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Prove the following statement: If $r_1$ and $r_2$ are real numbers whose quotient is irrational, then any real number $x$ can be approximated arbitrarily well by numbers of the form $z_{k_1,k_2}= k_1r_1+ k_2r_2, k_1, k_2$ integers; i.e., for every real number $x$ and every positive real number $p$ two integers $k_1$ and $k_2$ can be found such that $\left |x−(k_1r_1+k_2r_2)\right | < p$.

This question seems quite interesting, but I didn't know how to use the part about the quotient of $r_1$ and $r_2$ being irrational. How in the solution below did they get the conclusion that "all numbers $z_{m_1,m_2} = m_1r_1 + m_2r_2 (m_1, m_2 \in \mathbb{Z})$ are distinct"? Then the rest of the solution also seems a little bit fuzzy so if anyone can explain that would help.

Solution enter image description here

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  • $\begingroup$ to see they're distinct: what would happen if $m_1r_1+m_2r_2=m_3r_1+m_4r_2$? $\endgroup$ – symplectomorphic Jan 13 '16 at 3:55
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Suppose $m_1 r_1 + m_2 r_2 = n_1 r_1 + n_2 r_2$, for $m_i, n_i \in \mathbf{Z}$. Then,

$$(m_1 - n_1) r_1 + (m_2 - n_2 ) r_2 = 0 $$

Since $\frac{r_1}{r_2}$ is irrational, this in particular implies $r_i \ne 0$. Thus, divide by $r_2$ to get

$$(m_1 - n_1)\frac{r_1}{r_2} + m_2 - n_2 = 0 \Rightarrow \frac{r_1}{r_2}=\frac{n_2-m_2}{m_1-n_1}$$

Since $m_i,n_i \in \mathbf{Z}$, then so are their differences. Thus, we have expressed the quotient as a quotient of integers, i.e. a rational number. This is a contradiction.

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  • $\begingroup$ Where are they getting $[-n(|r_1|+|r_2|,n(|r_1|+|r_2|)]$ from? $\endgroup$ – user19405892 Jan 13 '16 at 4:01
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It may simplify matters by "rescaling" the problem without loss of generality by assuming that one of the numbers is $1$ (the density property you are seeking is clearly preserved). To show that integer combinations of $1$ and $r$ are dense, think of it as taking fractional part of integer multiples of $r$. This should also make clear why $r$ needs to be irrational (which is equivalent to the ratio of the original pair of numbers being irrational).

What you point out next is that the fractional part $\{nr\}$ for $n=1,2,3,\ldots$ are all points in the half-open interval $[0,1)$. What you need to know next is that these points are "evenly" distributed, i.e., they are dense in the interval. Talking about both $r_1$ and $r_2$ is a confusing before you understand the solution, but afterward you can write it down also in terms of $r_1$ and $r_2$.

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  • $\begingroup$ Where are they getting $[-n(|r_1|_|r_2|,n(|r_1|+|r_2|)]$ from? Can you explain the next part of the solution? $\endgroup$ – user19405892 Jan 13 '16 at 4:10

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