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 α\alpha and β\beta, such that 1α+1β=1\frac{1}{\alpha}+\frac{1}{\beta}=1, the two sequences, say A=aiA=a_i and B=bi B=b_i generated by an=nαa_n = \lfloor n \alpha \rfloor and bn=nβb_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 nn which doesn't lie in either sequence. Then let

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

In other words,

αi<n, n+1α(i+1)<α(i+1)\alpha i<n, \space n+1 \leq \lfloor \alpha (i+1) \rfloor < \alpha (i+1) and βj<n, n+1β(j+1)\beta j<n, \space n+1 \leq \beta (j+1)

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

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

αi<n, n+1<α(i+1)<α(i+1)\alpha i<n, \space n+1 < \lfloor \alpha (i+1) \rfloor < \alpha (i+1) and βj<n, n+1<β(j+1)\beta j<n, \space n+1 < \beta (j+1)

These in turn imply

i<nα, n+1α<i+1,j<nβ, n+1β<j+1i<\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 1α+1β=1\frac{1}{\alpha}+\frac{1}{\beta}=1, we have

i+j<n, n+1<i+j+2    i+j<n<i+j+1i+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=ain=a_i and n=bjn=b_j

This is the same as

nαi<n+1n \leq \alpha i<n+1 and nβj<n+1n \leq \beta j<n+1

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

n<αi<n+1n<\alpha i<n+1 and n<βj<n+1n<\beta j<n+1

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

nα<i<n+1α\frac{n}{\alpha}<i<\frac{n+1}{\alpha} and nβ<j<n+1β\frac{n}{\beta}<j<\frac{n+1}{\beta}

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


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,ϵ p, \epsilon \space \exists a positive integer qq such that {pq}<ϵ \{ pq \} < \epsilon where {x}=xx \{ 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 pp is irrational; the case where it is rational is left for you.

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

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

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

This can be proved by using the substitution ax=m+r1ax=m+r_1 and bx=n+r2bx=n+r_2 where m, nNm, \space n \in \mathbb{N} and the fact that {x}={xn} \{ x \} = \{ x-n \} (again, nn is an integer whereas xx is any real.

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

a1={p}a_1= \{ p \}

ai=kai11a_i=ka_{i-1}-1 such that (k1)ai1<1<kai1(k-1)a_{i-1}<1<ka_{i-1}

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

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

Secondly aia_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 aia_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>0L>0. Then we have 1z1>L1z\frac{1}{z-1}>L \geq \frac{1}{z} for an integer zz. (zz exists because z=1Lz= \lfloor \frac{1}{L} \rfloor works)

Therefore there exists an nn such that 1z1>ai>L\frac{1}{z-1}>a_i>L and therefore ai+1=zai1a_{i+1}=za_i-1 for all ini \geq n.

Now I shall prove by induction that an+x=zxanzx1z1a_{n+x}=z^x a_n - \frac{z^x-1}{z-1}.

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

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



=zx+1an+kzk1z11=z^{x+1} a_{n+k}-\frac{z^k-1}{z-1}-1

which is what my formula predicts.

Therefore for all xx we have

an+x=zxanzx1z11za_{n+x}=z^x a_n - \frac{z^x-1}{z-1} \geq \frac{1}{z}

Dividing by zxz^x yields

anzx1+zx2+...+1zx1zx+1a_n - \frac{z^{x-1}+z^{x-2}+...+1}{z^x} \geq \frac{1}{z^{x+1}}

which is

an1z+1z2+1z3+...+1zx+1a_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 1z1\frac{1}{z-1} as xx approaches infinity, but we have 1z1>an\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
4 years, 11 months ago

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

Sharky Kesa - 4 years, 11 months ago

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