Math 2534 Solution Homework 5 on Proof
Follow the outline for proof write up. Justify your steps and use complete sentences.
A note on the method of contradiction:
Contradiction means
(x, P( x)  Q( x))  x P( x  Q( x))
To do a proof of contradiction you start with
contradiction to the truth of P(x).
Q( x) and develop a
Problem 1: Use Direct Proof: (definitions only)
Theorem: If a, b and c are natural numbers and a c and b d then ab cd .
Proof: Assuming that a c and b d, then by definition of divisible there exist integers k and q so
that ak = c and bq = d. Now consider cd and using commutative and associative laws we have
that cd = (ak)(bq) = (ab)(kq) = (ab)h where h = kq is an integer. So by definition of divisibility
we have that ab cd .
Problem 2: Use Direct proof: (use former theorems)
Theorem: For all natural numbers, if a and b are each prime numbers greater than 2,
then the product ab is odd.
Proof: All prime numbers greater than 2 have been proven to be odd. Therefore the product ab
is also odd since the product of two odd numbers is always odd.
(Sorry about number confusion here_ I originally had more problems and forgot to
correct numbers. )
Problem 6: Use proof by Contradiction: ( definitions only)
Theorem: For all natural numbers n, 6 does not divide 6n – 5.
Proof by contradiction: Assume that there exist at least one natural n so that 6 does divide
6n – 5. Then by definition of divisible there exist an integer q such that 6q = 6n – 5. Using
algebra manipulation we have that 6q – 6n = – 5
And 6p = – 5 where p is also an integer since p = q – n . Now solve for p and we see that
p = – 5/6 can not be an integer. Therefore we have a contraction to the statement that p is an
integer and we conclude that 6 does not divide 6n – 5 for any value on n.
Problem 7: Use proof by Contraposition: (use Theorems)
Theorem: If 7n3 + 2 is odd then n is odd for all natural numbers.
Proof by contraposition: For all natural numbers n, If n is even then 7n3 + 2 is even.
By a previous theorem we know that n2 = (n)(n) is also even since the product of any two even
integers is always even. This theorem is also applied to n3 = n2n where n2 and n are each even.
Now consider 7n3 + 2 .Since n3 is even we know that 7n3 is also even since The product of an
even and odd number will be even. . By adding 2 we will have that 7n3 + 2 is also even since
the sum of two even numbers will always be even. Since the contrapositive is true, the original
conjecture is also true and when 7n3 + 2 is odd, n is odd.
Problem 8: Use proof by Contraposition: (Use definitions only)
Theorem: If a does not divide b + c, then a does not divide b or does not divide c.
Proof by contraposition: If a divides b and a divides c, then a divides (b + c).
By definition of divisible there exist integers m and n so am = b and an = c.
Now consider b + c = am + an = a(m + n) = af where f = m + n is also an integer. Therefore by
the definition of divisible a divides (b + c). Since the contraposition is true its equivalent form
(the original conjecture ) is also true .
Problem 9: Direct proof: (use former theorems)
Theorem: For all natural numbers n2  n is even.
Proof: consider the number n2  n  n(n  1)  (n  1)n . Notice that it factors into consecutive
integers. We know from previous theorems that consecutive integers will have opposite parity.
Therefore we have that the product of
(n – 1)(n) will be even since the product of an even number and odd number is always even.
Problem 10: Use proof by Contradiction: (use definitions only)
Theorem: The quotient of a non-zero rational number divided by an irrational number is
always irrational.
Proof by contradiction. We will assume that the quotient is rational for at least one non-zero
rational number r and one irrational number w. Since r is rational, by definition of rational
a
r  for integers a, b and b  0 and since we have assumed the quotient to be rational we have
b
a
  c
r c
b
that there exist integers c and d so that  where d  0 Therefore we have that   
w d
w
d
a
  ab m
b
which gives us that w    
where m = ab is and integer and n = bc  0 is also an

 c  bc n
 
d 
integer. So by definition on rational we see that w must be rational. But this contradicts that w is
given to be irrational and this is a contradiction. So the quotient must be irrational.