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My title was rejected a few times, here is what it was initially:

If you take two real numbers- one irrational and one possibly irrational - how close does their $\mathbb{Z}$ span come to any given real number?

I'm working on a problem for differential geometry, actually, and my approach thus far has brought me into a situation where I'm given a real number $M$, and hope to approximate it by combinations of the form $$ \mathbb{Z}2\pi \alpha + \mathbb{Z}2\pi $$ where $\alpha$ is a fixed irrational number. I would like to be able to pick $k$ and $n$ such that I get arbitrarily close to $M$, but can't find any particularly relevant information in any of the books I have, or on wikipedia. At first I thought I might have been looking for some kind of variation on the Euclidean algorithm of some sort, but I haven't managed to find anything yet. If someone could point me to some relevant literature (is there some kind of related Diophantine approximation that would help me out?), that would be most appreciated.

Todd
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    The $2\pi$ is irrelevant to this problem; the question you have is whether $\mathbb{Z}\alpha +\mathbb{Z}$ gets arbitrarily close to $\frac{M}{2\pi}$. If $\alpha$ is rational, then the answer is no in general. – vadim123 Nov 14 '13 at 05:15
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    For any irrational $\alpha$, the fractional parts of $n\alpha$ are dense in the interval $[0,1]$. That seems to be exactly what you need, and it is true if $\alpha$ is irrational. The proof is by a pigeonhole argument. – André Nicolas Nov 14 '13 at 05:18
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    In fact the proof is here. – vadim123 Nov 14 '13 at 05:19

3 Answers3

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Every additive subgroup of $\mathbb{R}$ is either dense or has a least positive element—see the answers to this question.

If the span of $\{\alpha,\beta\}$ has a least positive element, then that element must generate the subgroup and therefore must divide both $\alpha$ and $\beta$. So if $\alpha$ is irrational and $\beta$ is a nonzero rational number, the span of $\{\alpha,\beta\}$ must be dense.

More generally, the span of $\{\alpha,\beta\}$ is dense in $\mathbb{R}$ if and only if $\alpha$ and $\beta$ are nonzero and their ratio is irrational.

Community
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Trevor Wilson
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I wanted to add a bit to Trevor Wilson's answer. Another way to state the result is the following:

Theorem: If $L$ is a linear form in two variables that is not a multiple of a rational form, then the image of $\mathbb{Z}^2$ under $L$ is dense in $\mathbb{R}$.

While this is a standard problem in analysis, its generalizations lead to difficult number theoretic questions. What happens if we replace $\mathbb{Z}$ with the set of squares or higher powers? Do we still have a similar result? Let $S=\{1,4,9,16,25,\dots\}$ be the set of squares, and suppose that we have $\alpha,\beta>0$ that are not rational multiples of each other.

Conjecture: Is it true that $$\alpha\cdot S-\beta\cdot S$$ is dense in $\mathbb{R}$?

While this conjecture may seem reasonable at first, it is in fact false. There are irrational numbers, such as any quadratic irrational, which cannot be too well approximated by rationals, which means that the above form does not necessarily hit real numbers close to $0$. Specifically, if $\alpha=1$ and $\beta=3+2\sqrt{2}$, then $\alpha n^2-\beta m^2$ cannot be too small, since this would have implications about the rational approximations to $1+\sqrt{2}=\sqrt{3+2\sqrt{2}}.$ However, if a third variable is added, then there is an analogous result, but it is difficult.

Theorem: (Margulis) Let $Q$ be an indefinite non-degenerate quadratic form in $3$ variables, that is not a multiple of a rational form. Then the image of $\mathbb{Z}^3$ under $Q$ is dense in $\mathbb{R}$.

This is proven by looking at the orbits of points in $\text{SL}_3(\mathbb{R})/\text{SL}_3(\mathbb{Z})$ under a subgroup which corresponds to non-degenerate indefinite quadratic forms in $3$ variables. (Specifically $\text{SO}(2,1)$)

My reference for this topic is Ergodic Theory with a view towards Number Theory by Einsiedler and Ward.

Eric Naslund
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You don't need a variant of the Euclidean algorithm: you can apply it directly!

There are two cases:

  • The algorithm eventually terminates at the pair of numbers $(0, x)$
  • The algorithm never terminates, and produces a sequence of numbers converging to $0$

In the latter case, if you can produce numbers in the span arbitrarily close to 0, then you can get arbitrarily close to any real number by taking multiples.