Math 217 Solutions #3
1. Dene the relation E on the set of reals R by xEy i x y Z, where x, y R
and Z denotes the set of integers. Show that E is an equivalence relation.
Solution. For reexivity, if x R then x x = 0 Z and therefore xEx. For
symmetry, if x
December 7th, 2013
This nal exam has a total of 10 questions. Answer all questions.
This nal exam is three hours long.
To receive full credit you must explain your answers. You may use the back of the
Math 217 Solutions #2
1. List the elements of the following sets:
A = cfw_x Q : x2 = 2 and
B = cfw_n N : 2 < 3n + 1 < 20
where Q denotes the set of rational numbers and N = cfw_0, 1, 2, . . . denotes the
set of natural numbers.
Solution. If x2 = 2 then
Math 217 Solutions #1
1. Assuming that p and r are false and that q and s are true, nd the truth value
of each of the following statement forms.
(a) (p q ).
(b) (p q ) (q r).
(c) (p q ) r.
(d) (s (p r) (p (r q ) s).
(a) (p q ) = (F T) = T = F.
Math 217 Solutions #5
1. Suppose that n, q, m, r Z with n = qm + r. Using induction on k , show that
for all k N, there exists some s Z with nk = sm + rk .
Solution. We proceed by induction on k . In the base case, k = 0, we have
n0 = 1 = 0m + r0 . Thus,
Math 217 Solutions #7
1. Fill in the Cayley table below, given that G = cfw_e, a, b, c, d is a group.
Solution. First we show that e is the identity of G. Since ab = c, neither a nor
b can be the identity (e.g., if a is the i
Math 217 Solutions #6
1. Find all integers k 2 such that
(a) 3 k 2 (mod k ).
Solution. As k | k 2 , we need to solve 3 k 2 0 (mod k ). Since 3 0
(mod k ), k | 3 and hence k = 1 or k = 3 (as 3 is prime). We are asked to
nd k 2, so k = 3 is the only possibi
October 31, 2013
This midterm exam has a total of 5 questions. Answer all questions.
This midterm exam is two hours long.
To receive full credit you must explain your answers.
Calculators, notes, and oth
Math 217 Solutions #9
1. For each function : G H listed below, determine whether is a homomorphism. If so, identify its kernel Ker() and determine whether is injective or
(a) G = Z under addition, H = Zn , (a) = [a] for a Z.
Solution. Since (a
Math 217 Solutions #10
1. Let x = x1 x2 xn and y = y1 y2 yn B n be binary words of length n.
(a) Show that wt(x + y ) = wt(x)+wt(y ) 2 wt(x y ), where x y = z1 z2 zn
has digits zi = xi yi , where the product xi yi is taken in Z2 .
Solution. The left expre
Math 217 Solutions #4
1. Let : X Y and : Y Z be functions. Prove the following:
(a) If is one-to-one, then is one-to-one.
Solution. Take a, b X with (a) = (b). Since (a) = (b), we also
have ( )(a) = (a) = (b) = ( )(a). Since is injective,
we must have a =
Math 217 Solutions #8
1. Suppose that a group G has a subgroup of order 45 and a subgroup of order
75. If |G| < 400, determine |G|.
Solution. If H is a subgroup of G then |H | divides |G|. If we let n = |G|, then
45 | n and 75 | n. Since 45 = (3)2 (5) and