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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.

My doubt is regarding m. Why do we have to find an m which lies between -m Unassuming said: ↑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 m_{1}and m_{2}such that m_{1}>nx, m_{2}>-nx. Then

-m_{2}<nx<m_{1}.

Hence there is an integer m ( with -m_{2}<= m <= m_{1}) 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._{2}and m_{1}.

Why can't we directly say that there exists an m_{1}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 m_{1}which is such that nx lies between m-1 and m.Why do we have to show that -m_{2}is less than nx. Do we have to establish that nx lies in an interval which has both a lower and upper bound(-m_{2}and m_{1}respectively). Can't we simply establish that nx lies in an interval which has an upper bound ( i.e m_{1}) and then go on to show existence of m ? Why do we need m_{2}

*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. *

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

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# Rudin Theorem 1.20 doubt

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