Here, I am going to show a proof of why the sqrt(p) is irrational. The most popular example of this is the sqrt(2), but here I will apply that to all situations.
The hypothesis that the sqrt(p) is not rational is that for every rational number r that's an element of Q, there is no r^2 that equals p.
By contradiction, let us suppose that there exists an r that's an element of Q where r^2 that equals p. In doing so, we can put r into the simplified fraction of the integer m/n, where m and n have no common factors outside of 1. Because r^2 = p, r^2 = p = m^2/n^2. Thus, m^2 = pn^2. Thus, p is a factor of m^2. Because m and m^2 share prime common factors, p is a factor of m as well. In doing this, we also get n^2 = m^2/p^2, so p^2 is also divisible by n^2, and because of n and n^2 share common prime factors, p is a common factor of n as well. However, this contradicts the idea that m and n have no common factors because they both have factors of p, so the sqrt(p) is not irrational, so the sqrt(p) is irrational.
- Jarrod Kopczynski
