# Thread: There are rationals between any two reals? -- Baby Rudin

1. In Principles of Mathematical Analysis 1.20(b), Rudin proves that between any two real numbers x and y, there exists a rational number p. His strategy is to "blow up" x and y by some integer n (generated by the archimedian property) such that ny - nx > 1. Then he attempts to locate another integer m that is between nx and ny, such that m - 1 < nx < m < ny. From here we have

nx < m < ny

and when we multiply each term by (1/n) we can say that

x < m/n < y where p = m/n. Hence the proof is done.

Now my question is as follows: how can Rudin just pick an m such that m - 1 < nx < m < ny?

I certainly understand that we can use the archimedian property to find an m greater than nx, but how can we be SURE that this m is less than ny and even less than 1 unit distance greater than nx?

Here is the proof in full from the text in case it helps:

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.

2.

3. It seems pretty intuitive to me that if then we can find an integer between and . If and is an integer, then the next integer falls on and we can find no such . But if you think about it, the maximum distance from any real number to some integer is 1, so the condition guarantees that we have such an integer, and then it's clear also that .

EDIT: Also, notice the inequalities and in the quoted proof. Due to the in , we can choose if nothing else, and this works. Or, may be closer to than a distance of 1.

4. Originally Posted by gwsinger
Now my question is as follows: how can Rudin just pick an m such that m - 1 < nx < m < ny?
This easily follows from the fact that is a complete ordered field. I do not have the text from which you are working, but I DO know that Rudin works logically from first principles, in general. That is, I should be very surprised if you do not know what a complete ordered field is, or what the definition implies. If so, you should have no real difficulty with this.

If you do, come back and ask again

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