MATH 310C
SPRING 2007
SAMPLE EXAM I — SOLUTIONS
1. (a) Choose x∈(A∪C)−B. Then x∈A∪Cbut x /∈B. Since x∈A∪C,x∈Aor
x∈C. If x∈A, then we have that x∈Abut x /∈B, which means that x∈A−B,
which is a subset of (A−B)∪C. If x /∈A, then xmust be in C, which means that
x∈(A−B)∪C. Thus, every element of (A∪C)−Bis also in (A−B)∪C, which
means that (A∪C)−B⊆(A−B)∪C.
(b) As long as B∩Cis non-empty, equality will not hold. One example is A={1,2,3},
B={1,2}, and C={2,3,4}. Then, (A∪C)−B={3,4}but (A−B)∪C={2,3,4}.
2. (a) The negation is: There exists an x∈Rsuch that, for all y∈R,x+y /∈Z.
(b) The negation is: For all x∈Z, there exists y∈Zsuch that x > y and x2
y/∈N.
3. By hypothesis, ais divisible by 3. Suppose that a+bis also divisible by 3. Then a+b= 3k
for some integer kand a= 3k0for some integer k0. So, 3k=a+b= 3k0+b, which implies
that b= 3(k−k0). The integers are closed under subtraction, which means that k−k0is also
an integer. Thus, bis divisible by 3. We’ve shown that, if a+bis divisible by 3, then bmust
also be divisible by 3. Thus, the contrapositive is true: if bis not divisible by 3, then a+b
is not divisible by 3.
4. Base case: If n= 1, then (1 + x)n= 1 + xand 1 + 1 ·x= 1 + xand thus the conclusion is
true.
Induction step: Suppose that (1 +x)k≥1 + kx for some natural number k. We want to show
that (1+ x)k+1 ≥1 +(k+1)x. We start with the left-hand side: (1+x)k+1 = (1 + x)k(1+x)≥
(1+kx)(1 + x), by the induction hypothesis. We now have that (1+ x)k+1 ≥1+ kx +x+kx2,
which in turn is greater than or equal to 1 + kx +x, since kx2≥0. Thus, (1 + x)k+1 ≥
1+(k+ 1)x, which completes the induction step.
We’ve shown that (1 + x)n≥1 + nx for all natural numbers n.
5. f(x) is injective: Suppose f(x1) = f(x2). Then
x1+ 1
x1−1=x2+ 1
x2−1.
If we multiply both sides by (x1−1)(x2−1) to eliminate the denominators, we get:
(x2−1)(x1+ 1) = (x2+ 1)(x1−1).
Multiply out and combine like terms to see that x1=x2.
f(x) is not onto: There is no xsuch that f(x) = 1. If such an xdid exist, then
x+ 1
x−1= 1.
This would mean that x+1 = x−1, which would imply that 1 = −1. This is a contradiction.
Thus, there is an element of the target, 1 ∈R, that is not in the image of f.
Since f(x) is not onto, f(x) is not a bijection.