Here are some proofs that were covered in class or in the discussion. This file will be added to
throughout the semester
Note: The are often several different ways to do the same proof. I’ve only shown one way in
the examples below.
Prove If n is odd then n2 is odd. A direct proof will be used.
Proof: Let n be an odd number
Hypothesis
Then there is an integer k such that n = 2k + 1
Definition of odd number
n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2[2k2 + 2k] + 1
Algebra
2
Therefore, n is an odd number
Definition of odd number
Prove that if n + 1 separate passwords are issued to n students, then some student gets
at least two passwords. An indirect proof will be used.
Contrapositive: If each student gets at most one password then less than n + 1 passwords
were issued.
Proof: Suppose each student got at most one password.
Then, since there are n students the total number of passwords is n.
This contradicts the hypothesis that n + 1 passwords were issued.
Prove n3 - n is divisible by 3 for n  0. (Direct proof. We'll do it by induction later.)
Proof:
n3 - n = n(n2 - 1)
algebra
= n(n + 1)(n - 1)
algebra
Notice that n - 1, n, and n + 1 are three consecutive integers so one of them has a remainder
of 0 when divided by 3. Therefore, their product is divisible by 3.
For every integer n, 3(n2 + 2n + 3) – 2n2 is a perfect square. (Direct Proof)
3n2 + 6n + 9 – 2n2 = n2 + 6n + n = (n + 3)2
Algebra
The difference of two consecutive cubes is odd. (Proof by Cases)
Proof: Let x and x + 1 be consecutive numbers
Definition of consecutive
3
3
3
2
3
2
(x + 1) – x = x + 3x + 3x + 1 – x = 3x + 3x + 1
Algebra
Case 1: x is even
Case 2: x is odd
x2 is even
both 3x2 and 3x are odd so 3x2 + 3x is even
Both 3x2 and 3x are even so
Then, 3x2 + 3x + 1 is odd
3
3
(x+1) – x has the form 2k + 1 and is odd
If n is an integer and n3 + 5 is odd, then n is even (Indirect proof)
Proof. Suppose n is odd
Then, n = 2k + 1 for some integer k
Definition of odd
n3 + 5 = (2k + 1)3 + 5 = 8k3 + 12k2 + 6k + 6
Algebra
= 2[4k3 + 6k2 + 3k + 3]
Algebra
Therefore, n3 + 5 is even
Definition of even
If n is an integer and n3 + 5 is odd, then n is even (Proof by contradiction)
Proof. Suppose both n and n3 + 5 are odd
Assume p  q
If n is odd, then n3 is odd
Property of odd number
3
3
(n + 5) – n = 5
Algebra
But since odd – odd = even, this implies 5 is even, a contradiction.
For any two numbers x and y, |x + y|  |x| + |y|
(Proof by cases)
Case 1: x > 0 and y > 0
|x| = x
|y| = y
|x + y| = |x| + |y| = x + y
Case 2: x > 0, y < 0
|x| = x
|y| = -y
|x + y|  |x| < |x| + |y| or |x + y|  |y|  |x| + |y|
Case 3: y > 0, x < 0
analogous to case 2
Case 4: x < 0, y < 0
|x| = -x
|y| = -y
|x + y| = |-(x + y)| = |-x + -y|
= | |x| + |y| | = |x| + |y|
p. 77 # 73 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
Proof
(1)  (2): If n is even then n = 2k for some k  Z.
Then, n + 1 = 2k + 1 is an odd number
Definition of odd.
(2)  (3): Suppose n + 1 is odd.
Clearly, 2n is even.
Then n + 1 + 2n = 3n + 1 is odd since the sum of an even number and an odd number is odd.
(3)  (4): Suppose 3n + 1 is odd.
Then, 3n + 1 = 2j + 1 for some j  Z.
Definition of odd.
Subtracting 1 from both sides of the equation, we get 3n = 2j so 3n is even by the definition of
an even number.
(4)  (1): Let’s do this one by an indirect proof.
Assume n is odd.
Then n = 2r + 1 for some r  Z.
Definition of odd
Then, 3n = 3(2r + 1) = 6r + 3 = 2(3r + 1) + 1.
3n = 2s + 1 for some s  Z.
Therefore, 3n is odd.
Algebra
Algebra
Definition of odd
For every integer n there is an even integer x such that x > n (Constructive Proof)
Case 1: n  0
Case 2: n > 0
Let x = 2
Let x = 2n which must be even and > n
Since 2 > 0  n and x is even
For every positive integer n there is a prime number p such that p > n
(Nonconstructive Proof)
Let n be a positive integer, and consider the number x = n! + 1
Case 1: x is a prime number. Since x > n we’re done.
Case 2: x is a composite number. Thus there is a prime number q such that q is a divisor of x
Claim: q > n. Consider any prime number p < n. Do to the way x was constructed, if x is
divided by p then p divides n! evenly so that there is a remainder of 1 when x is divided by p.
Therefore, the prime number q that divides x must be greater than n
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For sets A and B, A  B if and only if B  A
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Part 1: Prove if A  B then B  A
Part 2: Prove if B  A then A  B
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Let x  B
Let y  A
Then, x  B
Defn of complement
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Thus, x  A
Given that A  B
Then, y  A
Defn of complement
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_
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Then, x  A
Defn of complement
Then, y  B
Given B  A
_ _
Then, y  B
Defn of complement
Therefore, B  A Definition of subset
Thus, A  B
Defn of subset
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Page 95 # 15 Prove A – B = A  B
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Part 1: Prove A – B  A  B
Part 2: Prove A  B  A – B
Let x  A – B
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x  A and x  B Defn. of difference
Let y  A  B
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x  A and x  B Defn of complement
x  A and x  B
Defn. of intersection
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x  A and x  B
Defn of complement
xAB _
Defn. of intersection
xA–B
Defn. of set difference
A–BAB
Defn of subset
ABA–B
Defn. of subset
Therefore the two sets are equal by definition of set equality.