Homework 1. Solutions.
m
∈ Q , where m, n ∈ N, and p2 = 12. Then we can
n
assume that m, n do not have common factors. From p2 = 12 get
Chap. 1 #2. Suppose p =
m2 = 12n2 .
(1)
Then m2 is divisible by 3.
We first prove that m is divisible by 3. Indeed, otherwise, either m = 3k + 1 or
m = 3k − 1 for some integer k. In both cases,
m2 = (3k ± 1)2 = 9k 2 ± 6k + 1 = 3(3k 2 ± 2k) + 1,
i.e. m2 is not divisible by 3, a contradiction.
Thus m = 3m1 . Then (1) becomes (3m1 )2 = 12n2 , so 4n2 = 3m21 .
We prove that n2 is divisible by 3. Indeed, otherwise either n2 = 3k + 1 or
n2 = 3k − 1, i.e. n2 = 3k ± 1. Then
4n2 = 12k ± 4 = 3(4k ± 1) ± 1,
i.e. 4n2 is not divisible by 3, a contradiction.
Thus n2 is divisible by 3. As we proved earlier, this implies that n is divisible by
3, which means that m and n have a common factor 3. A contradiction.
Chap. 1 #4.
Let x ∈ E. Then α ≤ x and x ≤ β. Thus, α ≤ β.
Chap. 1 #6. Fix b > 1.
(a) First we note that if m, n are positive integers, then
(bm )n = bmn .
(2)
Indeed, on the left-hand side, product of b by itself m times is repated n times, so
total mn multiplications, which is equal to the right-hand side.
Also, using that b0 = 1 and b−m = b1m for positive integer m, we obtain (2) for all
integer m, n.
Now let m, n, p, q be integers, n > 0, q > 0, and r = m
= pq . Then mq = np.
n
1
1
From Theorem 1.21, there exist real positive numbers α = (bm ) n and β = (bp ) q , i.e.
positive α, β such that αn = bm and β q = bp . Then, using (2),
αnp = (αn )p = (bm )p = bmp = (bp )m = (β q )m = β mq .
Since np = mq, it follows that if np = mq 6= 0, then α = β. If np = mq = 0, then
using that n > 0, q > 0, we obtain m = p = 0, and then αn = bm = b0 = 1, thus
1
1
α = 1. Similarly β = 1. Thus in both cases α = β, i.e. (bm ) n = (bp ) q .
1
(b) Let r = m
, s = pq where m, n, p, q be integers, n > 0, q > 0. Then r + s =
n
mq+np
. Using part (a) and (2) for integers, we have:
nq
1
1
1
1
mq
np
m
p
br+s = (bmq+np ) nq = (bmq bnp ) nq = (bmq ) nq (bnp ) nq = b nq b nq = b n b q = br bs .
(c) First we show that if r > 0 is rational, and b > 1 is real, then br > 1.
where m, n ∈ N. Then, multiplying inequality b > 1 by itself m
Indeed, r = m
n
1
1
times, we obtain bm > 1. It follows that (bm ) n > 1, since if 0 < (bm ) n ≤ 1, then,
multiplying this inequality by itself n times, we obtain bm ≤ 1, a contradiction. Thus,
1
br = bm/n = (bm ) n > 1.
Now we use this to show that
br1 > br2
if r1 , r2 ∈ Q
and r1 > r2 .
(3)
Indeed, then r1 − r2 > 0, then, as we showed above, br1 −r2 > 1. Multiplying both
sides by the positive number br2 and using part (b) of problem, we have
br1 = br1 −r2 br2 > 1 · br2 = br2 ,
i.e. br1 > br2 .
Now, if r ∈ Q, it follows that br is an upper bound of B(r): indeed, if t ∈ Q and
t ≤ r, then bt ≤ br . Also, since br ∈ B(r), it follows that br is the least upper bound
of B(r), i.e. br = sup B(r).
(d) We first show that bx+y ≥ bx by . Let s, r ∈ Q and s ≤ x, r ≤ y. Then t = s + r
satisfies t ∈ Q, t ≤ x + y, so from the definition of bx+y as sup B(x + y), we have
bx+y ≥ bt = bs+r . Then by part (b) of problem, get bx+y ≥ bs+r = bs br . From this,
fixing r and using that br > 0, obtain
bs ≤
That is,
we have
bx+y
br
bx+y
br
for each s ∈ Q, s ≤ x.
is an upper bound of B(x). Since bx is the least upper bound of B(x),
bx+y
.
br
This holds for any r ∈ Q, r ≤ y. Rewriting this as
bx ≤
br ≤
bx+y
bx
for each r ∈ Q, s ≤ y,
x+y
we see that b bx is an upper bound of B(y). Since by is the least upper bound of B(y),
x+y
we have by ≤ b bx , that is bx+y ≥ bx by .
Now we prove that bx+y ≤ bx by . Here√a difficulty is√the following: if x, y are
irrational, and x + y ∈ Q (for example, x = 2 and y = − 2), then bx+y ∈ B(x + y),
and bx+y > bs br for any s ∈ B(x) and r ∈ B(y). Thus we argue as following: If t ∈ Q
2
and t < x + y, then denoting ε = x + y − t, we have ε > 0. Then there exist s, r ∈ Q
with x − 2ε < s < x and y − 2ε < r < y. Moreover, r + s > x + y − ε = t, and we
obtain using (3):
bt < bs+r = bs br ≤ bx by ,
where we used that bs ≤ bx , which holds since bs ∈ B(x), and similarly br ≤ by ; also
bs > 0 and br > 0. The last inequality holds for each rational t < x + y, so
sup C(x + y) ≤ bx by ,
(4)
where, for any real z, C(z) is the set of all numbers br where r is rational and r < z.
Note that C(z) differs from B(z) only by the strict inequality in r < z. Clearly,
C(z) ⊂ B(z), so sup C(z) ≤ sup B(z). It remains to show that for any z ∈ R
sup C(z) = sup B(z),
(5)
then (4) implies bx+y ≤ bx by .
Thus it remains to show (5). Denote a = sup C(z). As we showed, a ≤
sup B(z) = bz . Suppose a < bz . Then we will show that for sufficiently large n ∈ N
1
bz− n > a.
Indeed, (6) can be rewritten as
bz
a
(6)
n
> b.
z
(7)
Since a < bz , then ba > 1. Thus in order to prove (7), we need to show that if c > 1,
then cn > b for sufficiently large n ∈ N. This last fact follows from the inequality
cn − 1 ≥ n(c − 1), which was shown in the proof of Theorem 1.21: indeed, we choose
b
n ≥ c−1
. Now (7) is proved, thus (6) is proved. Now we get a contradiction as
following: Choose any rational r such that z − n1 < r < z. Then br ∈ C(z), and
1
br > bz− n > a by (6). Thus a is not an upper bound of C(z), a contradiction. This
shows (5).
3