Prove: every irrational number $q$, given $e>0$, there exists natural numbers $N$ and $M$ such that $|Nq-M|<e$

Patatiniuh
2022-07-15
Answered

Prove: every irrational number $q$, given $e>0$, there exists natural numbers $N$ and $M$ such that $|Nq-M|<e$

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SweallySnicles3

Answered 2022-07-16
Author has **21** answers

Consider the quotient $\mathbb{R}/\mathbb{Z}$ which can be identified to the circle ${S}^{1}$. The irrationality of $q$ is equivalent to the injectivity of the quotient map $\pi :\mathbb{R}\to {S}^{1}$ restricted to $\mathbb{Z}q$. The assertion amounts to prove that for any open neighborhood $V\ni 0$ in ${S}^{1}$ there is always a $n\overline{q}\in V$ with $n\ne 0$, i.e. $\pi (\mathbb{Z}q)\cap (V\setminus (0))\ne \mathrm{\varnothing}$

Consider the translates ${V}_{n}=n\overline{q}+V\subset {S}^{1}$ for $n\in \mathbb{Z}$ (still open, of course). If you can find an $m\in \mathbb{Z}$ such that there exists $n\overline{q}\in {V}_{m}\setminus (m\overline{q})$ then $(n-m)\overline{q}\in V\setminus (0)$ and you are done.

If not, it means that for all $m\in \mathbb{Z}$, $\pi (\mathbb{Z}q)\cap {V}_{m}=\{m\overline{q}\}$. But this says that $\pi (\mathbb{Z}q)$ is discrete in ${S}^{1}$ and since ${S}^{1}$ is compact, this is in contradiction with its infinitness.

Consider the translates ${V}_{n}=n\overline{q}+V\subset {S}^{1}$ for $n\in \mathbb{Z}$ (still open, of course). If you can find an $m\in \mathbb{Z}$ such that there exists $n\overline{q}\in {V}_{m}\setminus (m\overline{q})$ then $(n-m)\overline{q}\in V\setminus (0)$ and you are done.

If not, it means that for all $m\in \mathbb{Z}$, $\pi (\mathbb{Z}q)\cap {V}_{m}=\{m\overline{q}\}$. But this says that $\pi (\mathbb{Z}q)$ is discrete in ${S}^{1}$ and since ${S}^{1}$ is compact, this is in contradiction with its infinitness.

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I suggested the following problem to my friend: prove that there exist irrational numbers a and b such that ab is rational.

Now, his inital solution was like this: let's take a rational number r and an irrational number i. Let's assume $a={r}^{i}$ b=\frac{1}{i}$$

So we have

${a}^{b}=({r}^{i}{)}^{\frac{1}{i}}=r$

which is rational per initial supposition. b is obviously irrational if i is. My friend says that it is also obvious that if r is rational and i is irrational, then ${r}^{i}$ is irrational. I quickly objected saying that $r=1$ is an easy counterexample. To which my friend said, OK, for any positive rational number r, other than 1 and for any irrational number i ${r}^{i}$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

Now, his inital solution was like this: let's take a rational number r and an irrational number i. Let's assume $a={r}^{i}$ b=\frac{1}{i}$$

So we have

${a}^{b}=({r}^{i}{)}^{\frac{1}{i}}=r$

which is rational per initial supposition. b is obviously irrational if i is. My friend says that it is also obvious that if r is rational and i is irrational, then ${r}^{i}$ is irrational. I quickly objected saying that $r=1$ is an easy counterexample. To which my friend said, OK, for any positive rational number r, other than 1 and for any irrational number i ${r}^{i}$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

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