Math 2112 Solutions
Assignment 1
3.4.10 a. 207div4 b. 207mod4.
Since 207 = 4(51) + 3, then 207div4 = 51 and 207mod4 = 3.
3.4.22 Use the result of exercise 20 to prove that the square of any
integer has the form 3k or 3k + 1 for some integer k.
Proof: Let n be any integer. From excercise 20, we know that either n = 3a or
n = 3a + 1 or n = 3a + 2 for some integer a.
Case 1: Suppose n = 3a. Then n2 = (3a)2 = 9a2 = 3(3a2 ). Thus
n = 3k for some integer k.
Case 2: Suppose n = 3a + 1. Then n2 = (3a + 1)2 = 9a2 + 6a + 1 =
3(3a + 2a) + 1. Thus n = 3k + 1 for some integer k.
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Case 3: Suppose n = 3a + 2. Then n2 = (3a + 2)2 = 9a2 + 12a + 4 =
3(3a + 4a + 1) + 1. Thus n = 3k + 1 for some integer k.
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Therefore, the square of any integer has the form 3k or 3k + 1 for some integer
k.
3.6.22 If a and b are rational numbers, b 6= 0, and r is an irrational
number, then a + br is irrational.
Proof: Assume by way of contradiction that a + br is rational. Since a and b
are rational, then there are integers m, n, p, s, t, u, s, t, u 6= 0 such that a = m
s ,
b = nt and a + br = up . Therefore,
a + br
m n
+ r
s
t
n
r
t
r
p
u
p
u
sp − mu
su
t(sp − mu)
nsu
=
=
=
=
Since b =6= 0, n 6= 0. Thus since n, s, u 6= 0, we have written r as an integer
over a nonzero integer, implying r is rational, a contradiction. Therefore a + br
is irrational.
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3.6.24 If a, b and c are integers and a2 + b2 = c2 , must at least one of a
and b be even? Justify your answer.
Yes, one of a and b must be even. Assume by way of contradiction that a and
b were both odd. Thus a = 2k + 1 and b = 2j + 1 for some integers k and j.
Since and odd times an odd gives you an odd number, and the sum of two odd
numbers is even, then c2 must be even. But c2 is even iff c is even. Thus c = 2m
for some integer m and hence c2 = 4m2 . On the other hand,
a2 +b2 = (2k+1)2 +(2j+1)2 = 4k 2 +4k+1+4j 2 +4j+1 = 4(k 2 +k+j 2 +j)+2 = c2 .
Thus, we may write c2 = 4n + 2 for some integer n. This is a contradiction
since 4n + 2 6= 4m when m and n are both integers. Thus, one of a and b must
be even.
3.7.11 If an integer is a perfect square, then its cube root is irrational.
Counterexample: Let n = 64. Then n is a perfect square (8 ∗ 8), but the cube
root of 64 is 4, which is rational.
3.7.16 Prove that every integer greater than 11 is a sum of two composite numbers.
Proof: Let n be and interger greater than 11. Note that n = (n − 4) + 4,
n = (n − 6) + 6, and n = (n − 8) + 8. Since 4,6 and 8 are all composite, if
we can show that one of n − 4, n − 6 and n − 8 is composite, then one of the
three previous equations must allow us to write n as the sum of two composite
numbers. Note also that n − 6, n − 7 and n − 8 are three consecutive numbers,
and thus one of them is always divisible by three. If n − 6 or n − 8 is divisible by three, then since n is greater than 11, the number which is divisible by
three must be composite. Suppose then, that n − 7 is divisible by three. But
then n − 4 = (n − 7) + 3, and hence n − 4 has the same remainder as n − 7
when dividing by 3, namely zero remainder. Thus n − 4 must be composite, as
required.
3.7.24 Prove that there is at most one real number b with the property that br = r for all real numbers r.
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Proof: Suppose that there exists real number a and b such that for all real
numbers r, we have ar = r and br = r. Consider ab. Since a is a multiplicative
identity, we know that ab = b. Also, since b is a multiplicative identity, we know
that ab = a. Thus a = ab = b, and hence a = b. Thus, there is at most one real
number which is a multiplicative identity.
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