Fractional Parts Proof

Question

OK, here are two questions out of Nathanon's Additive Number Theory from the section on fractional parts ($\S$4.4).

I think I'm missing something. I don't understand what there is to prove? Let $x\in \mathbb{R}$ and let

$$||x||:= \min(|n-x|:n \in \mathbb{Z}) $$

So, its just the distance from $x$ to the nearest integer. Questions 1 and 2 of $\S4.4$ are to show:

$$||\alpha +\beta|| \le ||\alpha||+||\beta||$$ and this little thing: $$||x|| = ||-x|| = ||x+n||, \forall n \in \mathbb{Z}$$ I've seen similar questions before on a course in ele number theory, and didn't do so well on them. As stated above, I don't know what there is to prove, or how I would do it. Do I actually need to go back to the definition of $||x||$? Thanks for any help on this..

Answer 1

Yes, as simple as it sounds: to show a theorem about $\Vert\cdot\Vert$, it is very useful to have a look at its definition. After all, the defining properties are a priori all you know about it.

Let $\alpha, \beta$ be real numbers. By definition of $\Vert\cdot\Vert$, there exists an integer $n\in \mathbb Z$ such that $\Vert\alpha\Vert = |n-\alpha|$. Similarly, there exists an integer $m\in\mathbb Z$ such that $\Vert\beta\Vert = |m-\beta|$. Since $n+m\in \mathbb Z$ and we take the minimum over all elements of $\mathbb Z$, surely $$\Vert\alpha+\beta\Vert\le|(\alpha+\beta)-(n+m)|\le |\alpha-n|+|\beta-m|=\Vert\alpha\Vert + \Vert\beta\Vert.$$

See if you can show the other thing yourself (hint: it suffices to show $\Vert -x\Vert\le \Vert x\Vert$ and $\Vert x\Vert\le \Vert x+n\Vert$ - why?).

Answer 2

For the first part:

Suppose the nearest integer to $\alpha$ is $m$ and the nearest integer to $\beta$ is $n$. Then \begin{align*} \| \alpha + \beta \| & \leq | \alpha + \beta - (m + n) | \\ &= |\alpha - m + \beta - n| \\ &\leq |\alpha - m| + |\beta - n| \\ &\leq \| \alpha \| + \| \beta \|. \end{align*}

Answer 3

Let $m$ be an integer then $$||x+m|| = \min(|n-x-m|:n \in \mathbb{Z}) = \min(|(n-m)-x|:n-m \in \mathbb{Z}) =||x||.$$

Answer 4

Yes, you do need to go back to the definition of $\|x\|$. To prove that $\|\alpha+\beta\|\le\|\alpha\|+\|\beta\|$ for all $\alpha,\beta\in\Bbb R$, for instance, you have to show that the integer closest to $\alpha+\beta$ cannot be larger than the sum of the integers closest to $\alpha$ and to $\beta$. Here’s one way in which you might approach the task.

Let $m$ and $n$ be the integers closest to $\alpha$ and $\beta$, respectively. (If either $\alpha$ or $\beta$ is halfway between two consecutive integers, you can use either of those integers.) Then $m+n$ is at least a reasonable candidate for the integer nearest to $\alpha+\beta$, so let’s have a look at the distance from $\alpha+\beta$ to $m+n$:

$$\begin{align*} |\alpha+\beta-(m+n)|&=|(\alpha-m)+(\beta-n)|\\ &\le|\alpha-m|+|\beta-n|\\ &=\|\alpha\|+\|\beta\|\;. \end{align*}$$

We wanted to show that $\|\alpha+\beta\|\le\|\alpha|+\|\beta\|$, so if we can show that $$\|\alpha+\beta\|\le|\alpha+\beta-(m+n)|\;,$$ we’re done. But this is immediate from the definition of $\|\cdot\|$:

$$\begin{align*} \|\alpha+\beta\|&=\min\{|\alpha+\beta-k:k\in\Bbb Z\}\\ &\le|\alpha+\beta-(m+n)|\;, \end{align*}$$

since the minimum of a set is always less than or equal to each member of the set.

Showing that $\|x\|=\|x+n\|$ for each $n\in\Bbb Z$ is easier, because it’s easier to come by the right intuition. Let $m=\lfloor x\rfloor$, so that $m\le x<m+1$; then the integer nearest to $x$ is either $m$ or $m+1$, unless $x$ is halfway between them, in which case it doesn’t matter which we choose, the distance from $x$ is $\frac12$. But then clearly $m+n\le x+n<m+n+1$, and the distances from $x+n$ to $m+n$ and to $m+n+1$ are identical to the distances from $x$ to $m$ and to $m+1$: we’ve simply translated the picture $n$ units. Thus, either $x\le m+\frac12$, and $$\|x+n\|=(x+n)-(m+n)=x-m=\|x\|\;,$$ or $x>m+\frac12$, and

$$\|x+n\|=(m+n+1)-(x+n)=(m+1)-x=\|x\|\;.$$

(There are more efficient ways to say this; I’m concentrating on how you might actually discover correct arguments.)

Finally, what about $\|-x\|=\|x\|$? Once again let $m=\lfloor x\rfloor$, so that $m\le x<m+1$. Observe that $-(m+1)<-x\le-m$, so $-m$ and $-(m+1)$ are the two integers nearest $-x$. Moreover, the distance from $-x$ to $-m$ is the same as the distance from $x$ to $m$, and the distance from $-x$ to $-(m+1)$ is the same as the distance from $x$ to $m+1$. (I’ll leave the algebraic verification of that to you.) Can you now finish the argument to show that $\|-x\|=\|x\|$?