Can finite sums of two numbers come arbitrarily close to zero?

Question

Given two real numbers $a$ and $b$, define an $a$-$b$-sum as a finite sum of $a$'s and $b$'s, i.e. a sum:

$$m\cdot a + n\cdot b$$ where $m,n$ are non-negative integers.

Is there a pair of numbers $a<0$ and $b>0$, such that for every $\epsilon>0$ there is an $a$-$b$-sum $S \in (0,\epsilon)$?

The claim is obviously untrue if $a$ and $b$ are integers, since in that case the sum is also integer so there is no $S \in (0,1)$.

It is also untrue if $a$ and $b$ are rationals, since in that case the sum is always an integer multiple of $\frac{1}{pq}$ (where $p,q$ are the denominators of $a,b$ respectively) so there is no $S\in (0,\frac{1}{pq})$.

Is there a pair of irrational numbers that makes the claim true?

I might be misreading something, but since the sum is finite isn't it equal to $ma + nb$ for some $m,n\in\mathbb N$? — Ennar, Dec 22 '15 at 10:03
@Ennar you are right. I simplified the question accordingly. — Erel Segal-Halevi, Dec 22 '15 at 10:22
This fact is relevant: The set $C$ of integer linear combinations $ma+nb$ is an additive subgroup of $\mathbb R$ and so is either discrete cyclic or dense. Which case holds depends on whether $\alpha:=\inf (C \cap \mathbb R^+)=0$ or not. — lhf, Dec 22 '15 at 10:36

score 6 · Accepted Answer · answered Dec 22 '15 at 10:11

6

Take $a=\sqrt{2}$ and $b=-1$ .

From Kronecker's Theorem there is a positive integer $c$ such that : $$\{c\sqrt{2}\} < \epsilon $$ (This case of the theorem can be proved by a simple pigeon-hole argument )

Now simply take the following $(a,b)$-sum :

$$c \cdot a + \lfloor c\sqrt{2} \rfloor \cdot b=c\sqrt{2}-\lfloor c\sqrt{2} \rfloor=\{c\sqrt{2} \}<\epsilon$$ as wanted .

answered Dec 22 '15 at 10:11

by ${x}$ you mean the fractional part of $x$? – Erel Segal-Halevi Dec 22 '15 at 10:23
@ErelSegal-Halevi, yes. – wythagoras Dec 22 '15 at 10:24
4

Yes. A pigeonhole argument works whenever $a$ and $b$ are linearly independent over $\Bbb{Q}$. For the specific pair $\sqrt2$ and $-1$ you can also take a look at this. Mariano's or lhf's answer in the same thread gives an argument that IIRC works whenever $a/b$ is irrational. – Jyrki Lahtonen Dec 22 '15 at 10:25

Can finite sums of two numbers come arbitrarily close to zero?

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