An assignment I am working on (Problem 4-4 in Lee's Introduction to Smooth Manifolds) reduces to the following problem. Given $\alpha, \beta, \varepsilon\in\mathbb{R}$, with $\alpha$ irrational and $\varepsilon > 0$, show there exist two integers, $m, n$ such that $$|m - \alpha \cdot n + \beta| < \epsilon.$$

If $\beta\in \mathbb{Z}$, then we could simply use Dirichlet's approximation theorem, but I've not been able to prove it in general.

One thing I've tried is approximating $\beta$ as a rational number $\tilde \beta = p/q \in \mathbb{Q}$, which leads to $|m - \alpha n + p/q | < \epsilon$ which can be manipulated into $|qm + p - \alpha q n | < \epsilon$. We can then let $\tilde m = qm + p$ and $\tilde n = q n$, and apply DAT to show $\tilde m$ and $\tilde n$ exist such that $|\tilde m + \alpha \tilde n| < q\epsilon$. One problem with this approach, though, is that when we solve for $m = \frac{\tilde m - p}{q}$ we aren't guaranteed to get an integer.

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    $\begingroup$ Search for Kronecker's density theorem. You only need the 1-dimensional case. The proof is pretty much the same as with Dirichlet's. You can find integers $m,n$ such that the fractional part of $m-n\alpha$ is in $(0,\epsilon)$. But then $2m-2n\alpha,3m-3n\alpha,\ldots$ make a sequence with increments less than $\epsilon$. Surely you can approximate anything in $(0,1)$ well enough with one of those. $\endgroup$ Commented Oct 25, 2019 at 5:48
  • $\begingroup$ Wow, that's oddly specific to what I need. Thanks! $\endgroup$
    – Paul Wintz
    Commented Oct 25, 2019 at 5:56
  • $\begingroup$ Actually, it looks like that does not apply. Using the notation from the Wikipedia article (en.wikipedia.org/wiki/Kronecker%27s_theorem), the condition only holds if $\beta r \in \mathbb{Z}$ for some $r \in \mathbb{Z}$, which is impossible if $\beta$ is irrational. $\endgroup$
    – Paul Wintz
    Commented Oct 25, 2019 at 6:06
  • $\begingroup$ That WP-page looks very weird to me. The version of Kronecker's density theorm I know of says that if the set $1,\alpha_1,\alpha_2,\ldots,\alpha_n$ of real numbers is linearly independent over $\Bbb{Z}$ (in particular all the $\alpha_i$ are irrational), then the fractional parts of the vectors $m(\alpha_1,\ldots,\alpha_n)$ are dense in $[0,1]^n$. $\endgroup$ Commented Oct 25, 2019 at 6:18
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    $\begingroup$ On Wolfram MathWorld, the theorem is stated in exactly the form I need. Thanks! mathworld.wolfram.com/KroneckersApproximationTheorem.html $\endgroup$
    – Paul Wintz
    Commented Oct 25, 2019 at 6:34

1 Answer 1


As suggested by Jyrki Lahtonen in the comments above, we can use Kronecker's Approximation Theorem, which, according to Wolfram MathWorld, is as follows for the one-dimensional case.

For any $\alpha, \beta, \epsilon \in \mathbb{R}$, with $\alpha$ irrational and $\epsilon > 0$, then there exists integers $m$ and $n$ with $n>0$, such that $$|m - \alpha n + \beta| < \epsilon.$$


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