Vtgyjfy

If $alpha$ is a real number and $n$ is a positive integer, there are integers $a$ and $b$ such that $1leq a leq n$ and $|alpha a - b| < frac{1}{n+1}$.

Here is an attempt of the proof. I'm more or less skeptical of the second case and on whether or not the cases are exhaustive. Ultimately, I am wondering if the proof is alright.

Notation

Let $x in mathbb{Z}$. Then I use $[x]$ to denote the floor of $x$.

A real number $rho$ can be expressed as the sum of its integer $[rho]$ and fractional parts ${rho}$ as $rho = [rho] + {rho}$ where $0 leq {rho} < 1$.

Proof

Consider the $n+2$ fractional parts ${jalpha}$ where $j = 0,cdots,n+1$.

Then partition the interval $[0,1)$ into the $n+1$ subintervals

$$Big[frac{k-1}{n+1},frac{k}{n+1}Big)$$

Then each fractional part lies in one of the subintervals.

Since there are $n+2$ numbers but only $n+1$ intervals, the Pigeonhole principle says that at least two of these numbers lie in the same interval. Further, because each interval has length $frac{1}{n+1}$ and does not include its right endpoint, each point in a given interval will be less than $frac{1}{n+1}$.

Hence, there are integers $0 leq p < q leq n+1$ such that $|{qalpha} - {palpha}| < frac{1}{n+1}$.

Upon manipulating the inequality of $p$ and $q$ we get

$$0 leq p leq q-1 leq n$$

First, assume that $p< q-1$. Then $1 leq q-1-pleq n$.

Therefore, let $a = q-1-p$ and $b = [qalpha]-[palpha]-alpha$.

Now, we have

$$begin{align}|aalpha-b| &= |(q-1-p)alpha - ([qalpha]-[palpha]-alpha)| \\ &= |(alpha q - [alpha q]) - (alpha p - [alpha p])| \\ &= |{alpha q} - {alpha p}| \\ &< frac{1}{n+1}end{align}$$

So the given choices for $a$ and $b$ suffice.

Now suppose that $q-1 = p$. Then $1 = q-p$, which satisfies the inequality $1 leq q-p leq n$. Hence, let $a = q-p$ and $b=[alpha q] - [alpha p]$. Then

$$begin{align}|a alpha - b| &= |(q-p)alpha - ([alpha q] - [alpha p])| \\ &= |(alpha q-[alpha q]) - (alpha p - [alpha p])| \\ &= |{alpha q} - {alpha p}| \\ &< frac{1}{n+1}end{align}$$

Thus the given choies for $a$ and $b$ suffice.

Upon exhausting all possibilities, we conclude that one of first $n$ multiples of a real number $alpha$ must be within $frac{1}{n+1}$ of an integer.

My comment is wrong: the problem in your proof is that $b$ must be an integer, and in your first case, where $p < q-1$, you have $b$ to be $([qalpha] - [palpha] - alpha)$, but since $alpha$ is a real number, b is not necessarily an integer.

Here is how I did the proof (it's really similar and only differs in one small aspect):

Consider the $n+2$ numbers: $0, {alpha}, {2alpha}, ..., {nalpha}, 1$.

Now divide the closed interval $[0,1]$ into $n+1$ partitions:

$[0, frac{1}{n+1}), [frac{1}{n+1}, frac{2}{n+1}), ..., [frac{n}{n+1}, 1]$

By the pigeonhole principle, since all $n+2$ numbers fall in the interval $[0,1]$, there must be one of the $n+1$ subintervals that contains two numbers. Now, there are two cases:

Case 1: The two numbers are in the first $n$ subintervals, subinterval $i$, $[frac{i-1}{n+1},frac{i}{n+1})$.

Let $p, q in mathbb{Z}$ such that $0leq p lt q leq n$ without loss of generality. We have $|{qalpha} - {palpha}| lt frac{1}{n+1}$

Thus, by definition of ${}$, we have $|qalpha - lfloor qalpha rfloor - (palpha - lfloor palpha rfloor)| lt frac{1}{n+1}$

Rearranging, we have $|(q-p)alpha - (lfloor qalpha rfloor - lfloor palpha rfloor)| lt frac{1}{n+1}$

Take $a = q-p$ and $b = lfloor qalpha rfloor - lfloor palpha rfloor$, both of which are integers, to satisfy the inequality.

Case 2: The two numbers are in the last subinterval, $[frac{n}{n+1}, 1]$. One of the numbers must be 1. Let the other number be ${palpha}$, for $p in [0,n]$.

Thus, we have $|{palpha} - 1| leq frac{1}{n+1}$. Note that it is less than or equal to because this is the only closed interval.

Simplifying as we did above, we get $|palpha - (lfloor palpha rfloor + 1)| leq frac{1}{n+1}$.

Take $a = p$ and $b = lfloor palpha rfloor + 1$, satisfying the constraints. QED.

Your proof is wrong particularly in the first case:

First, assume that $p<q−1$. Then $1 le q−1−p le n$.

This is clearly wrong, take $p=1$, $q=2$. Then, $q-1-p=0 < 1$.

You can intuitively understand why the proof is wrong by noticing that you can have $a>n$ when $p=0$ and $q=n+1$, which would invalidate the entire proof.