Math 42001, Homework Set 3, Solutions
Problems 1.5; 3, 4, 7, 10, 1.6; 20, 25, 26
September 20, 2006
p. 28, #3 Show that (ma; mb) = m (a; b) if m > 0:
Solution. Let c = (ma; mb) and d = (a; b). Then d j a and d j b: Consequently, md j ma and md j mb
and thus md j c. Choose integers x; y such that d = ax + by. Then md = amx + bmy and since
c j am; c j bm, we conclude that c j md. Since c; d; and m are positive, we conclude that c = md and
we are done.
p. 28, #4 Show that if a j m and b j m and (a; b) = 1 then (ab) j m.
Solution. Since a j m and b j m; there are integers u; v such that au = m = bv. As (a; b) = 1; there
is integers x; y such that ax + by = 1. Thus
m = axm + bym = axbv + byau = (xv + yu) ab
and so (ab) j m.
p. 28, #7 De…ne the least common multiple of positive integers m and n to be the smallest positive integer v
such that both m j v and n j v. Show that v =
mn
(m;n) :
Solution. Let d = (m; n). Since m j v and n j v; choose positive integers s and t such that ms = v = nt
and integers x and y such that mx + ny = d. Then, vd = vmx + vny = ntmx + msny = (tx + sy) mn
and so (mn) j (vd). On the other hand, choose positive integers k and l such that m = dk and n = dl.
Now, use the Division algorithm to write dkl = vq + r for some integers q and r with 0
Then r = dkl
vq = ml
msq = m (l
sq) and r = dkl
vq = nk
ntq = n (k
r < v.
tq). Hence m j r
and n j r and since r < v, it follows that r = 0. Therefore dkl = vq and so mn = d2 kl = vdq, forcing
(vd) j (mn) and the result is established.
p. 28, #10 To check that a given integer n > 1 is a prime, prove that it is enough to show that n is not divisible
p
n.
by any prime p with p
1
p
Solution. Equivalently, show that n is composite i¤ there is a prime p
n such that p j n.
(=)) : Suppose that n is composite. Let p be the smallest prime that divides n. Then n = pk for
some positive integer k > 1. Let q be the smallest prime that divides k. Then q j n and so p q.
p
In particular p2 pq pk = n and thus p
n.
p
((=) : Now suppose that there is a prime p
n such that p j n. Then n = pk for some positive
p
p
integer k. Hence n = pk k n forcing k
n p > 1. Thus n is the product of two integers
greater than one and as such, composite.
2
2
p. 38, #20 Prove that jz + wj + jz
2
2
wj = 2 jzj + jwj
.
Proof.
2
2
jz + wj + jz
wj = (z + w) (z + w) + (z
w) (z
w)
= (z + w) (z + w) + (z
w) (z
w)
= zz + zw + wz + ww + zz
2
2
2
zw
wz + ww
2
= jzj + jwj + jzj + jwj
2
2
= 2 jzj + jwj
p. 39, #25 The complex number
has order n and
k
Proof. Assume that
is said to have order n
1 if
r
= 1 and as 0
n
m
= 1 and
6= 1 for 0 < m < n. Show that if
= 1, where k > 0, then n j k.
has order n and
k
= 1, where k > 0. By the Division Algorithm, there are
positive integers q; r such that k = nq + r and 0
So,
:
r < n. Now 1 =
k
=
nq+r
=(
n q
)
r
= 1q
r
=
r
.
r < n, we conclude that r = 0. Thus k = nq and so n j k.
p. 39, #26 Find all complex numbers
Solution. First note that if
having order n. (These are the primitive nth roots of unity.)
n
= cos ' + i sin ' for some 0
n
= 1 then j j = 1 and so j j = 1. Thus writing in polar form we obtain
' < 2 . Thus by De Moivre’s theorem 1 =
n
= cos n' + i sin n'.
By solving the trigonometric equation cos n' = 1 on the interval [0; 2 ) we obtain the solution set
2k
n
n 1
.
k=0
That is
= cos 2kn + i sin 2kn for k = 0; 1; : : : ; n
n? Of course not. We claim that the order of
(=)) : Suppose that the order of
0 < m < n we have that
m
1. Do all these numbers have order
is n i¤ (n; k) = 1:
= cos 2kn + i sin 2kn for some 0
k<n
1 is n. Then for any
= cos 2km
+ i sin 2km
6= 1 which implies that cos 2km
6= 1 and
n
n
n
thus n 6 j km. Let d = (n; k) and write n = dx and k = dy for some positive integers x and y.
2
Observe that kx = dyx = ny and so n j kx. Since 0 < x
n and n 6 j km for all 0 < m < n, we
conclude that x = n and so d = 1.
((=) : Conversely, suppose that (n; k) = 1 and suppose that
m > 0 (to show that m
that n j m. Hence, m
m
= cos 2km
+ i sin 2km
= 1 for some
n
n
= 1 and thus n j km. As (n; k) = 1, we conclude
n). Then cos 2km
n
n and we are done.
3