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Math 266 - Final Exam Practice Questions
- Let p be a prime and a ∈ Z. Here is a logical statement: “If p divides a^2 , then p divides a.” What is the contrapositive statement?
Solution: If p does not divide a, then p does not divide a^2.
- Let P and Q be propositions. Find a statement logically equivalent to P ∨ Q using only negation (∼) and conjuction (∧) operations.
Solution: ∼ (∼ P ∧ ∼ Q)
- Prove that there is a rational number between any two unequal real numbers.
Solution: This problem is hard, because it can’t be done with simple arithmetic. Instead, you need to use a fact about real numbers that can only be proved from the definition of real numbers (which we didn’t cover). The fact you need is that for any real number x, there is an integer n with x < n. Or, equivalently, you can use the greatest integer function. Try again to prove it using this hint. Here is a proof: Suppose x, y ∈ R and x < y. Choose an integer n > (^) y−^1 x , so that (^1) n < y − x. Then ny − nx > 1, so there is an integer m with nx ≤ m ≤ ny (you could let m be bnxc + 1). Then x ≤ mn ≤ y and we have shown that the rational number mn is between x and y.
- Define, for x, y ∈ R:
max(x, y) =
x if x ≥ y; y if y > x. Prove, for any z ∈ R, that if z ≥ x and z ≥ y, then z ≥ max(x, y).
Solution: Suppose z is as described. If x ≥ y then z ≥ x = max(x, y). If y > x, then z ≥ y = max(x, y). In both cases, z ≥ max(x, y).
- For x, y ∈ R, give a precise definition of min(x, y), the minimum of x and y. Prove that min(x, y) = − max(−x, −y).
Solution: Define, for x, y ∈ R:
min(x, y) =
y if x ≥ y; x if y > x.
We handle three cases. If x = y, min(x, y) = x = −(−x) = − max(−x, −y). If x > y, then −x < −y, so − max(−x, −y) = −(−y) = y = min(x, y). If x < y, then −x > −y, so − max(−x, −y) = −(−x) = x = min(x, y).
- Let {An}∞ n=1 and {Bn}∞ n=1 be two families of sets. Prove
( (^) ∞ ⋂
n=
An
n=
Bn
⋂^ ∞
n=
An ∪ Bn.
Solution: Let x ∈ (
n=1 An)^ ∪^ (
n=1 Bn). Then^ x^ ∈^ (
n=1 An) or^ x^ ∈^ (
n=1 Bn). If x ∈ (
n=1 An) then^ x^ ∈^ An^ for all^ n, therefore^ x^ ∈^ An^ ∪^ Bn^ for all^ n, and so x ∈
n=1 An^ ∪^ Bn. If^ x^ ∈^ (
n=1 Bn) then^ x^ ∈^ Bn^ for all^ n, therefore^ x^ ∈^ An^ ∪^ Bn for all n, and so x ∈
n=1 An^ ∪^ Bn.
- Let R be a relation on a set A, and define a new relation S on A by xSy iff xRy or yRx. Prove that S is symmetric.
Solution: Suppose xSy. Then xRy or yRx. Therefore yRx or xRy. So ySx, and so S is symmetric.
- Define an equivalence relation on Z 7 (the integers modulo 7) by xRy if x ≡ y +1 (mod 7) or x ≡ y + 2 (mod 7). Draw the digraph for R.
Solution: (a) Yes. Suppose f (r 1 ) = f (r 2 ). Let f (r 1 ) = (p 1 , q 1 ) and f (r 2 ) = (p 2 , q 2 ). Then r 1 = p 1 /q 1 = p 2 /q 2 = r 2 and f is one-to-one.
(b) No. For example, there is no r with f (r) = (2, 2). (Since that would require r = 2/2, which is not in lowest terms).
- Let f : R → R by f (x) = sin(x).
(a) Find f ([− 14. 000001 , 182 .632]). (b) Find f −^1 ([− 14. 000001 , 182 .632]).
Solution: (a) [− 1 , 1].
(b) R.
- A function f : R → R is called even if f (−x) = f (x) for all x ∈ R.
(a) Give three examples of even functions. (b) Prove: For any function g : R → R, the function f (x) = g(x^2 ) is even. (c) Prove: If f : R → R is even, then there is a function g : R → R with f (x) = g(x^2 ).
Solution: (a) For example, f (x) = x^2 , f (x) = cos(x), f (x) = |x|.
(b) f (−x) = g((−x)^2 ) = g(x^2 ) = f (x).
(c) For x ≥ 0, let g(x) = f (
x) (and we can let g(x) = 0 or anything for x < 0). Then g(x^2 ) = f (
x^2 ) = f (±x) = f (x), since f (−x) = f (x).
- A class a ∈ Zm is called a quadratic residue modulo m if there is b ∈ Z with b^2 ≡ a (mod m). Find the quadratic residues modulo 13.
Solution: They are (the equivalence classes of) 0, 1, 3, 4, 9, 10, 12.
- Find integers a, b, and c so that:
- c 6 ≡ 0 (mod 35)
- ac ≡ bc (mod 35)
- a 6 ≡ b (mod 35).
Solution: Since (a − b)c ≡ 0 (mod 35), you must have c ≡ 5 or c ≡ 7. For example, c = 5, a = 8, b = 1 will work.
- Prove that if a ≡ b (mod m) and n|m, then a ≡ b (mod n).
Solution: If a ≡ b (mod m), then m|b−a. Since n|m, n|b−a and so a ≡ b (mod n).
- Prove that 2^44 − 1 is divisible by 89.
Solution: 211 = 2048 ≡ 1 (mod 89), so 2^44 = 2^11 · 211 · 211 · 211 ≡ 1 (mod 89).
- Suppose a is not divisible by 17. Prove that either a^8 + 1 or a^8 − 1 is divisible by 17.
Solution: (This problem requires a very clever idea). By Fermat’s Little Theorem, a^16 − 1 ≡ 0 (mod 17). Factoring, we have (a^8 − 1)(a^8 +
- ≡ 0 (mod 17). So, 17|(a^8 − 1)(a^8 + 1). Since 17 is prime, by Euclid’s lemma, 17 |(a^8 − 1) or 17|(a^8 + 1).
- Define the numbers cn by c 1 = 1, c 2 = 1, and for n ≥ 1, cn+2 = 1/(cn + cn+1). Prove that 1 / 2 ≤ cn ≤ 1 for all n ∈ N.
Solution: The statement is clearly true for n = 1, 2. Suppose 1 / 2 ≤ cn ≤ 1 and (^1) / 2 ≤ cn+1 ≤ 1. Then 1 ≤ cn + cn+1 ≤ 2, so that 1 ≥ 1 /(cn + cn+1) ≥ 1 /2, so 1 ≥ cn+2 ≥^1 / 2. By strong induction, the statement is true for all n.
- Let f (x) = (1 − x)−^1. Prove by induction that the nth^ derivative of f is given by f (n)(x) = n!(1 − x)−n−^1.
