Proof of Rudin Theorem 1.20 (b)

Asked Jun 17 '19 at 11:20

Active Jun 17 '19 at 12:18

Viewed 298 times

This is different to the linked question as the answers to that question do not address the actual process of finding m1 and m2.

The theorem states:

(a) If x ∈ R, y ∈ R and x >0, then there exist a positive integer n such that nx >y.

(b) If x ∈ R, y ∈ R, and x < y, then there exists a p ∈ Q such that x < p < y.

I understand part (a) of the theorem. Proof of part (b) opens as follows:

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

This part I understand. But I'm unsure how this next part is done:

Apply (a) again, to obtain positive integers m1 and m2 such that m1 > nx, m2 > -nx. Then -m2 < nx < m1

How are the integers m1 and m2 obtained? I know that it involves the same process used in part a, I just can't figure out the steps.

edited Jun 12 '20 at 10:38

Community

asked Jun 17 '19 at 11:20

yerman

1

Are we all supposed to know what is the statement of that theorem? – José Carlos Santos Jun 17 '19 at 11:24
I'll add it now – yerman Jun 17 '19 at 11:25
Use LaTeX please. – Michael Rozenberg Jun 17 '19 at 11:26
Thanks for marking it as a duplicate... The answer you linked skips the part that I'm asking for clarification on. – yerman Jun 17 '19 at 11:29
And sorry @MichaelRozenberg but I don't know how to use it. Beginner here. – yerman Jun 17 '19 at 11:36
For every real number $r$, you can find an integer $n_r$ such that $n_r>r$. $nx$ is a real number, so there exists an integer $m_1$ such that $m_1>nx$. $-nx$ is a real number, so there exists an integer $m_2$ such that $m_2>-nx$. Multpily the last inequality by $-1$. Combine the two inequalities, to get $-m_2<nx<m_1$. – uniquesolution Jun 17 '19 at 11:53
Thanks @uniquesolution, that makes more sense now. – yerman Jun 17 '19 at 11:56
@JoséCarlosSantos - Regarding your question: No, not all of us. However, those of us who do not, should not rush to close the question down. No need to be mean to new users. – uniquesolution Jun 17 '19 at 12:03
@uniquesolution I closed the question for one and only one reason: because it is a duplicate. – José Carlos Santos Jun 17 '19 at 12:09
@JoséCarlosSantos I have explained above how it is not a duplicate. – yerman Jun 17 '19 at 12:09
@rory_c Now it is a duplicate. The question that I've put a link to contains an answer to your question. – José Carlos Santos Jun 17 '19 at 12:19

Proof of Rudin Theorem 1.20 (b)

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