Textbook pp. 152 - 154
2 (a) It is not a function because it is not single valued.
(c) It is not a function because f(-2) and f(2) are not defined.
4 (b) Domain: The set of positive integers Range: The set of natural numbers greater than 1.
(c) Doamin: The set of binary numbers Range: The set of natural numbers
12 (b) Not one-to-one because for example, 2 is mapped to by 1 and -1.
(d) Not one-to-one because for example f(1) = f(2) = 1.
20 (a) f(x) = x2
(c) f(x) = x + 1 if x is even and f(x) = x - 1 if x is odd.
22 (b) Not a bijection
(c) R is not defined at x = -2. So it is not a function over R.
40(a) To prove f(S U T) = f(S) U f(T)
Claim 1: f(S U T) is a subset of f(S) U f(T).
Proof: If y is in f(S U T), then there is x in S U T such that y = f(x).
Hence x is in S or x is in T.
If x is in S, then y is in f(S). Hence y is in f(S) U f(T).
If x is in T, then y is in f(T). Hence y is in f(S) U f(T).
Hence f(S U T) is a subset of f(S) U f(T).
Claim 2: f(S) U f(T) is a subset of f(S U T).
Proof: If y is in f(S) U f(T), then y is in f(S) or y is in f(T).
If y is in f(S), then there is x in S such that y = f(x).
Hence x is in S U T. Hence y is in f(S U T).
Similarly for when y is in f(T).
Hence f(S) U f(T) is a subset of f(S U T).
By Claims 1 and 2 f(S U T) = f(S) U f(T).
Textbook pp. 216 - 217
2 (a) It is O(x2).
(c) It is O(x2).
(e) It is not O(x2).
8 (b) n = 5.
14 (d) Yes, x3 is O(x2 + x4).
(e) Yes, x3 is O(3x).
(f) Yes, x3/3 is O(x3).
26 (b) 6n (= 2n3n)
(c) nnn!