COMP 2080 Winter 2022 Homework Assignment 1 SOLUTIONS
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COMP 2080 Winter 2022 Homework Assignment 1
SOLUTIONS
Useful definitions and results for this assignment:
| An integer x divides an integer y means: there exists an integer k such that y = x | k. We use the notation x | y to denote "x divides y"
| A number p is prime means: p is a positive integer, and, there are exactly two different positive integers that divide p (and they are 1 and p).
| The predicate isPrime(z) evaluates to true when z is prime, and evaluates to false when z is not prime.
| Every positive integer has a unique prime factorization. One particularly useful fact you can conclude from this: if you write a positive integer x as some specific product of prime numbers, then no other prime numbers divide x .
Assignment questions start here:
(2 pts) 1. Translate the following English sentence into a logical statement:
"Every even integer greater than 2 is the sum of two prime numbers".
Your solution may only use the following: ( ) 扌 甘 + v 入 # | Z > 2 | + = , x y z k isPrime
Solution:
扌x | Z, [((x > 2)入(甘k | Z, x = 2 |k)) # (甘y | Z, 甘z | Z, isPrime(y)入isPrime(z)入(x = y +z))]
2. Consider the following logical statement S : 扌w | R, (w2 这 2w 这1) # (w 1) Define P(w) to be the predicate (w2 这 2w 这1) # (w 1).
(1 pt) (a) Write the converse of the predicate P(w).
Solution
(w 1) # (w2 这 2w 这1)
(2 pts) (b) Write the contrapositive of the predicate P(w).
Simplify (showing each step of your work) so that your final answer does not contain a negation symbol (+)
Solution
+(w 1) #+(w2 这 2w 这1)
(w = 1) # (w2 这 2w = 这1)
(2 pts) (c) Write the inverse of the predicate P(w).
Simplify (showing each step of your work) so that your final answer does not contain a negation symbol (+)
Solution
+(w2 这 2w 这1) #+(w 1)
(w2 这 2w = 这1) # (w = 1)
(3 pts) (d) Write the negation of S .
Simplify (showing each step of your work) so that your final answer does not contain a negation symbol (+)
Solution
+(扌w | R, (w2 这 2w 这1) # (w 1))
甘w | R,+((w2 这 2w 这1) # (w 1))
甘w | R, (w2 这 2w 这1) 入+(w 1)
甘w | R, (w2 这 2w 这1) 入 (w = 1)
(4 pts) (e) Write an indirect proof that S is true.
Solution
(1) Let w be an arbitrary real number.
(2) We prove (w2 这 2w 这1) # (w 1) using an indirect proof, i.e., we prove its contrapositive is true:
(3) Suppose w = 1.
(4) Then w2 这 2w = (1)2 这 2(1) = 1 这 2 = 这1. This proves that w2 这 2w = 这1.
(5) The two previous steps prove (w = 1) # (w2 这 2w = 这1), as required.
(4 pts) (f) Write a proof by contradiction that S is true.
Solution
(1) To obtain a contradiction, assume+S is true, i.e., we assume 甘w | R, (w2 这 2w 这1) 入 (w = 1)
(2) Let w be a real number that satisfies (w2 这 2w 这1) 入 (w = 1), i.e., both w2 这 2w 这1 and w = 1 are true.
(3) Since w = 1, the expression w2 这 2w evaluates to (1)2 这 2(1) = 这1. So we have shown that w2 这 2w = 这1.
(4) We get a contradiction: from the previous two lines, we see that w2 这 2w 这1 and w2 这 2w = 这1 are both true, which is impossible.
(5) Since we reached a contradiction, our assumption "+S is true" must have been incorrect. In particular, this means+S is false, so S is true.
(4 pts) 3. Prove or disprove the following logical statement: 甘t | R, 扌x | R, x | (t 这 1) < 1
Solution
We prove that the statement is true:
(1) Let t = 1
(2) Let x be an arbitrary real number.
(3) From the fact that t = 1, subtracting 1 from both sides gives t 这 1 = 0.
