
Upper Bound
S = {x  x in R , x>=0, x^2 < c}
a. Show that c+1 is an upper bound for S and therefore, by completeness axiom, S has a least upper bound that we denote by b.
b. Show that if b^2>c, we can show choose a suitably positive number r such that br is also an upper bound for S, thus contradicting the choice of b as an upper bound.
c. If b^2< r, then we can choose a suitable positive number r such that b+r belongs to S, thus contradicting the choice of b as an upper bound of S.
All I could do here was to show that
if c<=1, then x<1<1+c. So 1+c is an upper bound for S
and
if c>1, x<Sqrt(c)<c<1+c. So 1+c is an upper bound for S.
How is the rest of part(s) proved?!

b) If for some positive number d.
Now
Note that if , then we have .
Since , there exists a positive r such that .

Suppose that then . WHY?
What is wrong with that picture?
How does that contradiction show that is an upper bound on ?