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I found this statement in my textbook and I wasn't able to prove it:

$$\forall x \in \mathbb{R} ; \exists (k, r) \in \mathbb{Z}\times [0, 1) / x = k + r $$

I tried proving it by contradiction and succeeded in doing it for numbers in $\mathbb{Z}$, and now I am stuck at proving it in $\mathbb{R}- \mathbb{Z}$. I would like if someone could give me a nudge or a hint .Note that this statement is useful when solving equations involving the floor function such as $(\lfloor x\rfloor)^2=\lfloor x^2 \rfloor $

Oh and please this is my first question on the forum, I have a precalculus level, and if I've violated any rule or done something incorrectly let me know because somehow my question got closed.

HP Impact
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    For $x$ not an integer, look at the set of natural numbers greater than $|x|$. This set is non-empty, by the Archimedean property. Since the natural number are well-ordered, there is a minimum $n$ of that set. If $x>0$, you can take $k=n-1, r=x-k$. If $x<0$, you can take $k=-n, r=x-k$. – plop Sep 21 '22 at 20:26
  • I don't understand what it means when you say you succeeded for "doing it" for numbers in Z. Surely to prove something like this you have to start out at the axioms for the construction of the real numbers, so you have only got 1 and 0 and other unspecified numbers, then you have to show that every number consists of $\pm(1+1+1+...+1)$ and a part which is less than one? – Suzu Hirose Sep 21 '22 at 22:32
  • I meant that I was able to prove the statement for all integers because every integer is written in the form x = k + r with r = 0 and k an integer, but I didn't see how I would prove it for non integers, maybe my approach is a bit weird that's why I am asking for suggestions here – HP Impact Sep 21 '22 at 22:47
  • When you're asked to prove something like this, which seems to be a statement of the obvious, it usually means that you're being asked to go back to the axioms for constructing the real numbers and show that they imply that this is true. In other words this "obvious" thing is not an axiom but something which can be derived from the axioms. If it's in your textbook then surely the textbook also discusses the construction of the real numbers. – Suzu Hirose Sep 21 '22 at 22:56

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For every real we have that $[x]\leq\,x\,\leq\,[x]+1$ and clearly $x=[x]+r$ where

$0\leq\,r<1$. The integer part is well defined as the maximum of integers less or equal than $x$.

Because every upper bounded set has a supremum. (Axiom of Completeness or Least Upper bound property).

Since the set is a subset of integers the sup is a max.

Also $x<[x]+1$ because if $[x]+1\leq\,x$ then $[x] +1$ would be an element of the set greater than the max. I think this is a sufficient demonstration.

  • Doesn't this actually assume the result we are trying to prove? – Suzu Hirose Sep 21 '22 at 22:30
  • No, because we don't use that [x] is already defined. We define it ! –  Sep 21 '22 at 22:42
  • Well right there at the start you say "and clearly $x=[x]+r$" so I think you have assumed the result, I don't think this is at all a valid proof. – Suzu Hirose Sep 21 '22 at 22:45
  • I don't understand your point. Do you need a proof that if $N\leq,x,<N+1$ then $x=r+N$ where $0\leq,r<1$?? Just define$,, r=x-N$. –  Sep 21 '22 at 22:54
  • Again, in your comment, you're just assuming the result but you need to show that it follows from the axioms for constructing the reals. – Suzu Hirose Sep 21 '22 at 22:58
  • That is exactly what I did my friend. I used the least upper bound property of the reals! What else do you want? To construct the Real Numbers by Dedekind Cuts or by Cauchy sequences?? This is a question not a thesis! –  Sep 21 '22 at 23:01