Here are some examples that will be covered in the discussion 4 this week. You may
want to look them over before going to discussion so you can ask questions if needed.
Any examples you don’t get to will be carried over to next week’s discussion.
Example 1: Prove that if n is an odd integer then there is an integer m such that n = 4m + 1 or
n = 4m + 3.
Example 2: Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd.
Example 3: If m + n is odd, then m is even or n is even.
Example 4: Use a contrapositive proof to show that the sum of a rational number and an
irrational number is irrational. That is, prove if i is irrational and r is rational, then s = i + r is
irrational.
Proof. Let r be a rational number and let i be an irrational number. (given)
Assume that r + i = s and that s is a rational number.
Then,  a, b, c, d  Z, c 0 and d  0 such that r = a/b and s = c/d defn. rational number
Thus, a/b + i = c/d
(substitution)
i = c/d - a/b = (bc - ad)/bd
(algebra)
Since the closure property tells us the product and difference of integers are integers, this
implies i is rational, a contradiction of the hypothesis.
Example 5: Prove the square of an odd integer equals 8k + 1 for some integer k.
Proof: Let n be an odd integer.
(given)
Then, n = 2r + 1 for some r  Z
(defn. of odd)
2
2
2
Then, n = (2r + 1) = 4r + 4r + 1 = 4(r)(r + 1) + 1
(algebra)
Either r or r + 1 must be even since they are consecutive integers
If r is even, then n2 = 4(2j)(r + 1) + 1 = 8j(r + 1) + 1 (defn. even, substitution, algebra)
If r+1 is even then n2 = 4(r)(2s) + 1 = 8rs + 1
(defn. even, substitution, algebra)
In either case, n2 has the form 8k + 1
Example 6: The sum of the squares of two odd integers cannot be a perfect square.
Note this is a proof by contradiction i.e. assume p and q and arrive at a contradiction.
Proof: Suppose m and n are odd and m2 + n2 = k2.
There are integers s and t such that m2 = 8s + 1 and n2 = 8t + 1
Previous example
Then, m2 + n2 = 8s + 8t + 2 = 2(4s + 4t + 1) = k2.
algebra
If k2 is even then k is even
previous proof
Then, k2 must have a factor of 4
square of even number has factor of 4
So, 2(4s + 4t + 1) = 4x or 4s + 4t + 1 = 2x
algebra
Thus, 4s + 4t + 1 is even, a contradiction.
contradiction is that a number can’t be
be both even and odd
Example 7: Prove both parts
a. Prove that if a|b and a|c then a|(b+c).
b. Prove that if a|b and c|d then ac|bd
Example 8: Prove there does not exist a smallest real number.
Example 9: Prove there does not exist a smallest positive real number.
Example 10: Use mathematical induction to prove 1•1! + 2•2! + … + n•n! = (n + 1)! –1 for n  1
Example 11: Use mathematical induction to prove that 1/(1•2) + 1/(2•3) + … + 1/[n(n+1)] = n/(n + 1)
(Substitution of the hypothesis is shown in dark red.)
Basis: n = 1 lhs = 1/1•2 = ½ rhs = 1/(1+1) = ½
Hypothesis: Assume 1/(1•2) + 1/(2•3) + … + 1/[k(k+1)] = k/(k + 1) for k > 1
Induction step: 1/(1•2) + 1/(2•3) + … + 1/[k(k+1)] + 1/[(k+1)(k+2]
= k/(k + 1) + 1/[(k+1)(k+2]
= [k(k+2) + 1]/[(k+1)(k+2)]
= [k2 + 2k + 1]/ [(k+1)(k+2)]
= (k+1)2/[(k+1)(k+2)] = (k+1)/(k+2)