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

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..

• You can always do things like $\sqrt{2} - (\sqrt{2} - 3) = 3$. Oct 22, 2015 at 7:00
• Sure, you definitely know it and you can also write it out. $x=\pi -3$. Oct 22, 2015 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. Oct 22, 2015 at 13:26
• @JeppeStigNielsen: Each addend is irrational. Oct 23, 2015 at 3:25
• Oct 23, 2015 at 11:09

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? Oct 22, 2015 at 7:07
• @user3256725, see the highlighted parts... Oct 22, 2015 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.. Oct 22, 2015 at 7:14
• Yes, thank you clement, i truly wish i could up vote more than once Oct 22, 2015 at 7:24
• @ElliotG If the answer is correct, we don't even know that $\pi\mathrm{e}$ is irrational, let alone transcendental! Oct 23, 2015 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.... Oct 22, 2015 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, 2015 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. Oct 22, 2015 at 16:58

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