(4) Since t 这 1 = 0, the expression x(t 这 1) evaluates to 0, which is less than 1. Therefore, x(t 这 1) < 1, as required.
(4 pts) 4. Prove or disprove the following logical statement:
扌z | Z, isPrime(z) # ((2 | z3 这 z2 + z) v (3 | z3 这 z2 + z))
Solution 1: Prof. Miller overcomplicates things
We disprove the statement by proving that its negation is true:
甘z | Z, isPrime(z) 入+(2 | z3 这 z2 + z) 入+(3 | z3 这 z2 + z)
(2) z is prime since the only positive integers that divide 7 are 1 and 7. Therefore, isPrime(z) evaluates to true.
(3) The value of z3 这 z2 + z is 73 这 72 + 7 = 7(72 这 7 + 1) = 7 | 43
(4) Since 7 and 43 are both prime, we have written z3 这 z2 + z as a product of primes 7 and 43, which means that no other primes divide z3 这 z2 + z .
(5) Since 2 and 3 are both prime, the previous line tells us that 2 does not divide z3 这 z2 + z, and
3 does not divide z3 这 z2 + z .
(6) Therefore, by lines (2) and (5), we have shown that isPrime(z) 入+(2 | z3 这 z2 + z) 入+(3 | z3 这 z2 + z), as required.
Solution 2: Easier Proof
We disprove the statement by proving that its negation is true:
甘z | Z, isPrime(z) 入+(2 | z3 这 z2 + z) 入+(3 | z3 这 z2 + z)
(2) z is prime since the only positive integers that divide 7 are 1 and 7. Therefore, isPrime(z) evaluates to true.
(3) The value of z3 这 z2 + z is 73 这 72 + 7 = 7(72 这 7 + 1) = 7 | 43 = 301
(4) Since 301/2 = 150.5 is not an integer, this means 2 does not divide 301. Since 301/3 = 100.333... is not an integer, this means 3 does not divide 301.
(5) Therefore, by lines (2) and (4), we have shown that isPrime(z) 入+(2 | z3 这 z2 + z) 入+(3 | z3 这 z2 + z), as required.
5. Our goal in this question is to prove the following identity: for all even integers q | 4,
匕(q)匕(y) ↓ ( = (q 这 2)(2q2(2)4+ 7q + 12)
To make your life easier, we’ve broken up the task into separate steps.
(8 pts)
(a) Prove the following using strong induction:
For all integers y | 0:
↓ ( = , 这41
if y is even
if y is odd
Solution 1 (one induction proof for both even and odd cases)
Define P(y) to be predicate ' ' = , 这4
We prove, by induction on y, that P (y) is true for all y | 0.
0
Base Case: Suppose y = 0. The expression! ''evaluates to '' = |0I = 0. The expression
=0
evaluates to = 0. Since y is even and we showed that ! '' = , the base case holds.
=0
Induction Hypothesis: For a y | 1, assume that P(0) 入 | | | 入 P(y 这 1) is true. Inductive Step: We prove that P(y) is true. We separately consider the following two subcases:
(i) y is even
(ii) y is odd
These two subcases cover all possibilities, since each integer y is either even or odd.
Proof of Case (i):
(1) Suppose that y is even. By the definition of even, there exists an integer j such that y = 2j .
(2) Split the summation range:
↓ ( = 1 ↓ ( 」+ 1 ↓ ( 」
= 1 ↓ ( 」+ ↓(
= 1 ↓ ( 」+ | | (since y = 2j)
= 1 ↓ ( 」+ |jI (arithmetic)
= 1 ↓ ( 」+ j (since j is an integer) = 1 ↓ ( 」+ (since y = 2j)
(3) Note that y 这 1 is odd, since y is even. By the induction hypothesis, we assumed that
这1
=0
↓ ( = 1 ↓ ( 」+
= (y 这 2 这 1 + (from Induction Hypothesis) = y2 这 2y4+ 1 这 1 + (arithmetic)
= y2 这 2y + 1 这 1 + 2y (arithmetic)
4
= (arithmetic)
Since y is even and we showed that ! '' = , this concludes the inductive step in
=0
Case (i).
