Archimedean property help

[stiffy]

Q: If y is in R, y>0, then there is a positive integer n such that 1/n<y.

A: All I can think of is to use the fact that N is unbounded, thus for any value of y I can creat a smaller fraction with a larger value of n.

Also,

Q: Consider the set A={1/n : n is in R}. Prove that inf(A)=0.

Any help understanding how to do these would be much appreciated. Thank you.

 

[daon]

Yes. If y>0, then 1/y > 0.

You have probably proved the following: For every real number x, here is a unique integer n such that n-1 <= x < n.

Then let n be the positive integer such that n-1 <= 1/y < n. Then (you may need to take this step by step algebraically) 1/n < y. You may use the unboundedness of N to get the same result (if the theorem states such a thing for reals and naturals).

For the second, "obviously" 0 is a lower bound. You need to show that: if L is any lower bound of A, then L <=0. Thus 0 is the greatest lower bound.