# subtraction of two irrational numbers to get a rational [duplicate]

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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..

## marked as duplicate by Martin Sleziak, Ofir Schnabel, N. F. Taussig, jameselmore, MicahOct 23 '15 at 17:41

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• You can always do things like $\sqrt{2} - (\sqrt{2} - 3) = 3$. – levap Oct 22 '15 at 7:00
• Sure, you definitely know it and you can also write it out. $x=\pi -3$. – cr001 Oct 22 '15 at 7:13
• 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. – Jeppe Stig Nielsen Oct 22 '15 at 13:26
• @JeppeStigNielsen: Each addend is irrational. – user2357112 Oct 23 '15 at 3:25
• – Martin Sleziak Oct 23 '15 at 11:09

## 3 Answers

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...

• I'm very interested in that theorem you described? Would you be willing to put a quick proof? – user3256725 Oct 22 '15 at 7:07
• @user3256725, see the highlighted parts... – Clément Guérin Oct 22 '15 at 7:10
• 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.. – Elliot G Oct 22 '15 at 7:14
• Yes, thank you clement, i truly wish i could up vote more than once – user3256725 Oct 22 '15 at 7:24
• @ElliotG If the answer is correct, we don't even know that $\pi\mathrm{e}$ is irrational, let alone transcendental! – David Richerby Oct 23 '15 at 10:50

$\pi - \pi = 0$ which is rational.

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

• You sneaky math people.... – user3256725 Oct 22 '15 at 7:06
• Haha. We are trained to look for the most trivial counter examples and build up more interesting ones from there ;) – CPM Oct 22 '15 at 7:07
• @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. – hBy2Py Oct 22 '15 at 16:58

.223456789101112131415... - .123456789101112131415... = .1