- #1

gwsinger

- 18

- 0

**There are rationals between any two reals? -- Rudin 1.20**

In Rudin 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.