Homework Help: Need help with introductory analysis proof

1. Sep 14, 2009

michael.wes

The question is in Wade's "Introduction to Analysis" 3rd. edition.

Question 1.3.4
Prove that for all real numbers 'a' and natural numbers 'n' there exists a rational number r_n such that |a - r_n| < 1/n.

(my) Proof:

|a - r_n| < 1/n
iff -1/n < a-r_n < 1/n
iff -1/n - a < -r_n < 1/n - a
iff a - 1/n < r_n < a + 1/n

Certainly, a - 1/n < a + 1/n, and both are real numbers by closure properties of R. Since I already proved that the rational numbers are dense in R, there exists a rational number between a - 1/n and a + 1/n, and we are done.//

Is this a valid proof or am I missing something? If it is correct, it seems that this is no more than an alternate statement of the density of Q in R, but I sense that I could be missing a subtler proof or point. Any help is appreciated!

2. Sep 14, 2009

HallsofIvy

Well, one problem is that you are starting with "|a-r_n|< 1/n" which is what you want to prove. You can do that if it is clear that everything you do can be reversed. That is, if this works (I'll let you decide that), then a better proof would be

Given a real number, a, and a natural number, n, since n> 0, 1/n> 0 so a+ 1/n> a and -1/n< 0 so a- 1/n< a. That is, a-1/n< a+ 1/n.

Now, do you have the theorem that "between any two distinct real number there exist another real number"? If so that's exactly what you would need- let r be that number.

3. Sep 14, 2009

michael.wes

That's exactly the answer I needed; and yes, I do have the result. Thanks HallsofIvy!