Are the irrationals + zero an additive group?

Question

Is it true that the only way two irrationals can sum to a rational is if they sum to zero?

Thanks!

I do not understand the downvotes to this perfectly fine question. Would anyone care to enlighten me? — user1729, Sep 08 '13 at 14:59
See also http://math.stackexchange.com/questions/157245/is-the-sum-and-difference-of-two-irrationals-always-irrational and http://math.stackexchange.com/questions/867569/are-there-any-non-trivial-counterexamples-to-the-non-closure-of-the-irrational-n — Chris Culter, May 29 '15 at 18:02

score 11 · Accepted Answer · 2013-09-07T19:30:46.847

11

(Artificial) Counterexample: $\sqrt{2} + (1 - \sqrt{2}) \in \Bbb Q$.

edited Sep 07 '13 at 19:30

answered Sep 07 '13 at 19:10

6

Of course, every counterexample can be written similarly. – Sep 07 '13 at 19:14
3

What got downvoted? – Doug Sep 07 '13 at 19:26
it should be artificial right? – clark Sep 07 '13 at 19:30
@clark Yes, thank you. – Sep 07 '13 at 19:30
I am sorry.. :/ It certainly wasn't me. I appreciated your answer. – Doug Sep 07 '13 at 19:32
2

Why sweat the downvote (which looks like it's been removed anyway)? You got eight (eight!) upvotes, which your answer didn't deserve in my opinion. Take the rough with the smooth! – TonyK Sep 07 '13 at 19:59

score 5 · Answer 2 · answered Sep 07 '13 at 19:23

5

No. All counterexamples will be of form $\alpha + (q- \alpha)$, $\alpha \notin \mathbb{Q}$, $q \in \mathbb{Q}$. (Can you prove this?)

answered Sep 07 '13 at 19:23

Fernando Pessoa

1

Your comment is really the result I'm looking for.. I was trying to prove that $k = (1/\pi) \pm \sqrt{2m/\pi}$ has no solutions for $k, m$ integers. I'll give it a shot. – Doug Sep 07 '13 at 19:29
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@Dan use the fact that $\sqrt{\frac{1}{\pi}}$ is transcendental. – Dan Brumleve Sep 07 '13 at 19:32
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Note that algebraic numbers do form an additive group. – Dan Brumleve Sep 07 '13 at 19:32
@Fernando: Suppose $\alpha \notin \mathbb{Q}$, $p \in \mathbb{Q}$, and $\beta \in \mathbb{R}$ such that $\alpha + \beta = p$. Then $\beta = p - \alpha$. That was easy! :) – Doug Sep 07 '13 at 19:49
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@Dan first show that $\sqrt{\frac{2 \cdot m}{\pi}}$ is transcendental and then write your equation as a polynomial in $\sqrt{\frac{2 \cdot m}{\pi}}$. I don't think you should need much beyond the definition of transcendental. – Dan Brumleve Sep 07 '13 at 20:00
$0 = (1/2m)x^2 \pm x - k$ cannot have the transcendental $\sqrt{2m/\pi}$ as a solution. Got it! Thanks so much, Dan! – Doug Sep 07 '13 at 20:15
I am confused. I had accepted another answer. – Doug May 29 '15 at 18:01

score 1 · Answer 3 · answered Sep 07 '13 at 19:12

1

No. $p/q + i$ minus $i$ is rational. Where $i$ is irrational.

answered Sep 07 '13 at 19:12

Doug

3 Answers3