Prove that every number can be written in following form. Let $$\alpha>0 $$ Prove that every number x can be written in following from $$x=k\alpha +x_1$$ where k is an integer, and $0 \le x_1 < \alpha$.
I have tried using archimedianty by using the fact that every rational is between some consecutive integers,then I let k be negative or positive  one and let alpha be absolute x(this is for when x is rational).I am not sure if that works.If someone could supply proper proof or a hint I would be thankful
Reference: Spivak Calculus chapter 8 exercise 10
 A: Consider the set $A=\{n\in\mathbb Z\mid n\alpha \le x\}$. It is bounded from above by $x/\alpha$, and is not empty (since $n\alpha\to-\infty$ when $n\to-\infty$, so for sufficiently large negative $n$, $n\alpha$ will be less than $x$).
A nonempty set of integers that is bounded above has a maximal element. Our $k$ is the maximal element of $A$.
Now let $x_1=x-\alpha k$. Since $k\in A$ we have $k\alpha\le x$ and so $x_1\ge 0$.
On the other hand, if $x_1\ge \alpha$ then $(k+1)\alpha\le x$, and thus $k+1$ would have been in $A$, contradicting the fact that $k$ was the maximal element in $A$.
Therefore $x=k\alpha+x_1$ with $k\in\mathbb Z$ and $0\le x_1 <\alpha$.

Exercise: There's an appeal to Archimedes' axiom hidden somewhere in this argument. Find it!

If you know that every real $y$ is in $[k,k+1)$ for some $k\in\mathbb Z$, you can proceed a little quicker by setting $y=x/\alpha$ and getting $k$ from that.
A: Proof: Fix $\alpha = 2$ All integers are either even or odd: If this integer is even they take the form $k \alpha$ and if they are odd they take $k\alpha + 1$. 
We restrict $x_1 \in [0,1)$.
From this, any real number can be obtained by adding some $x_1$ to $k \alpha$ or $k\alpha +1$. 
Clearly $\alpha> x_1\geq0$
