Solutions to the Homework # 3

1)  As the set of natural numbers is not bounded above (otherwise if 
M is the least upperbound and n is a natural number > M - 1/2, then 
n+1 is a natural number > M), there is a natural number n > y/x.  As 
x > 0, this is equivqlent to nx > y.

2)  Let w = 2/(y-x) and choose a natural number n > w.  Then 10^n > n 
so, 1/10^n < (y-x)/2.  There is an integer m such that m/10^n < x. 
This is clear if x > 0 as we may take m = 0.  If x < 0 and l is a 
natural number such that l/10^n > -x (possible by problem 1), then 
-l/10^n < x.  Let k be the smallest integer such that k/10^n > x. 
Then k/10^n < y for otherwise (k-1)/10^n > x too.  We take z to be 
k/10^n.

3) As x approaches + infinity log(x) aproaches + infinity too, so 
7/log(x) approaches zero and (1 + 7/log(x)pi approaches pi.  Hence 
its cosine approaches - 1.  Since cosine is always greater than or 
equal to -1, -1 is the greatest lower bound of S.