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Math 347 C
HOUR EXAM I
6 July 2011
SOLUTIONS
- Show that for the real number x, satisfying |x| > 3, it follows that
x^2 + 2 x^2 + 1
SOLUTION Assume x
(^2) + x^2 +1 ≥^1 .1 and derive a contradiction. Thus x
(^2) + x^2 +1 ≥^1 .1 implies that^ x
(^2) + 2 ≥ 1 .1(x (^2) + 1) since x (^2) + 1 > 0. Collecting terms we get,. 9 ≥. 1 x^2 , whence 9 ≥ x^2. This gives 3 ≥ |x|, a contradic- tion.
- Let f : R → R be the function defined by
f (x) = x^3 + x + 1.
a) What does it mean to say that f is injective? b) Is f is injective? Prove your answer. SOLUTION a) f is injective if whenever a, b ∈ R and f (a) = f (b), then a = b. b) Let a, b ∈ R such that f (a) = f (b) and assume that a 6 = b. Thus a^3 +a+1 = b^3 +b+1 and so a^3 −b^3 = b−a. We know that a^3 −b^3 = (a−b)(a^2 + ab+b^2 ). Therefore a^3 −b^3 = b−a can be written as (a−b)(a^2 +ab+b^2 ) = b−a. Since we are assuming a 6 = b, a−b 6 = 0 and so we can cancel a−b. Hence a^2 +ab+b^2 = − 1 or a^2 +ab+b^2 +1 = 0. This can be thought of as a quadratic equation in a. Applying the Quadratic Formula, we get
a = −b ±
b^2 − 4(b^2 + 1) 2
Clearly b^2 −4(b^2 +1) < 0 and so the equation has no solution. Hence our assumption is wrong and a = b. Therefore f is injective.
- Suppose that {A 0 , A 1 , A 2 , A 3 , A 4 , A 5 } are subsets of [5] = { 1 , 2 , 3 , 4 , 5 } satisfying:
A 0 ⊂ A 1 ⊂ A 2 ⊂ A 3 ⊂ A 4 ⊂ A 5.
(Recall: B ⊂ C means that B is a subset of C, but B 6 = C.) a) Give an example of such a collection of subsets: A 0 , A 1 , A 2 , A 3 , A 4 , A 5.
b) How many such collections of subsets are there? SOLUTION a) One such example is A 0 = ∅, A 1 = { 1 }, A 2 = { 1 , 2 }, A 3 = { 1 , 2 , 3 }, A 4 = { 1 , 2 , 3 , 4 }, A 5 = { 1 , 2 , 3 , 4 , 5 }. b) Since Ai is a proper subset of Ai+1 it follows that Ai+1 consists of Ai and AT LEAST ONE more element. Since there are exactly 5 steps in going from one of these sets to the next, then each set contains EXACTLY one more element than its predecessor. (Note: It follows that A 0 must be the empty set and that A 5 = [5].) Thus each such sequence {Ai} is completely determined by the sequence of added elements which consists of all of the elements in [5]. Hence there are as many such sequences as there are permutations of [5], namely 5!.
- Prove: ∑n
i=
i(i − 1) =
(n − 1)n(n + 1) 3
SOLUTION Use induction on n. Let P [n] be the statement:
∑n i=1 i(i^ −^ 1) =^
(n−1)n(n+1)
P [1] states that
i=1 i(i^ −^ 1) =^
(1−1)1(1+1) 3 which is clearly true, since both sides are zero. Assume that P [k] is true and consider P [k + 1], which asserts:
k∑+
i=
i(i − 1) =
k(k + 1)(k + 2) 3
Then
∑k+ i=1 i(i^ −^ 1) = (k^ + 1)(k) +^
∑k i=1 i(i^ −^ 1) and by the induction assumption, (k + 1)(k) +
∑k i=1 i(i^ −^ 1) = (k^ + 1)(k) +^
k(k^2 −1)
But (k + 1)(k) + k(k
(^2) −1) 3 =^
3(k+1)k+k(k^2 −1) 3 =^
k(k+1)(3+k−1) 3 =^
k(k+1)(k+2)
Thus
∑k+ i=1 i(i^ −^ 1) =^
k(k+1)(k+2) 3 ,^ which is precisely the statement^ P^ [k^ + 1]. Hence the formula follows by the Theorem of Mathematical Induction.
- Let S = {x ∈ R | x^2 > x + 6} and T = {x ∈ R | x > 3 }. Determine which of the following statements are true. a) S ⊆ T. b) T ⊆ S.
SOLUTION a) x ∈ S, if and only if x^2 − x − 6 > 0. That is, (x − 3)(x + 2) > 0, whence, x − 3 and x + 2 have the same sign. We examine two cases. Case 1. x−3 and x+2 are both positive. Then x > 3 and x > − 2. This is equivalent to x > 3 and so x ∈ T. Hence in this case S ⊆ T.
