Why do we need an m between -m2 and m1 to prove Theorem 1.20?

[Khichdi lover]

https://www.physicsforums.com/showthread.php?t=253563

There is this post on this theorem. But the issue I wanted to discuss isn't discussed there fully(so far as I could follow the thread). So I am creating this thread for that issue, so as to avoid necro-post in that thread.

Pg 9

If x in R, y in R, and x<y, then there exists a p in Q such that x<p<y.

Proof:
Since x<y, we have y-x>0, and the archimedean property furnishes a positive integer n such that
n(y-x)>1.

Apply the archimedean property again, to obtain positive integers m₁ and m₂ such that m₁>nx, m₂>-nx. Then

-m₂<nx<m₁.

Hence there is an integer m ( with -m₂<= m <= m₁) such that

m-1 <= nx < m

If we combine these inequalities, we obtain

nx<m<= 1 + nx < ny.

Since n>0, it follows that

x < m/n < y. This proves (b) with p = m/n.

My doubt is regarding m. Why do we have to find an m which lies between -m₂ and m₁.

Why can't we directly say that there exists an m₁ which is greater than nx (by Archimedian property).
And then go on to say that there would be an m less than or equal to m₁ which is such that nx lies between m-1 and m.Why do we have to show that -m₂ is less than nx. Do we have to establish that nx lies in an interval which has both a lower and upper bound(-m₂ and m₁ respectively). Can't we simply establish that nx lies in an interval which has an upper bound ( i.e m₁) and then go on to show existence of m ? Why do we need m₂

*I am sorry if I am assuming anything which is not obvious and hence needed to be proven. I am new to analysis , hence I am not too sure of what is obvious and what is not. *

He might have needed the m1 and m2 because we cannot just pick any number. We therefore used archimedian property to find m1 and m2, and then again to find a m in between them but a little to the "right" of nx.
.

is the above , the correct reason ? Still I am having difficulty understanding the need for -m2.

 

[micromass]

The thing is that we have to construct m such that

m-1\leq nx&lt;m

For this, we choose

m=\min\{a\in \mathbb{N}~\vert~nx&lt;a\}

But in order for that minimum to exist, we need the set \{a\in \mathbb{N}~\vert~nx&lt;a\} to have a lower bound. This is exactly what -m_2 provides...

 

[Khichdi lover]

ok got it . Thanks.

But 1 doubt still remains though .
in this set you pointed out -

m=\min\{a\in \mathbb{N}~\vert~nx&lt;a\}

shouldn't 'a' belong to the set of integers instead of the set of naturals.
As any set of naturals will have a minimum. But any set of integers may not have a minimum.

 

[micromass]

ok got it . Thanks.

But 1 doubt still remains though .
in this set you pointed out - {a\in \mathbb{N}~\vert~nx&lt;a\} ,

shouldn't 'a' belong to the set of integers instead of the set of naturals.
As any set of naturals will have a minimum. But any set of integers may not have a minimum.

Yes, of course. It needs to be the integers

 

[Khichdi lover]

Thanks. Sorry didn't get the latex code correct.
DOUBT resolved!