# Difficulty in understanding corollary of Archimedean property.

Question

prove that$$\forall x \in R \exists n\in Z$$ such that $$n \leq x < n+1$$ (such $$n$$ is unique)

My Attempt The prove given in my book(see the corollary 5) is very brief.

I did not understand what it has to do with Archimedean property. As Archimedean property is established upon completeness I would rather try to prove it from that.

Consider $$A=\{m|m Now a is bounded hence sup exists(say$$\alpha$$. So $$\forall \epsilon>0 \exists a\in A$$ such that $$\alpha-\epsilon < a \leq \alpha$$ But taking $$\epsilon=1$$ does not really prove the required statement. We have to show that $$\alpha$$ and $$\alpha-1$$ to be integers. So this approach failed.

My book's proof Proof consists of two parts

1)To show $$x \geq n$$

2) To show $$n+1>x$$

It proves (1) with constructing a set $$\{m:m and says that it is bounded above so it has suprema say $$n,n\in Z$$ thus $$x\geq n$$.

2) To prove (2) it does not give any explanation. I think it has been done using Archimedean property ie.

$$\forall x\in R \exists m\in Z$$ such that $$m>x$$

But my question is how to prove $$m=n+1$$ ?

I need following help

1) Please can you write the second part of proof given in my book in a detailed or explanatory manner ?

2) I want to know whether my approach was reasonable or not.

Assume $$0 \leq x.$$
By the Archimedian property, exists integer $$n$$ with $$x < n$$.
As the nonnegative integers are well ordered, there is a least integer$$k$$with $$x < k.$$ Thus $$k - 1 \leq x < k.$$

If $$x < 0$$, then $$0 < -x$$ and exists integer $$k$$ with $$k - 1 \leq -x < k.$$ Thus $$-k < x \leq -k + 1.$$ If $$x = 1 - k,$$ then $$1 - k \leq x < 2 - k.$$ If $$x < 1 - k, then -k \leq x < 1 - k.$$

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• Can you elaborate how in first case $x \geq 0$( I got the $n>x$ from archi property) but how $k-1 \leq x$ ? Please explain a bit. – M Desmond Apr 8 '19 at 12:48
• If $k-1>x$ then $k$ would not be the least integer greater than $x$ – J. W. Tanner Apr 8 '19 at 13:15

(1). The order-completeness of $$\Bbb R$$ implies that $$\forall x\in \Bbb R\,\exists n\in \Bbb Z^+\;(x< n).$$

Proof: By contradiction suppose $$x\in \Bbb R$$ and $$\forall n\in \Bbb Z^+ \,(x\ge n).$$ Let $$U$$ be the set of upper bounds for $$\Bbb Z^+.$$ For any $$y\in U$$ we have $$\forall n\in \Bbb Z^+ \,( y\ge 2n),$$ which implies $$\forall n\in \Bbb Z^+\,(y/2\ge n),$$ which implies $$y/2 \in U.$$ But $$y\in U\implies y\ge 1> 0 \implies y>y/2\in U,$$ so $$y\ne \min U.$$

That is, the non-empty set $$\Bbb Z^+,$$ which has an upper bound $$x$$, does not have a minimum (least) upper bound. This contradicts completeness.

(2). It is immediate from (1) that $$\forall x\in \Bbb R\,\exists n\in \Bbb Z^+\,(x\le n).$$

(3). If $$0\le x\in \Bbb R$$ then by (1) and by the well-ordering of $$\Bbb Z^+$$ that $$\{n\in \Bbb Z^+: x has a least element $$n_0.$$ Since $$n_0> n_0-1\ge 0$$ we have either $$n_0-1=0\le x$$ or (by definition of $$n_0$$) that $$n_0-1\in \Bbb Z^+$$ and $$x\ge n_0-1.$$ In either case we have $$n_0-1\in \Bbb Z$$ and $$n_0-1\le x

(4). If $$0>x\in \Bbb R$$ then $$0<-x\in \Bbb R^+.$$ By (2) and by the well-ordering of $$\Bbb Z^+ ,$$ the set $$\{n\in \Bbb Z^+: -x\le n\}$$ has a least element $$n_1.$$ Since $$n_1>n_1-1\ge 0$$ we have either $$n_1-1=0<-x$$ or (by definition of $$n_1$$) that $$n_1-1\in \Bbb Z^+$$ and $$-x>n_1-1.$$ In either case we have $$n_1-1<-x\le n_1 ,$$ so $$-n_1-1\in \Bbb Z$$ and $$-n_1-1\le x< -n_1.$$

Remarks: In an ordered field $$R$$ the interaction between the order and the arithmetic is (by definition) that $$\forall x,y,z\in R\,(x>y\implies x+z>y+z)$$ and $$\forall x,y,z\in R\, ((x>y\land z>0)\implies xz>yz).$$ Any ordered field $$R$$ can be extended to a larger ordered field $$R^*$$ which has members that are larger than any member of $$R.$$ But $$R^*$$ cannot be order-complete. And if $$x\in R^*$$ is greater than every $$r\in R$$ then $$0<1/x for every $$s\in R^+.$$ A lack of a rigorous definition of $$\Bbb R$$ resulted in about 3 centuries of endless debate from the time of Galileo (well before Newton) to the early 1800's, about infinitesimals ("indivisibles") among mathematicians, philosophers, and theologians.

• I suspect it might be shorter to prove that $\forall x\in \Bbb R\,\exists n\in \Bbb Z \,(x\in [n,n+1]).$ So if $n+1\ne x\in [n,n+1]$ then $x\in [n,n+1)$ while if $n+1=x$ then $x\in [n+1,n+2).$ – DanielWainfleet Apr 8 '19 at 23:35