# Irrationals and integers

So basically this note is just about relations between irrational numbers and integers. This is also my first note so any comments about LaTeX, formatting or anything like that is appreciated. Thanks.

So basically there's the really beautiful theorem in maths, probably one of my favourites, which is Beatty's theorem. This theorem states that given any two positive irrationals $$\alpha$$ and $$\beta$$, such that $$\frac{1}{\alpha}+\frac{1}{\beta}=1$$, the two sequences, say $$A=a_i$$ and $$B=b_i$$ generated by $$a_n = \lfloor n \alpha \rfloor$$ and $$b_n = \lfloor n \beta \rfloor$$ have the property that any positive integer is in exactly one of these two sequences. The proof of this comes in two parts, both of which are proofs by contradiction and are extremely similar; skip ahead if you don't need convincing:

Step 1: Every integer lies in at least one of the two sequences

Suppose that there exists an integer, say $$n$$ which doesn't lie in either sequence. Then let

$$a_i<n, \space n+1 \leq a_{i+1}$$ and $$b_j<n, \space n+1 \leq b_{j+1}$$

In other words,

$$\alpha i<n, \space n+1 \leq \lfloor \alpha (i+1) \rfloor < \alpha (i+1)$$ and $$\beta j<n, \space n+1 \leq \beta (j+1)$$

The first part comes from the fact that $$n$$ is an integer, and thus if $$n>\lfloor k \rfloor$$ then that means that $$n>k$$. The second one comes from $$\lfloor x \rfloor \leq x$$ for reals $$x$$.

Actually, equality cannot occur in the second and fourth inequalities as $$\alpha, \beta$$ are both irrational so any positive integral multiple of them is never equal to an integer. Therefore

$$\alpha i<n, \space n+1 < \lfloor \alpha (i+1) \rfloor < \alpha (i+1)$$ and $$\beta j<n, \space n+1 < \beta (j+1)$$

These in turn imply

$$i<\frac{n}{\alpha}, \space \frac{n+1}{\alpha}<i+1, j<\frac{n}{\beta}, \space \frac{n+1}{\beta}<j+1$$

Now adding corresponding inequalities and making use of the fact that $$\frac{1}{\alpha}+\frac{1}{\beta}=1$$, we have

$$i+j<n, \space n+1<i+j+2 \implies i+j<n<i+j+1$$

Hopefully you can see where the contradiction is.

Step 2: No integer lies in both sequences

Suppose an integer lies in both sequences, that is

$$n=a_i$$ and $$n=b_j$$

This is the same as

$$n \leq \alpha i<n+1$$ and $$n \leq \beta j<n+1$$

However equality can never occur due to the same reasoning as above so

$$n<\alpha i<n+1$$ and $$n<\beta j<n+1$$

Lastly we divide by $$\alpha, \space \beta$$ respectively to get

$$\frac{n}{\alpha}<i<\frac{n+1}{\alpha}$$ and $$\frac{n}{\beta}<j<\frac{n+1}{\beta}$$

Adding and using the fact that $$\frac{1}{\alpha}+\frac{1}{\beta}=1$$, we have

$$n<i+j<n+1$$

Again the contradiction should be quite obvious. $$\square$$

Main point of the article

Now I was just thinking about irrationals and stuff, and I came up with an intuitive property which is

For any positive $$p, \epsilon \space \exists$$ a positive integer $$q$$ such that $$\{ pq \} < \epsilon$$ where $$\{ x \} = x- \lfloor x \rfloor$$

For example, {3.14}=0.14, {2.5}=0.5, etc.

This proof is original and so is the problem so if you've seen this elsewhere it's just a coincidence that my proof is the same as someone else's. Great minds think alike

Now for the proof. Firstly, here I'm only dealing with the case where $$p$$ is irrational; the case where it is rational is left for you.

Lemma 1: If $$m, \space n \in \mathbb{N}$$ then there exists $$k \in \mathbb{N}$$ such that $$mx+nx=kx$$

I'll leave the proof of this as a (mental) exercise.

Lemma 2: $$\{ \{ ax \} + \{ bx \} \} = \{ (a+b)x \} \space \forall \space a \in \mathbb{N}$$ and $$x \space \in \mathbb{Z}$$

This can be proved by using the substitution $$ax=m+r_1$$ and $$bx=n+r_2$$ where $$m, \space n \in \mathbb{N}$$ and the fact that $$\{ x \} = \{ x-n \}$$ (again, $$n$$ is an integer whereas $$x$$ is any real.

Now with these two lemmas let's attack the main problem. Define a sequence

$$a_1= \{ p \}$$

$$a_i=ka_{i-1}-1$$ such that $$(k-1)a_{i-1}<1<ka_{i-1}$$

Now by our two lemmas for every $$a_i$$ there exists $$k$$ such that $$a_i= \{ kp \}$$. This means that we only need to find an $$a_i$$ such that $$a_i<\epsilon$$ to complete the proof.

Firstly, $$a_i$$ is a strictly decreasing sequence. See if you can prove this; it should follow directly from the definition.

Secondly $$a_i$$ is bounded. Again this is really straightforward.

Now by the monotone bounded sequence theorem, which states that

If a sequence is bounded and monotonic then it converges to a limit,

we have that $$a_i$$ converges. If it converges to 0 then we are done. Therefore we need to prove that this sequence converges to zero.

Suppose it converges to $$L>0$$. Then we have $$\frac{1}{z-1}>L \geq \frac{1}{z}$$ for an integer $$z$$. ($$z$$ exists because $$z= \lfloor \frac{1}{L} \rfloor$$ works)

Therefore there exists an $$n$$ such that $$\frac{1}{z-1}>a_i>L$$ and therefore $$a_{i+1}=za_i-1$$ for all $$i \geq n$$.

Now I shall prove by induction that $$a_{n+x}=z^x a_n - \frac{z^x-1}{z-1}$$.

For $$a_n, \space a_{n+1}$$ it works by just checking.

Now suppose my formula works for $$x=k$$. Then for $$x=k+1$$:

$$a_{n+k+1}$$

$$=za_{n+k}-1$$

$$=z^{x+1} a_{n+k}-\frac{z^k-1}{z-1}-1$$

which is what my formula predicts.

Therefore for all $$x$$ we have

$$a_{n+x}=z^x a_n - \frac{z^x-1}{z-1} \geq \frac{1}{z}$$

Dividing by $$z^x$$ yields

$$a_n - \frac{z^{x-1}+z^{x-2}+...+1}{z^x} \geq \frac{1}{z^{x+1}}$$

which is

$$a_n \geq \frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...+\frac{1}{z^{x+1}}$$

which is a contradiction as the RHS approaches $$\frac{1}{z-1}$$ as $$x$$ approaches infinity, but we have $$\frac{1}{z-1}>a_n \square$$

Sorry if you weren't able to follow the parts which I didn't fully prove but LaTeX is really annoying to write; ask me in the comments if you didn't understand why something was true.

I hope you enjoyed this!

Note by Wen Z
2 years, 5 months ago

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Very nice note on the irrational numbers. Relevant problem for the converse of Beatty's Theorem.

- 2 years, 5 months ago