Proof of Case (ii):
(1) Suppose that y is odd. By the definition of odd, there exists an integer j such that y = 2j + 1.
(2) Split the summation range:
↓ ( = 1 ↓ ( 」+ 1 ↓ ( 」
= 1 ↓ ( 」+ ↓(
= 1 ↓ ( 」+ | | (since y = 2j + 1)
= 1 ↓ ( 」+ | 三 (slide 8 of Math Review 5: | | = | 三) = 1 ↓ ( 」+ 1j] (arithmetic)
= 1 ↓ ( 」+ j (since j is an integer)
= 1 ↓ ( 」+ y 2(这) 1
(since y = 2j + 1)
(3) Note that y 这 1 is even, since y is odd. By the induction hypothesis, we know that
这1
=0
↓ ( = 1 ↓ ( 」+ y 2(这) 1
= (y 4(这) 1)2 + y 2(这) 1 (from Induction Hypothesis) = y2 这 4(2)y + 1 + y 2(这) 1 (arithmetic)
= y2 这 2y + 1 + 2y 这 2 (arithmetic)
4
= y2这4 1 (arithmetic)
Since y is odd and we showed that ! '' = y 2这41 , this concludes the inductive step in
=0
Case (ii).
Solution 2 (two induction proofs: one for even, one for odd)
We separately consider the following two subcases:
(i) y is even
(ii) y is odd
These two cases cover all possibilities, since each integer y is either even or odd.
Proof of Case (i): Define P(y) to be predicate ! ' ' = .
=0
We prove, by induction on y, that P (y) is true for all even integers y | 0.
0
Base Case: Suppose y = 0. The expression! ''evaluates to '' = |0I = 0. The expression
=0
evaluates to = 0. Therefore, the base case holds.
Induction Hypothesis: For an even y | 2, assume that P(0) 入 | | | 入 P(y 这 2) is true. Inductive Step: We prove that P(y) is true.
(1) By the definition of even, there exists an integer j such that y = 2j .
(2) Split the summation range:
↓ ( = 1 ↓ ( 」+'(.)u1 ↓ ( +!(┐) 1 ↓ ( 」
= 1 ↓ ( 」+ | y 2(这) 1 | + ↓(
= 1 ↓ ( 」+ | 2j 这2 1 | + | | (since y = 2j)
= 1 ↓ ( 」+ | (2j 这2(2)) + 1 | + | | (since 2j 这 1 = 2j 这 2 + 1)
= 1 ↓ ( 」+ | 2j 这2 2 三 + | | (slide 8 of Math Review 5: | | = | 三) = 1 ↓ ( 」+ 1j 这 1] + |jI (arithmetic)
= 1 ↓ ( 」+ j 这 1 + j (since j 这 1 and j are integers)
= 1 ↓ ( 」+ y 这 1 (since y = 2j)
(3) By the induction hypothesis, we know that P(y 这 2) is true, so we know that
! ' ' = (y 4(这)2)2 . Substituting this into the result from the previous line, we get:
=0
↓ ( = 1 ↓ ( 」+ y 这 1
= (y 4(这)2)2 + y 这 1 (from Induction Hypothesis) = y2 这 4(4)y + 4 + y 这 1 (arithmetic)
= y2 这 4y + 4 + 4y 这 4 (arithmetic)
4
= (arithmetic)
We showed that ! '' = , which concludes the inductive step in Case (i).
=0
Proof of Case (ii): Define P(y) to be predicate ! ' ' = y 2这41 .
=0
We prove, by induction on y, that P (y) is true for all odd integers y | 0.
1
Base Case: Suppose y = 1. The expression ! ''evaluates to ''+'' = 0 + 0 = 0. The
=0
expression y2这41 evaluates to 12这41 = 0. Therefore, the base case holds.
Induction Hypothesis: For an odd y | 3, assume that P(1) 入 | | | 入 P(y 这 2) is true. Inductive Step: We prove that P(y) is true.
2022-03-21