Math 235 - Dr. Miller - HW #5: Proof by Contradiction - SOLUTIONS
1. [15 pts - 5 each] Prove each of the following claims by contradiction. Be sure among other things to
set up your initial assumptions, acknowledge your contradiction, and conclude the proof correctly.
(a) Use the definition of > or < to prove that if a and b are negative real numbers, then a + b ≤ 0.
(You should use the fact that for all real numbers x, if x is negative, then −x is positive.)
Assume to the contrary that there exist negative real numbers a and b for which a + b > 0. By
definition of greater/less than, there exists a positive real number c for which 0 + c = a + b,
which becomes −a + c = b via algebra. But now, again using the fact that c is positive, the
definition of less than asserts that −a < b. However, because a is negative, −a is positive; the
fact that b is negative creates a contradiction, for we cannot have a positive number be less than
a negative number. Therefore, if a and b are both negative real numbers, it must be that a+b ≤ 0.
(b) Prove that for every prime number p,
√
p is irrational.
√
Suppose there exists a prime number p for which p is rational. Then by definition, there exist
√
integers a and b with b 6= 0 for which p = ab . Algebraically, this asserts that pb2 = a2 , which is
a positive integer greater than 1 because p > 1 and b is a non-zero integer. Therefore, pb2 can be
prime factored. Yet upon such factorization, the expression yields an odd number of prime factors p on the left-hand side but an even number of them on the right-hand side, contradicting the
Fundamental Theorem of Arithmetic: the prime factorization of pb2 must be unique. Therefore,
√
such a prime cannot exist and we conclude that for every prime number p, p is indeed irrational.
(c) Prove that log12 9 is irrational. (You should use the fact that log12 9 is positive.)
Proof: Suppose to the contrary that log12 9 is rational; we know that it is positive. Then there
exist positive integers a and b with b 6= 0 for which log12 9 = ab . By definition of logarithm, that
means that 12a/b = 9, so that algebra yields 12a = 9b . Because a and b are positive, 12a and
9b are both positive integers greater than 1, so we may prime factor them to obtain 22a3a = 32b .
Because a is positive, at least two 2s appear on the left-hand side, but none on the right-hand
side. This violates the Fundamental Theorem of Arithmetic, for we do not have a unique prime
factorization for the number 12a = 9b . Therefore, log12 9 must be irrational, as desired.
2. [3 pts] Try to prove that there exist infinitely many prime numbers of the form k 2 + 1 where k is an
integer via the same argument as our in-class proof of the infinitude of primes. At what stage does this
proof attempt break down? (You may use the fact that the product of any finite number of integers is
also an integer.)
Attempted Proof. Suppose not; i.e., suppose that there exist only finitely many primes of the indicated
form, and list them as p1 , p2 , . . . , pn where n is a fixed positive integer. (The subscript n is positive,
for the prime number 5 is of the form 22 + 1 where 2 is an integer.) Consider the number
!
!2
n
n
Y
Y
q=
pi + 1
or
q1 =
pi
+1
i=1
i=1
Because the product of any finite number of integers is also an integer and all our primes pi are positive,
this new number is a positive integer greater than 1. This means that it must have a prime factor, due
to the FTA. HERE IS WHERE WE MEET DIFFICULTY: the prime factor indicated need NOT be
one of those in our list, since our list does not include all possible primes, but only those of a special
form. Therefore, we are NOT forced into the contradiction of producing a prime number that ends up
being a factor of the number 1 (because we won’tQbe forced into using a prime that is already one of the
n
pi and therefore divides both q and the product i=1 pi ). As an aside, if you created the expression q,
you CANNOT conclude that q is not prime; there are perhaps many primes that are NOT of the form
k 2 + 1, and q is indeed NOT of that form. If you created the expression q1 , you CAN conclude that q1
is not prime, for it is of the form k 2 + 1, yet is larger than any listed prime of that form.