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Source. Laczkovich (1984), printed pp. 114–115 (PDF pp. 6–7), equations (11)–(15). The source uses density of $I(\alpha)$ and solves the two integer equations explicitly. This page proves the density fact and writes the same solution directly in the additive group.

Statement

For every irrational real $\alpha$, the subgroup $G_\alpha=\mathbb Z\alpha+\mathbb Z$ is dense in $\mathbb R$. If $t\in G_\alpha$ is positive and $N$ is a positive integer, there are $c,d\in G_\alpha$, both positive, such that

$$Nc+(N+1)d=t. \tag{1}$$

Thus, if $t=b-a$ with $a,b\in G_\alpha$, the two successive arithmetic progressions with steps $c$ and $d$ run from $a$ to $b$ and stay in $[a,b]\cap G_\alpha$.

Dependencies. The pigeonhole principle and the Archimedean property of the real numbers. No continued-fraction approximation is needed for this page.

Bears on. Problem 1125, through Theorem 2.

Proof

For an integer $m\ge1$, place the $m+1$ fractional parts of $0,\alpha,\ldots,m\alpha$ into $m$ half-open intervals of length $1/m$. Two lie in the same interval. Their difference has absolute value strictly between $0$ and $1/m$: it is nonzero by irrationality. Its absolute value belongs to $G_\alpha$, since the difference is an integer multiple of $\alpha$ minus an integer, and the group is closed under negation.

Therefore $G_\alpha$ has positive elements as small as desired. For any real $u<v$, choose $h\in G_\alpha$ with $0<h<v-u$. If $k=\lfloor u/h\rfloor+1$, then

$$u<kh\le u+h<v.$$

This proves density.

Now choose

$$z\in G_\alpha\cap\left(\frac{t}{N+1},\frac{t}{N}\right)$$

and set

$$c=(N+1)z-t,\qquad d=t-Nz.$$

Both are in $G_\alpha$ and positive, and direct expansion gives (1). The source writes $t$ as $n\alpha+k$ and $z$ as $t\alpha+u$, using its own integer letter $t$; then $c=p\alpha+q$ and $d=r\alpha+s$ with precisely its integer solutions $p=(N+1)t-n$, $r=-Nt+n$, $q=(N+1)u-k$, and $s=-Nu+k$.

Since $c,d>0$, the ordered points

$$a,a+c,\ldots,a+Nc,\ a+Nc+d,\ldots,a+Nc+(N+1)d=b$$

are all in the stated interval and subgroup.