If a is a prime number, prove that √a is not a rational number. (You may assume the uniqueness of prime factorization.)
Per the text: A positive integer a is said to be prime if a > 1 and whenever a is written as the product of two positive integers, a=n*m, one of the factors is a.
[I assume what this means is that if a=n*m, then n=a or m=a.]
The Attempt at a Solution
Proof by contradiction. Assume √a is rational. Then √a = (b/c), where b and c are positive integers with no common factors. So a=(b/c)^2=(b/c)*(b/c). Using the definition of a prime from above, a=(b/c) and a*c=b. Using substitution, a=((a*c)/c)*((a*c)/c), which reduces to a=a*a. Multiply both sides by (1/a) to get 1=a, which contradicts the definition of a prime that a > 1. So √a is not rational.
I have seen other proofs that the square root of a prime is irrational, but they usually end up with a*(c^2)=(b^2) and use the language "we know a divides b^2, and therefore a must divide b," or something similar. I'm trying to stay close to the examples and materials presented so far in my text, and nothing about assuming an integer divides another integer has come up, so I'm not sure if that's something I can assume yet. Is the above proof, where I say a=n*m means a=n or a=m, correct? I don't think I used the uniqueness of prime factorization as mentioned in the question, so I'm worried I'm missing something.
Thank you for reading, and for any insight you may provide.