Proof: For every $x \in \mathbb R\,\exists n\in\mathbb Z,\,r\in[0,1):x=n+r.$ I still do not understand how to approach proofs.  Any help would be appreciated.

For every real number $x$, there exists an integer $n$ and a real number $r\in [0,1)$ such that $x = n + r$.
Hint:  there is an integer $x$ call the floor of $x$, such that $x-1<\lfloor x\rfloor\leq x$.

 A: HINT:
To see why this is true, note that if $A\subseteq\Bbb Z$ which is bounded from above, then it has a maximal element. Now if $x\in\Bbb R$ consider $A=\{k\in\Bbb Z\mid k\leq x\}$.
A: Hint: $\exists n\in\mathbb{N}$ such that $n\le x<n+1$ (why?). Note that $0\le x-n<1$
If you want to include the floor function, note that this $n=\lfloor x \rfloor$
A: Let $x\in \mathbb R$, $n\in\mathbb Z$. By definition of the floor of $x$: $x-1<\lfloor x\rfloor\leq x\Rightarrow\lfloor x\rfloor+1> x> \lfloor x\rfloor$. Let $r\in\mathbb R$ be such that $0\geq r=x-\lfloor x\rfloor>1$.
A: Some of the answers posted earlier take the definition of $\lfloor x \rfloor$ for granted, so in my opinion they circumvent the actual answer, which is to prove the existence of $\lfloor x \rfloor$. 
To this end fix a real number $x$. I would take as obvious that there are integers $m,k$ with $m<x<k$ (proving this will require to dig too deep in the definition or real numbers). Look at $m,m+1,m+2,...,k$. Since $m<x$ but $k>x$ we could walk up this finite sequence till we reach the largest number $n$ with $n\le x$. (Clearly such a number $n$ exists, and $m\le n<k$.) Then $n+1>x$, for if $n+1\le x$ then $n$ would not have been the largest number in the sequence being $\le x$. Hence $0≤x−n<1$. We could take $r=x-n$, and this argument simultaneously defines $\lfloor x \rfloor = n$. 
