MATH 501
FALL 2006
Discussion of Chapter 1, #6
c) First note that if γ < br then γ cannot be an upper bound for B(r) since br ∈ B(r).
Thus it remains to show that br is an upper bound for B(r). If t ∈ Q, t ≤ r then we
can write r = np , t = m
for some p, m ∈ Z, n > 0, m ≤ p. Then (bt )n = bm ≤ bp = (br )n
n
which can only be true if bt ≤ br , as needed.
d) Let r, s ∈ Q, r ≤ x, s ≤ y. Then
br bs = br+s ≤ bx+y
since r + s ≤ x + y. It follows that
br ≤
bx+y
bs
bx ≤
bx+y
bs
bs ≤
bx+y
bx
by ≤
bx+y
bx
and so
From this we get
which implies
or
bx by ≤ bx+y
Next, if bx by < bx+y there must exist t ∈ Q such that t < x + y and bx by < bt . We
can choose rational numbers r, s such that
t−y <r <x s=t−r
Since r < x, s < y we have
bt = br bs ≤ bx by
a contradiction. Thus bx+y = bx by .