CS 30
Methods of Proof
Prove the following statements: 1. Gene is taking Stat 130, which has three tests, each worth 100 points. She would like to achieve a test average of over 80. However, she is very uncomfortable with the material on the first test. Show that if Gene gets at most a 40 on the first test, then she can achieve an average of at most 80. Proof: Let t1 , t2 , t3 be Gene’s test scores in the first, second and third tests respectively. Thus, from the premise, t1 ≤ 40, t2 ≤ 100 and t3 ≤ 100. But t1 + t2 + t3 40 + 100 + 100 Average(t1 , t2 , t3 ) = ≤ = 80 3 3 by the laws of inequality. That is, Gene can achieve an average of at most 80. 2. Show that there is a set A such that |℘(A)| = |A × A|. Proof: Let A = {1, 2}. Then ℘(A) = {∅, {1}, {2}, {1, 2}} and A × A = {(1, 1), (1, 2), (2, 1), (2, 2)}. Hence, |℘(A)| = 4 and |A × A| = 4. Therefore, there is a set A such that |℘(A)| = |A × A|. 3. Show that for all sets A, B, C, (A ∩ B)\C = (A\C) ∩ (B\C). Proof: We first show that (A ∩ B)\C ⊆ (A\C) ∩ (B\C). x ∈ (A ∩ B)\C ⇒ x ∈ (A ∩ B) ∧ x 6∈ C ⇒ (x ∈ A ∨ x ∈ B) ∧ x 6∈ C ⇒ (x ∈ A ∧ x 6∈ C) ∨ (x ∈ B ∧ x 6∈ C) ⇒ (x ∈ A\C) ∨ (x ∈ B\C) ⇒ x ∈ (A\C) ∩ (B\C)
definition of \ definition of ∩ De Morgan’s Law definition of \ definition of ∩
Hence, (A ∩ B)\C ⊆ (A\C) ∩ (B\C). Next we show that (A\C) ∩ (B\C) ⊆ (A ∩ B)\C. x ∈ (A\C) ∩ (B\C) ⇒ (x ∈ A\C) ∨ (x ∈ B\C) ⇒ (x ∈ A ∧ x 6∈ C) ∨ (x ∈ B ∧ x 6∈ C) ⇒ (x ∈ A ∨ x ∈ B) ∧ x 6∈ C ⇒ x ∈ (A ∩ B) ∧ x 6∈ C ⇒ x ∈ (A ∩ B)\C
definition of ∩ definition of \ De Morgan’s Law definition of ∩ definition of \
Hence, (A\C) ∩ (B\C) ⊆ (A ∩ B)\C. Therefore, (A ∩ B)\C = (A\C) ∩ (B\C). 4. Erik is struggling and just wants to pass Physics 71. His grade is based entirely on the average of four tests, each worth 100 points. A minimum grade of 60 is required to pass. Show that if Erik is to pass the class, then at least one of his test grades must be 60 or higher. Proof: The contrapositive of the statement is : If all of Erik’s test grades is below 60, then Erik will not pass the class. Let t1 , t2 , t3 , t4 be Erik’s test scores in the first, second, third and fourth tests respectively. If t1 , t2 , t3 , t4 < 60 then Average(t1 , t2 , t3 , t4 ) =
t1 + t2 + t3 + t4 60 + 60 + 60 + 60 < = 60 4 4
by the laws of inequality. That is, Erik’s average will not reach 60, meaning he will not pass the class. 5. Show that ∀n ∈ Z, (n3 − n) mod 3 = 0. Proof: By the quotient-remainder theorem, we can write n = 3q + r for some q, r ∈ Z such that r ∈ {0, 1, 2}. Then n3 − n = (3q + r)3 − (3q + r) = (3q)3 + 3(3q)2 r + 3(3q)r2 + r3 − (3q + r) = 27q 3 + 27q 2 r + 9qr2 + r3 − 3q − r = 3(9q 3 + 9q 2 r + 3qr2 − q) + (r3 − r)
by algebra
Let m = 9q 3 + 9q 2 r + 3qr2 − q. Note that m ∈ Z since q, r ∈ Z and Z is closed under addition and multiplication. That is, n3 − n = 3m + (r3 − r), for some m ∈ Z. Case 1. If r = 0, then r3 − r = 0. Hence, n3 − n = 3m for some m ∈ Z. Then (n3 − n) mod 3 = 3m mod 3 = 0, by definition of mod. Case 2. If r = 1, then r3 − r = 0. Hence, n3 − n = 3m for some m ∈ Z. Then (n3 − n) mod 3 = 3m mod 3 = 0, by definition of mod.
