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 $A=\{m|m<x,\forall m\in R\}$ 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 $$\{m:m<x,\forall m\in Z\}$$ 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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  • $\begingroup$ 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. $\endgroup$ – M Desmond Apr 8 '19 at 12:48
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    $\begingroup$ If $k-1>x$ then $k$ would not be the least integer greater than $x$ $\endgroup$ – 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 $\{n\in \Bbb Z^+: x<n\}$ 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 $n_0-1\le x<n_0.$

(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 $0<1/x<s$ 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.

  • $\begingroup$ 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 $x\in [n+1,n+2).$ $\endgroup$ – DanielWainfleet Apr 8 '19 at 23:35

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