Download CSCI 2824 Spring 2013: Assignment # 4

CSCI 2824 Spring 2013: Assignment # 4
Due Date: Friday, Feb 22, 2013
Problem 1 (30 points) Evaluate the following expressions (try not to use a calculator :-) Show
your working or write a one/two line justification.
1. 1000! mod 10.
2. 2100 mod 100.
3. 340000 mod 16.
4. 2257! mod 256.
Solution
1. 1000! mod 10 = 0. Reason: 1000! = 10 × .....
2. 2100 mod 100 = (1024)10 mod 100 = (2410 ) mod 100 = 5765 mod 100 = 765 mod 100 =
76.
3. 340000 mod 16 = (34 )10000 mod 16 = 8110000 mod 16 = 110000 mod 16 = 1.
4. 2257! mod 256 = 0. Reason: 256 = 28 divides 2257! .
Problem 2 (80 points) We have below the statements and attempted proofs of theorems. For each
of these (a) is the theorem valid? (b) is the proof valid, if not point out the flaw in the proof?
and (c) write down a correct proof if the theorem is valid but the proof invalid. Your proof will be
judged for clarity and correctness. Writing lengthy, overly complicated proofs will result in points
being deducted.
Thm 1: For every integer n, if n is odd, then n2 leaves a remainder of 1 modulo 4.
Proof: We check the result for n = 1, 3, 5, 7, 9. Their squares 1, 9, 25, 49, 81 all leave a remainder
of 1 modulo 4. Therefore, the theorem looks like it is true.
Solution:
Theorem is VALID. Proof is INVALID since it is a proof by example.
Proof (fixed): Let n be an odd number. We can write n = 2k + 1 for some k ∈ N. Therefore,
n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 4l + 1 for l = k 2 + k. We conclude that n2 mod 4 = 1. (QED)
Thm 2: For every natural number n, if n is even, then its cube is divisible by 8.
Proof: Let n be any given even number. We can write n = 2j for some natural number j.
Therefore, n3 = 8j 3 is divisible by 8.
Solution:
Theorem is VALID. Proof is VALID. Nothing more needs to be done for this problem.
Thm 3: If two natural numbers m, n both leave a remainder of 1 when divided by 4, then m − n
is divisible by 4.
Proof: Let m, n be such that m mod 4 = n mod 4 = 1. Then m = 4j + 1 for some j ∈ N and
similarly for n. Therefore (m − n) = (4j + 1) − (4j + 1) = 0, which is divisible by 4.
Solution:
Theorem is VALID. Proof is INVALID. The letter j is reused in the proof.
1
Proof (fixed): Let m, n be such that m mod 4 = n mod 4 = 1. Then m = 4j + 1 for some j ∈ N
and similarly let n = 4k + 1 for some k ∈ N. Therefore (m − n) = (4j + 1) − (4k + 1) = 4(j − k),
which is divisible by 4.
Thm 4: For any two natural numbers p, q if both p, q are prime then pq + 1 is also prime.
Proof: Take any two prime numbers p, q. Normally pq is not prime, obviously because it is divisible
by both p and q. But take m = pq + 1. It leaves a remainder of 1 when divided by p and 1 when
divided by q. So it has to be prime.
Solution:
Theorem is INVALID. Take p = 5 and q = 3, we have pq + 1 = 16 is not prime.
Proof is INVALID. The last line of the proof “So it has to be prime” does not follow from the
last but one line. Just because a number leaves a remainder of 1 with two primes does not make it
prime. The example above illustrates.
Thm 5: For any prime number p, (p!) + 1 is prime.
Proof: Let p be any prime number and let n = p! + 1. We have that for all prime numbers k
ranging from 2 to p, n mod k = 1. Therefore, n does not have prime factors between 2 and p.
Therefore, n is prime.
Solution:
Theorem is INVALID. Take p = 5, we have p! + 1 = 121 is composite.
Proof is INVALID. The last line does not follow from the last but one line. Just because a
number p! + 1 does not have prime factors between 2 and p, we cannot conclude that p! + 1 is
prime. The counterexample above illustrates this.
Thm 5: For all integers p, q, r, if p2 + q 2 + r2 = 0 then p = q = r = 0.
Proof: Let p = q = r = 0. We have p2 + q 2 + r2 = 0 + 0 + 0 = 0.
CANCELLED for this assignment.
Thm 6: For all integers p, q, if pq = 1 then p = q = 1.
Proof: Let p, q be any given integers such that pq = 1. Clearly, p 6= 0 and q 6= 0. We split cases
on p.
Case -1: p = 1. We immediately have q = 1 since pq = 1.
Case -2: p > 1. In this case, we cannot have pq = 1 since q = p1 is not an integer.
Therefore, Case-1 is the only case possible and gives us p = q = 1.
Solution:
The theorem is INVALID. We omit p = q = −1.
The proof is INVALID. The case split does not consider all cases.
Thm 7: For any natural number n ≥ 4, n2 − 3n + 2 is a composite number.
Proof: Let n be any given natural number. We can write m = n2 − 3n + 2 into two factors
(n − 2) × (n − 1). This, at once, shows that m cannot be prime.
Solution:
The theorem is VALID. The proof is INVALID. The proof shows that m = (n − 2) × (n − 1)
but it does not show that (n − 2) 6= 1 and (n − 1) 6= 1. Therefore, m is not necessarily a composite
number.
Proof (fixed): Let n be any given natural number such that n ≥ 4. We can factorize m =
n2 − 3n + 2 as m = (n − 2) × (n − 1). Since n ≥ 4, we have n − 2 ≥ 2 and n − 1 ≥ 3. Therefore,
combining these facts we have n−2 6= 1 and n−1 6= 1. The factorization shows that m is composite.
