Philosophy 4005-2 (Spring 2011)
Chris Pincock
Two mathematical proofs
I. Every triangle has interior angles summing to 180 degrees.
This proof assumes two things about angles involving parallel lines: (i) corresponding angles of parallel
lines are equal and (ii) alternate interior angles are equal. To see what this means consider parallel lines
A and B which are crossed by line C:
Angles a and a’ are examples of corresponding angles. So are b and b’. b and d’ are alternate interior
angles. So are c and a’.
Now consider any triangle ABC. We can extend side BC to D and draw a line CE parallel to AB at point C:
The total angle BCD is 180 degrees as this is a line. But angle ABC equals angle ECD (corresponding
angles). And angle BAC equals angle ACE (alternate interior angles). So, given that BCA + ACE + ECB = 180
and ABC = ECD and BAC = ACE, it follows that BCA + BAC + ABC = 180. This construction works for any
triangle, so we have shown our claim holds for all triangles.
II. There are infinitely many prime numbers.
The natural numbers are 1, 2, 3, … Prime numbers are natural numbers which are only divisible without
remainder by 1 and themselves. For example, 5 is a prime number, but 6 is not a prime number because
it is divisible by 2 and 3. Our proof assumes that every number is a product of prime numbers.
We show that there are infinitely many prime numbers. Our proof is a proof by contradiction. We first
assume that there are only finitely many prime numbers and show that this leads to a contradiction.
This shows that there are not only finitely many prime numbers, so there infinitely many prime
numbers.
Assumption: There are finitely many prime numbers. Call them p1, p2, …, pn. Consider the product of all
these prime numbers: p1 x p2 x … x pn. Call it p. This number p is clearly greater than all the primes on
our list. Now consider p+1. We want to show that if p1, p2, …, pn are all the primes, then the only
numbers which divide into p+1 without remainder at 1 and p+1 itself. To see why notice that each of the
p1, p2, …, pn leave a remainder of 1 when they are divided into p+1. What about some other number?
By assumption, it would not be a prime number, so it would be a product of some prime numbers on
our list. So, again, we would get some remainder. This shows that there is a number p+1 which is
divisible without remainder by only 1 and p+1. That is, p+1 is prime. But p1, …, pn were assumed to be
all the primes, so p+1 is not prime. Contradiction!
So, assuming there are only finitely many primes leads to a contradiction. Therefore, there are infinitely
many prime numbers.