Download Comparing Contrapositive and Contradiction Proofs

Comparing Contrapositive and Contradiction Proofs
Prove that if n is an integer and n3 + 5 is odd, then n is even.
Contrapositive Proof:
Proof by contradiction:
Another example of a proof by contradiction:
If a number added to itself gives itself, then the number is 0.
Proof:
Proving the equivalence of three or more statements
Equivalence Proof
Prove that if n is an integer, then the following four statements
are equivalent:
(1) n is even
(2) n + 1 is odd
(3) 3n + 1 is odd
(4) 3n is even
(1)  (2):
(2)  (3):
(3)  (4):
(4)  (1):
Additional Proof Examples
Example 1: p. 93 # 16 For every integer n, 3(n2 + 2n + 3) – 2n2 is
a perfect square.
Example 2: p. 93 # 27 The difference of two consecutive cubes
is odd.
Example 3: p. 93 # 31 For any two numbers x and y, |x + y|  |x| + |y|
Case 1: x > 0 and y > 0
Case 2: x > 0, y < 0
Case 3: y > 0, x < 0
Case 4: x < 0, y < 0
Example 4: Prove that if n + 1 separate passwords are issued to
n students, then some student gets ≥ 2 passwords.
More Examples
Example 5: 1 + 2 + … + n = n(n + 1)/2
Case 1: n is even.
Case 2: n is odd.
Example 6: Prove n3 – n is divisible by 3.
Example 7: Prove or disprove there exist three consecutive odd
primes.
Example 8: Prove or disprove that given a positive integer n
there exist n consecutive odd positive integers that are prime (n
 n consecutive odd primes).
Modification of Table 2.2 p. 91 Gersting
Proof Technique
Exhaustive
Proof
Direct Proof
How to prove P  Q
Show P  Q for all
possible cases
Assume P, deduce Q
Comments
Only works for a finite
number of cases.
The standard approach
to try.
Contrapositive
Assume Q', deduce P'
Use if Q' as a
(Indirect) Proof
hypothesis seems to
give more information
to work with.
Contradiction
Assume P Λ Q', deduce Try this approach when
a contradiction
Q says something is
not true.
Proof by Cases
Break the domain into
Try this for proving
two or more subsets
properties of numbers
and prove PQ for the where odd and even or
elements in each such positive and negative
numbers require
subset.
different proofs.
Counterexample Prove statement false
Used to show that PQ
by finding one value for is false.
P such that Q is false.
Section 2.2 Mathematical Induction
How to do an induction proof
1.
2.
3.
Mathematical Induction Examples
Example 1: Prove that 1 + 2 + … + n = n(n+1)/2
Basis: P(1)
k
Hypothesis: Assume  i = k(k + 1)/2 for k  1
i=1
Induction step: Prove P(k)  P(k + 1)
k+1
k
i=1+2+…+k+k+1=i+k+1
i=1
i=1
Therefore, by the principle of mathematical induction, the
statement is true for all n.
Example 2: Prove 1 + 3 + 32 + … + 3n = (3n+1 – 1)/2 for n  0.
Basis: k = 0
Hypothesis: Assume 1 + 3 + 32 + … + 3k = (3k+1 – 1)/2 for k  0.
Induction step: Prove P(k)  P(k + 1)
Examples (continued)
Algebraic Proof of example 2
Example 3: p. 105 # 11
1•3 + 2•4 + … + n(n + 2) = n(n + 1)(2n + 7)/6 for n  1
Basis: k = 1
k
Hypothesis: Assume  i(i + 2) = k(k + 1)(2k + 7)/6 for k  1
i=1
Induction step: 1•3 + 2•4 + … + k(k + 2) + (k + 1)(k + 3) =
Example 4: You have invested $500 at 6% interest compounded
annually. If you make no withdrawals, prove that at the end of n
years the account contains (1.06)n•500 dollars.
Basis: At the end of year 1, you have
Hypothesis: At the end of k years the account contains
(1.06)k•500 dollars.
Induction step:
Proving an inequality
Example 5: p. 106 # 30 Prove that n2 < 2n for n  5.
Basis: k = 5
Hypothesis: Assume that k2 < 2k for k  5
?
Induction step: Prove P(k)  P(k + 1). P(k + 1) is (k+1)2 < 2k+1
Proof:
Example of a product
Example 6: Prove that (1 – 1/22)•(1 – 1/32)• … •(1 – 1/n2) = (n + 1)/2n
Basis:
k
Hypothesis: Assume π (1 – 1/i2) = (k + 1)/2k
i=2
Induction step:
Example 7: p. 107 #43 Prove that 7n – 2n is divisible by 5 for n  1
Basis:
Hypothesis: 7k – 2k is divisible by 5 for k  1
Induction step: 7k+1 – 2k+1 = 7•7k + 2k+1
Example 8: Prove that xn – yn is divisible by x - y for n  1
Basis: n = 1
Hypothesis: xk – yk is divisible by x - y for k  1
Induction step:
Example 9: Use mathematical induction to prove that a set with
n elements has n(n - 1)/2 subsets of size 2.
Basis: n = 2.
Hypothesis: Assume if S is a set and |S| = k, then S has k(k-1)/2
subsets of size 2.
Induction step: Let S be a set such that |S| = k + 1 and let x  S.