Download MATH 139 W10 Test 1 Solutions 1 1. Write the negation for the

MATH 139 W10 Test 1 Solutions
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1. Write the negation for the following statements:
a) ∀ real numbers x, if x > 4 then x2 > 16.
Solution: ∃ real number x such that x > 4 and x2 ≤ 16.
b) ∃ real number x such that x is irrational and x2 is rational.
Solution: ∀ real numbers x, x is rational or x2 is irrational.
2. State the arguments for modus ponens and modus tollens.
Solution: Modus ponens argument:
∀x, P (x) → Q(x)
P (a) (for a particular a)
Therefore, Q(a).
Modus tollens argument:
∀x, P (x) → Q(x)
¬Q(a) (for a particular a)
Therefore, ¬P (a).
3. Determine which of the arguments below is valid. For valid arguments, state which
type of argument is used (eg. modus ponens). For invalid arguments, state which type
of error occurs (eg. converse).
a) All healthy people drink milk every day.
Cathy drinks milk every day.
Therefore, Cathy is a healthy person.
Solution The argument is of the general form:
∀x, P (x) → Q(x)
Q(a)
Therefore, P (a).
This is an invalid argument. It would be valid if we replace ∀x, P (x) → Q(x) by
its converse ∀x, Q(x) → P (x). Therefore, a converse error occurred.
b) All honest people pay their taxes.
John does not pay taxes.
Therefore, John is not honest.
Solution: This argument has the general form:
∀x, P (x) → Q(x)
¬Q(a)
Therefore, ¬P (a).
This is just (universal) modus tollens. Thus the argument is valid.
4. Prove or disprove the following statements:
MATH 139 W10 Test 1 Solutions
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a) For all integers a, b if 12 divides ab then 12 divides a or 12 divides b.
Solution: The statement is false. Let a = 3 and b = 4. Then ab = 12 and hence
12 divides ab. However 12 divides neither a nor b.
b) For any nonnegative integers a, b if a is prime and b < a, then gcd(a, b) = 1.
Solution The statement is false. For example, take a = 5 and b = 0. Then a is
prime and b < a, but gcd(5, 0) = 5.
c) For all integers a, b if a divides b then a2 divides b2 .
Solution The statement is true. Suppose a divides b. Then b = ka for some integer
k. Then b2 = k 2 a2 . Since k 2 is also an integer, it follows that a2 divides b2 .
5.
a) Find 43 div 9 and 43 mod 9.
Solution 43 = 4 · 9 + 7. Therefore 43 div 9 = 4 and 43 mod 9 = 7.
b) Prove or disprove: for all real numbers x and y, dxye = dxebyc.
Solution The statement is false. Take x = 1 and y = 0.5. Then dxye = d0.5e = 1.
However, dxebyc = 1 · 0 = 0.
6. Prove the following statement by contradiction: There is no smallest positive, rational
number.
Solution. By contradiction. Assume there is a smallest positive rational number r. Then
r
is positive and it is also rational. Furthermore, 2r < r. However, this contradicts the
2
fact that r is the smallest positive rational. Therefore there can be no smallest positive
rational number.
7. For each integer n ≥ 1 let P (n) be the formula:
1 + 3 + 5 + · · · + 2n − 1 = n2
Using the steps below, give a proof by induction to show that P (n) is true for all
integers n ≥ 1.
a) State what one needs to prove in the base step, and carry out the proof.
Solution We need to show that P (1) is true. When n = 1, the left side of the
equation equals 1. The right side equals 12 = 1. Thus P (1) is true.
b) State (exactly) what one needs to prove in the inductive step, including any
assumptions one will make.
Solution We need to show that if P (n) is true for some n ≥ 1, then P (n + 1) is
true. That is, assuming 1 + 3 + 5 + · · · + (2n − 1) = n2 , we need to show that
1 + 3 + 5 + · · · + (2(n + 1) − 1) = (n + 1)2 .
c) Carry out the proof of the inductive step.
Hint: (n + 1)2 = n2 + 2n + 1 = n2 + 2(n + 1) − 1.
Solution As stated in b) assume 1 + 3 + 5 + · · · + (2n − 1) = n2 . Now 1 + 3 + 5 +
· · · + (2n + 1) = 1 + 3 + 5 + · · · + (2n − 1) + (2(n + 1) − 1) = n2 + (2(n + 1) − 1).
From the hint, n2 + (2(n + 1) − 1) = (n + 1)2 . Thus P (n + 1) is also seen to be
true.
MATH 139 W10 Test 1 Solutions
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8. Let the universal set be the set R of real numbers and let
A = {x ∈ R − 3 ≤ x ≤ 0}, B = {x ∈ R − 1 < x < 2}.
Find each of the following sets:
a) A ∩ B. Solution {x ∈ R − 1 < x ≤ 0}
b) Ac ∪ B. Solution {x ∈ R x < −3 or x > −1}
c) B c − A. Solution {x ∈ R x ≥ 2 or x < −3}
9. Suppose A, B, and C are subsets of a universal set U.
(3pts)
a) Use a Venn diagram to illustrate the set (A ∪ B)c ∩ C c .
b) Disprove the following: Ac ∩ B c = (A ∩ B)c .
Solution The equation doesn’t hold in general. For example, take U = {a, b, c, d}, A =
{a, b}, B = {b, c}. Then Ac ∩ B c = {d} but (A ∩ B)c = {a, c, d}.
10. We have the following set B = {a, b} and operations + and ◦ defined by the following
tables:
+
a
b
a b
a b
b b
◦ a b
a a a
b a b
a) Show that the commutative law holds for both + and ◦.
Solution From the tables we have a + b = b = b + a and a ◦ b = a = b ◦ a. Therefore
the operations + and ◦ are both commutative.
b) Show that each element x ∈ B has a complement x.
Solution The ’0’ element is seen to be a and the ’1’ element is seen to be b. From
the tables a + b = b = b + a and a ◦ b = a = b ◦ a. Since 1 = b and 0 = a, it follows
that a = b and b = a.
Bonus Show that 1 · (1!) + 2 · (2!) + · · · + n · (n!) = (n + 1)! − 1 for all integers n ≥ 1.
MATH 139 W10 Test 1 Solutions
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Solution By induction. The assertion is true for n = 1 since in this case the left side of the
equation equals 1 · 1! = 1 and the right side equals 2! − 1 = 1.
Assume that the assertion is true for some n ≥ 1. That is, assume 1 · (1!) + 2 · (2!) +
· · · + n · (n!) = (n + 1)! − 1. We shall show that the assertion is true for n + 1. We have
1 · (1!) + 2 · (2!) + · · · + n · (n!) + (n + 1)(n + 1)! = (n + 1)! − 1 + (n + 1)(n + 1)!. Now the
right side of the equation equals (n + 1)!(1 + (n + 1)) − 1 = (n + 1)!(n + 2) − 1 = (n + 2)! − 1.
Thus the assertion is seen to be true for n + 1.