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I am trouble getting this proof and idea behind it. I have taken example of a and b as (3,4). I have problem with understanding the part where author writes " first step is to choose n large enough so that increments of sizes $\frac {1}{n}$ are too close to step over (3,4). Now this will happen if increments are smaller than size of interval, which he writes in next line using archimedian property.

After that he says now choose m to be smallest natural number greater than na. I don't get this.. Hope someone explains more clearly

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Robert Z
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J. Deff
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1 Answers1

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Let $a<b$ (you may ignore the condition that $a\geq 0$). Then $\frac{1}{b-a}>0$ and by the Archimedian Property, there is $n\in\mathbb{N}$ such that $n>\frac{1}{b-a}$, which implies that $nb-na>1$. It remains to show that there is $m\in\mathbb{Z}$ such that $na<m<nb$, that is the interval $(na,nb)$ whose size is greater than 1 contains at least an integer $m$. Now take the largest integer $M\in \mathbb{Z}$ such that $M\leq na$ (see Every non-empty subset of the integers which is bounded above has a largest element.) then $$na<M+1\leq na+1<nb$$ and we can take $m=M+1$.

In other words, we "expand" the real line by a positive integer factor $n$, so that the interval $(a,b)$ becomes $(na,nb)$ with $nb-na>1$. Then some integer $m$ will get trapped in $(na,nb)$.

Robert Z
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  • Is it that the increments of size $\frac{1}{n} $ or $n$ is done to insure that size of interval ($na, nb$ ) is greater than $1$ ? – J. Deff Mar 12 '19 at 07:20
  • Yes, the choice of $n$ is made in order to have that the size of interval $(, )$ is greater than 1. – Robert Z Mar 12 '19 at 07:22
  • Can you explain what the author has done ? i.e increments of size $1\n$ – J. Deff Mar 12 '19 at 07:28
  • The sentence "The first step...the interval $(a,b)$" means "Take $n\in\mathbb{N}$ such that $b-a>\frac{1}{n}$" (i.e. $n>\frac{1}{b-a}$ or $nb-na>1$). I think that the argument is more clear when you consider the "expanded interval" $(na,nb)$. – Robert Z Mar 12 '19 at 07:37
  • so whole idea of proof is that first we increase interval length to greater than 1 and then we automatically have one N in that ??? – J. Deff Mar 12 '19 at 07:40
  • Yes, we "expand" the real line by $n$ so that the interval $(a,b)$ becomes $(na,nb)$ with $nb-na>1$. Then some integer $m$ will get trapped in $(na,nb)$. – Robert Z Mar 12 '19 at 07:43
  • But what does author mean "the consecutive intervals of size 1/n step over (a,b)..whats need for writing this ? – J. Deff Mar 12 '19 at 07:50
  • @J.Deff If $n$ is not large enough then the consecutive intervals have size $1/n$ not small enough that they will "jump" over the interval $(a,b)$. – Robert Z Mar 12 '19 at 08:02
  • I see that, but i am saying that if we are using archimedian property to amplift the inteval then whats the need of doing this ? or how is that "The first step is ..." related to amplification of interval ? – J. Deff Mar 12 '19 at 08:09
  • There is no need of saying that. Just an "explanation" of the reason why we choose $n$ greater than $1/(b-a)$. – Robert Z Mar 12 '19 at 08:26
  • Ok..Thanks..... – J. Deff Mar 12 '19 at 08:29