Download Here are some proofs that were covered in class or in the discussion

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
_
Thus, x  A
Given that A  B
Then, y  A
Defn of complement
_
_
_ _
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
_
x  A and x  B Defn. of difference
Let y  A  B
_
_
x  A and x  B Defn of complement
x  A and x  B
Defn. of intersection
_
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.