John Pinto of Saddle Brook, New Jersey, USA, offers the following input concerning material on pages 226-233 of Math Proofs Demystified.

1) The Proof of the GCD Theorem seems to be a proof that if a GCD of x and y (call it g) does exist then it must have the following property:

1 <= g <= min (x,y)

However, it does not really prove that one actually exists. I think what needs to be done is the following:

a) Change the proof's premise and conclusion to be that any two positive integers (say x and y) have at least one common divisor and that any common divisor g must have the property

1 <= g <= min (x,y)

The proof in the book would remain almost unchanged to prove this.

b) Offer a new proof of the GCD theorem which makes use of the previous result and the fact (perhaps a new axiom that every non-empty finite set of integers has a unique greatest member)

2) In the proof of the Prime Factor Theorem there are two steps where the complete reasoning is not stated. In fact, they would make good candidates for "You'll get to justify this on the quiz"! The steps are the ones where you are trying to show that y and z are less than x. In both steps you are trying to show why they cannot be greater than x. The reason currently says (for the y case) "If y were greater than x, then x/y would be between 0 and 1, and would not be a positive integer". (It's similar for the z case.)

I think it would have been better to say something like "If y were greater than x, then x/y would be between 0 and 1. This would mean that, because z = x/y, z would be between 0 and 1, which cannot be the case because z is a positive integer." (Similarly for the other case.)

3) In the proof of Prime Factor Corollary, you skipped over a case that needs to be dealt with separately. In the proof you state that the number of prime factors m is greater than 1. However, after dividing the composite number n by the first prime factor p1, you immediately declare that the result is a product of primes. That is true only when m > 2. You left out the situation where m = 2. In that case, after dividing m by p1, you are left with a single prime, p2. That shows that in this case, m is divisible by a prime without any remainder.