MATH 310C
SPRING 2007
EXAM II SOLUTIONS
1. (a) i, iii, and iv are TRUE. ii is FALSE.
(b) i. Proof: Suppose fand gare injective and h(x1) = h(x2). Then g(f(x1)) = g(f(x2)).
Since gis injective, this means that f(x1) = f(x2) and, since fis injective, this
means x1=x2. Thus, his injective.
iii. Proof: Suppose that his surjective and choose c∈C. Since his surjective, there
is an ain Asuch that h(a) = c. Then g(f(a)) = cand f(a) is the element bin B
such that g(b) = c. Thus, gis onto (surjective).
iv. Proof: Suppose that gand hare bijective. Since g:B→Cis a bijection, g−1:C→
Bexists and is also a bijection. For each a∈A,h(a) = g(f(a)). This means that
f(a) = g−1(h(a)). In other words, f=g−1◦h. Since g−1and hare both bijections
and the composition of two bijections is a bijection, fis also a bijection.
(c) Counterexample: Let A={1,2},B={3,4,5}and C={α, β}. Let f(1) = 3
and f(2) = 4. Let g(3) = α,g(4) = βand g(5) = β. Then h(1) = g(3) = αand
h(2) = g(4) = β. So, his one-to-one (since hmaps each element of Ato a different
element of C), but gis not (since g(3) = g(5) but 3 6= 5).
2. No! An injective function ffrom an infinite set to itself is not necessarily surjective. Consider
f:N→Ndefined by f(n)=2n. This function is injective since f(n1) = f(n2) implies that
2n1= 2n2, which implies that n1=n2. But fis not surjective since there is no nin Nsuch
that f(n) = 5, for example.
(NOTE: If Ais finite and f:A→Ais injective, then fmust also be surjective.)
3. Proof: Since d1= gcd(a, c) and d2= gcd(b, c), there exist integers m1,n1,m2, and n2such
that
m1a+n1c=d1and m2b+n2c=d2.
Then
d1d2= (m1a+n1c)(m2b+n2c)
=m1m2ab +n1m2bc +m1n2ac +n1n2c2.
If c|ab, then cdivides every term on the right-hand side of this equation and, hence, c|d1d2.
4. If a|b, then there exists an integer ksuch that b=ak. If b|a, then there exists an integer k0
such that a=bk0. Combining these two equations gives: a=bk0=akk0. Since a6= 0, this
means kk0= 1. The only integers that divide 1 are 1 and −1. Moreover, since kk0= 1, kand
k0must have the same sign. That is, either k=k0= 1 (in which case, a=b) or k=k0=−1
(in which case, a=−b). So, |a|=|b|.
5. First, find a particular solution using the Euclidean algorithm or by inspection. One such
solution is:
8(−25) + 10(25) = 50.
