MATH104 Exam May 2006, Solutions
All questions except q.1 are standard homework examples
1. sigma, chi, ψ,β. (2 marks each.)
2.
a) Zcontains −1, 0, and 6 but not 9.5.
b) {x∈R|x > −1 and x2<20}contains 0 but not −1 or 9.5 or 6.
c) {1
2n2−3|n∈Z}contains −1 (take n= 2) and 9.5 (take n= 5) but not 0 (there is no n∈Z
with n2= 6) or 6 (there is no n∈Zwith n2= 18).
(2 marks for a), 4 marks for b) and 6 marks for c).)
3.
a) a≤1 and a≥ −2 (or: −2≤a≤1) (2 marks)
b) x > y or y > z (3 marks)
c) There exist xand ywith x > y and f(x)≤f(y) (3 marks)
d) There exists x∈Rsuch that for all y∈R, x ≤yor f(x)> f (y) (4 marks)
4.
a) fis not injective if there exist x, y ∈Rsuch that f(x) = f(y) but x6=y. (2 marks)
b) f(x) is injective. For let x, y ∈Rand suppose f(x) = f(y). Then 8x+ 12 = 8y+ 12 and
rearranging gives 8x= 8yso x=y. (4 marks)
c) x3−x= 0 for x= 0 and for x= 1. So let x= 0, y = 1. Then f(x) = f(y) = 0 but x6=y.
Hence fis not injective. (4 marks)
5.
For integers mand nthe statement m|nmeans that there exists an integer kwith n=km.
a) Ris not an equivalence relation. For not(−1R−1) since −1 + (−1) is not >0. Hence
property i) fails.
b) Ris an equivalence relation. For let x,y, and zbe any integers. Then
i) x−x= 0 and 4|0 since 0 = 0 ×4, so x R x.
ii) If x R y then x−y= 4k, for some k∈Zso y−x= 4(−k), and −k∈Zso y R x.
iii) If x R y and y R z then x−y= 4kand y−z= 4lfor some integers k, l so x−z=
(x−y) + (y−z) = 4(k+l), i.e. x R z.
c) Ris an equivalence relation. For let x,y, and zbe any elements of X. Then
i) x|x(true for any integer x).
ii) If x R y then x=y, since no element of Xdivides any other element. Hence y|x.
iii) As in ii), If x R y and y R z then x=yand y=zso x=zand x|z.
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