P(x):=(x-\pi)(x-e)=x^2-(\pi+e)x+\pi e$ is a polynomial with rational coefficients. This implies that both $\pi$ and $e$ are algebraic numbers which cannot be true...

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  • $\begingroup$ I'm very interested in that theorem you described? Would you be willing to put a quick proof? $\endgroup$ Oct 22, 2015 at 7:07
  • $\begingroup$ @user3256725, see the highlighted parts... $\endgroup$ Oct 22, 2015 at 7:10
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    $\begingroup$ Unrelated, but it always blows my mind that seemingly easy facts are so difficult to prove. Sometimes a theory sounds ridiculous and is simple to prove; other times it takes us centuries to prove whether $\pi e$ is transcendental or not.. $\endgroup$
    – pancini
    Oct 22, 2015 at 7:14
  • $\begingroup$ Yes, thank you clement, i truly wish i could up vote more than once $\endgroup$ Oct 22, 2015 at 7:24
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    $\begingroup$ @ElliotG If the answer is correct, we don't even know that $\pi\mathrm{e}$ is irrational, let alone transcendental! $\endgroup$ Oct 23, 2015 at 10:50
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$\pi - \pi = 0$ which is rational.

Edit: $\pi - (\pi -1) = 1$ which is a difference of two different irrationals which is rational. How's that?

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    $\begingroup$ You sneaky math people.... $\endgroup$ Oct 22, 2015 at 7:06
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    $\begingroup$ Haha. We are trained to look for the most trivial counter examples and build up more interesting ones from there ;) $\endgroup$
    – CPM
    Oct 22, 2015 at 7:07
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    $\begingroup$ @CPM This makes me feel a lot better, actually -- that there's training in how to find (most-)trivial counterexamples. Explains why I don't see them nearly so readily. $\endgroup$
    – hBy2Py
    Oct 22, 2015 at 16:58
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.223456789101112131415... - .123456789101112131415... = .1

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