MATH104 Exam May 2008, Solutions
All questions except q.1 are standard examples similar to classwork or homework.
1. sigma, psi, φ,γ. (1 mark each.)
2.
(a) f(n) = −1−6n.
(b) f(n) = 6 + 3n
(c) 2 ∈S(take n= 0); −6∈S(take n=−2); 4 ∈/ S since this would require n∈Nwith
n3= 2; 10 ∈S(take n= 2).
(2 marks for (a), 4 marks for (b) and 6 marks for (c).)
3.
(a) x6=−4 and x≤2. The statement (a) has a free variable x. (3 marks)
(b) There exist real numbers a, b, c such that ab =ac and b6=c. The original statement is false
and this negation is true: take a= 0, b = 1,c = 2; then ab =ac = 0 and b6=c. (4 marks)
(c) There exist x > 1 and y > 2 with f(x)≤0 or f(y)≤4 (3 marks)
(d) ∃N∈N,∀x∈R, f(x)≤N. (3 marks)
4.
(a) fis not injective if there exist x, y in the domain of fsuch that f(x) = f(y) but x6=y. (2
marks)
(b) Let x, y ∈Rand suppose f(x) = f(y). Then 2 + 4x= 2 + 4yand rearranging gives 4x= 4y
so x=y. (3 marks)
(c) Let x= 0, y = 2. Then f(x) = f(y) = 0 but x6=y. Hence fis not injective. (3 marks)
(d) Let x, y ∈R, x > 1, y > 1, Suppose that f(x) = f(y). Then x2−2x=y2−2y, which gives
x2−y2−2(x−y) = 0, that is (x−y)(x+y)−2(x−y) = 0, that is (x−y)(x+y−2) = 0.
But x > 1 and y > 1 so that the second factor is >0. Hence the first factor is 0, giving
x=y. (5 marks)
5.
Sclosed under addition means that, for all x∈Sand y∈S, it follows that x+y∈S.
(1 mark)
For the given S, let x, y ∈Sso that there exist integers k, ` with 3x=k, 3y=`. Thus
3(x+y) = k+`and since k+`∈Zwe deduce that x+y∈S. Thus Sis closed under addition.
(3 marks)
(a) Ris not an equivalence relation. Prop erty (i) fails, since for example 1 R1 is false. (2 marks)
(b) Ris an equivalence relation. For let x,y, and zbe any integers. Then
i) x−x= 0 and 6|0 since 0 = 0 ×6, so x R x.
ii) If x R y then x−y= 6k, for some k∈Zso y−x= 6(−k), and −k∈Zso y R x.
1