(QED).
2
Thm 8: For all integers x, y, z and natural numbers p, xp + y p 6= z p .
Proof: This is precisely the statement of Fermat’s last theorem that follows from a proof of the
modularity theorem for so-called semi-stable elliptic curves proposed by Andrew Wiles in collaboration with Richard Taylor.
Solution:
Theorem is INVALID. It fails to hold for p = 2 (Pythagorean triples) or x = y = z = 0. These
cases must be ruled out before the theorem can be said to hold.
Proof is INVALID. It is a proof by authority of a statement that is not that of Fermat’s last
theorem.
Problem 3 (40 points) Write down the converse and contrapositive for each of the statements of
the theorems in problem 2 that are implications. If a theorem’s statement is not an implication,
indicate it in your answer.
Solution:
1. Converse: For every integer n, if n2 leaves a remainder of 1 modulo 4 then n is odd.
Contrapositive For every integer n, if n2 mod 4 6= 1 then n is even.
2. Converse: For every natural number n, If n3 is divisible by 8 then n is even.
Contra: For every natural number n, If n3 is not divisible by 8 then n is odd.
3. Converse: For every natural number m, n, If m − n is divisible by 4 then both m, n leave a
remainder of 1 modulo 4.
Contra: For every natural number m, n, If m − n is not divisible by 4 then m mod 4 6= 1
or n mod 4 6= 1.
4. Converse: For any two natural numbers p, q if pq + 1 is prime then both p and q are prime.
Contra: For any two natural numbers p, q if pq + 1 is not prime then p is not prime OR q is
not prime.
5. Converse: For any natural number p, If p! + 1 is prime then p is prime.
Contra: For any natural number p, If p! + 1 is not prime then p is not prime.
6. Converse: For all integers p, q, if p = q = 1 then pq = 1
Contra: For all integers p, q, if p 6= 1 or q 6= 1 then pq 6= 1.
7. Converse: For all natural numbers n, If n2 − 3n + 2 is a composite number then n ≥ 4.
Contrapositive: For all natural numbers n, If n2 − 3n + 2 is not a composite number then
n < 4.
8. Not an implication.
Problem 4 (50 points) Prove the following results using induction. Write down the form of
induction (weak/strong) and clearly state the base case/induction hypothesis.
1. For every natural number n, if n ≥ 5 then n! mod 5 = 0.
3
Proof. Proof is by strong induction on n.
Base Case: n = 5. We have n! mod 5 = 120 mod 5 = 0.
Ind. Hyp. If for all 5 ≤ m ≤ n, m! mod 5 = 0 then (n + 1)! mod 5 = 0.
Proof: We have (n + 1)! mod 5 = n! × (n + 1) mod 5 = ((n! mod 5) × (n + 1 mod 5)) mod 5.
But by induction hypothesis, we have n! mod 5 = 0.
Therefore, (n + 1)! mod 5 = 0. (QED).
Note: The problem above can also be solved using weak induction since we are simply using the
induction hypothesis on n when proving for n + 1.
2. Let Fn be the nth Fibonacci number. For every natural number n, F0 + F1 + F2 + · · · + Fn =
Fn+2 − 1.
Proof. Proof is by weak induction on n.
Base Case: n = 0. We have F0 = 1 = F2 − 1 = 2 − 1. Therefore base case is verified.
Ind. Hyp. For all n ∈ N, If F0 + · · · + Fn = Fn+2 − 1 then F0 + · · · + Fn+1 = Fn+3 − 1.
Proof: We can write
F0 + · · · + Fn+1 =
=
=
=
F0 + · · · + Fn + Fn+1
(Fn+2 − 1) + Fn+1 (∵ induction hypothesis on n)
Fn+1 + Fn+2 − 1
Fn+3 − 1
3. For every natural number n, F02 + F12 + · · · + Fn2 = Fn Fn+1 .
Proof. Proof is by strong induction on n.
Base Case: n = 0. We have F02 = 1 = F0 F1 = 1 × 1. Therefore base case is verified.
2 =F F
2
2
Ind. Hyp. For all n ∈ N, If for all 0 ≤ m ≤ n, F02 + · · · + Fm
m m+1 then F0 + · · · + Fn+1 =
Fn+1 Fn+2 .
Proof: We can write
2
F02 + · · · + Fn+1
=
=
=
=
2
F02 + · · · + Fn2 + Fn+1
2 (∵ induction hypothesis on n)
Fn Fn+1 + Fn+1
Fn+1 (Fn + Fn+1 )
Fn+1 Fn+2
4. For every natural number n ≥ 1, we have
1
1
1
n(n+1) = n − n+1 ).
Pn
1
j=1 j(j+1)
= 1−
1
n+1 .
(Hint: You can write
Proof. Proof is by weak induction on n.
Base Case: n = 1. We have LHS = 21 = 1 − 21 = RHS. This is verified.
P
Pn+1 1
1
1
Ind. Hyp. For all n ∈ N, If nj=1 j(j+1)
= 1 − n+1
then j=1
j(j+1) = 1 −
4
1
n+2
Proof: We can write
Pn+1
1
j=1 j(j+1)
=
=
=
=
=
Pn
1
1
1
1
1
1
j=1 j(j+1) + (n+1)(n+2)
1
1
− n+1
+ (n+1)(n+2)
(∵
1
n+2−1
− n+1 ( n+2 )
1 n+1
− n+1
n+2
1
− n+2
5
induction hypothesis for n)