Final answer:
If gcd(a, m) > 1, then a and m share a common factor greater than 1, which prevents the existence of an integer x such that ax ≡ 1 (mod m). Therefore, a is not invertible modulo m since it is not coprime with m.
Step-by-step explanation:
To prove that if gcd(a, m) > 1, then a is not invertible modulo m, consider the properties of the greatest common divisor (gcd) and the concept of multiplicative inverses. According to number theory, an integer a is invertible modulo m if there exists an integer x such that ax ≡ 1 (mod m). In other words, a and x are inverses if their product modulo m is 1.
However, for an integer to have an inverse modulo m, it is necessary that it is coprime to m, meaning that gcd(a, m) = 1. If gcd(a, m) > 1, then a and m share a common factor greater than 1, which implies that any product ax will also share this factor, thus ax cannot be congruent to 1 modulo m. Therefore, a cannot have a multiplicative inverse in this modulo system, confirming that a is not invertible modulo m.