Case 3. If r = 2, then r3 − r = 6. Hence, n3 − n = 3m − 6 for some m ∈ Z. Then (n3 − n)
mod 3 = (3m − 6)
mod 3 = 3(m − 2)
mod 3 = 0,
by definition of mod. ∴ For all possible cases, we have shown that ∀n ∈ Z, (n3 − n) mod 3 = 0 6. Show that ∀n ∈ Z, if 4|n, then b n+2 c = n4 . 4 Proof: Note that, 4|n ⇒ n = 4k for some k ∈ Z, by definition of |. That is, k = n4 ∈ Z and n+2 = 4k+2 = 4k + 24 = k + 12 for some k ∈ Z, by algebra. 4 4 4 Since 0 < 12 < 1, then k <k+ n 4
<
1 2
n+2 4
<k+1 < n4 + 1
by the laws of inequality by substitution
Therefore, b n+2 c = n4 , by definition of b c. 4 √ 7. Show that 2 is irrational. Proof: √ Suppose √ that 2 is rational. a That is, 2 = for some a, b ∈ Z and b 6= 0. b a Without loss of generality, let us suppose is in lowest terms (i.e., gcd(a, b) = 1). b By algebra, we have √ a 2 = b a 2 √ 2 2 = b 2
=
a2 b2
2b2
=
a2
By the definition of even numbers, a2 is even. Note that, if an integer is odd, then its square is also odd. {Let n be an odd integer. By definition of odd, n = 2k +1 for some k ∈ Z. Hence, n2 = (2k +1)2 = 4k 2 +4k +1 = 2(2k 2 +2k)+1 where 2k 2 +2k ∈ Z. By the definition of odd numbers, n2 is odd.} Thus, by contraposition, if the square of an integer is even, then the integer must be even. In that case, since a2 is even, then a must be even. By the definition of even integers, a = 2q for some q ∈ Z. We now have,
2b2 2b2 b2
= = =
a2 (2q)2 = 4q 2 2q 2
By the definition of even numbers, b2 is even. And since b2 is even, then b must be even. That is, b = 2r for some r ∈ Z. This means that, gcd(a, b) = gcd(2q, 2r) ≥ 2. A contradiction to our assumption that gcd(a, b) = 1. ∴
√ 2 must be irrational.
8. Show that
√ 7+ 2 5
is irrational.
Proof: √ Suppose that 7+√5 2 is rational. That is, r = 7+5 2 ∈ Q. √ From algebra, we see that 2 = 5r − 7. However, √ 5r − 7 ∈ Q since Q is closed under addition and multiplication. Also, 2 6∈ Q from the previous exercise. This √is a contradiction. ∴ 7+5 2 is irrational. 9. Show that log3 5 is irrational. Proof: Suppose that log3 5 is rational. a That is, log3 5 = for some a, b ∈ Z and b 6= 0. b We know that log3 5 > 0, by the property of logarithms. Without loss of generality, we may assume that a, b > 0. a By the definition of logarithms, 3 b = 5. Raising both sides of the equation to the power b, we have 3a = 5b . Since a > 0, we see that 3 | 5b , by definition of |. That is, 3 | 5. This is a contradiction. ∴ log3 5 is irrational. √ 10. Let x ∈ R. Show that if x is irrational, then x is irrational. Proof: √ By contraposition, we must show that if x is rational then x is rational. √
√
a x = for some a, b ∈ Z and b 6= 0 b a2 ⇒ x = 2 for some a2 , b2 ∈ Z and b2 6= 0 b ⇒ x is rational
x is rational ⇒
by defn of rational by closure of Z and zero property by defn of rational
Disprove the following statements: 1. For all sets A, B, C, if A ⊆ B ∪ C, then A ⊆ B and A ⊆ C. Counterexample: Let A = Z, B = Z and C = ∅. We have Z ⊆ Z ∪ ∅ but Z 6⊆ ∅. 2. Z × N = N × Z. Counterexample: (−1, 1) ∈ Z × N but (−1, 1) 6∈ N × Z 3. There exists an integer m ≥ 3 such that m2 − 1 is prime. We need to prove that ∀m ≥ 3, m2 − 1 is not prime. Suppose we have m ≥ 3 and m2 − 1 is prime. Then by definition of prime, m2 − 1 has no divisors other that 1 or itself. But m2 − 1 = (m − 1)(m + 1). So either m − 1 = 1 or m + 1 = 1. That is, either m = 0 or m = 2. But in both cases, m < 3. This contradicts m ≥ 3. ∴ m2 − 1 must not be prime. 4. If p is a prime number, then 2p − 1 is also prime. Counterexample: Let p = 11. 211 − 1 = 2047 = 23 · 89 is not prime. What is wrong with the proof? 1.
shown true only for k = 2 and not for all integers k > 0 2.
shown only for successive integers 2k and 2k + 1 and not for any pair of odd integer and even integer ... should have let n = 2k + 1 and m = 2r for some k, r â&#x2C6;&#x2C6; Z 3.
the statement to be proved (m ¡ n is even) is already being used as a premise in the body of the proof ... this is wrong!
4.
shown only for two equal even numbers! should have let m = 2r and n = 2s for some r, s â&#x2C6;&#x2C6; Z 5.
shown only for two equal rational numbers! should have let r = a/b and s = c/d for some integers a, b, c, d where b, c are nonzero 6.
you are already using the statement to be proved as a premise in the body of the proof!