If p is prime prove that p^1/2 (root p) is irrational.
Generalize the proof that sqrt 2 is irrational. You can find several proofs here:
http://www.cut-the-knot.org/proofs/sq_root.shtml
You should be able to apply the proof to all square roots of prime integers.
To prove that the square root of a prime number, p, is irrational, we can generalize the proof for the irrationality of √2. The key idea is to assume that √p is rational and derive a contradiction.
Assume √p is rational, which means it can be expressed as a fraction a/b, where a and b are positive integers with no common factors other than 1. We can further assume that a/b is in its simplest form, meaning that a and b share no common factors.
Using this assumption, we have (√p)^2 = p = (a/b)^2. By squaring both sides, we get p = a^2/b^2.
Since p is prime, it cannot be expressed as the square of any integer other than 1 or p itself. Therefore, a^2/b^2 = p can only be true if a and b are both equal to 1 or both equal to p.
If a and b are both equal to p, then p^2 = p, which implies p = 1. However, p is a prime number, and by definition, a prime number is greater than 1. Hence, this case is not possible.
If a and b are both equal to 1, then p^2 = 1, which implies p = 1. However, we know that p is prime, and 1 is not considered a prime number. Therefore, this case is also not possible.
In both cases, we arrive at a contradiction, which shows that our assumption that √p is rational is false.
Hence, we conclude that the square root of a prime number, p, is irrational.
To prove that √p is irrational for a prime number p, we can use a proof by contradiction.
Assume that √p is rational. This means that it can be expressed as a fraction m/n, where m and n are positive integers and do not share any common factors other than 1.
Squaring both sides of the equation, we get p = (m/n)^2. Simplifying further, we have p = m^2/n^2.
From this equation, we can conclude that p is divisible by n^2. Since p is a prime number, its only divisors are 1 and itself. Therefore, n^2 must equal 1 or p (because n^2 cannot be less than 1 since n is a positive integer).
If n^2 = 1, then n = 1. But if n = 1, then m^2 = p, which implies that m^2 is equal to a prime number. This contradicts the fact that the square of an integer cannot be a prime number (except when m = 1 or m = -1).
If n^2 = p, then m^2 = p^2. However, this means that m^2 is divisible by p, and since m^2 = p^2, it implies that p is also a divisor of m^2. Again, this contradicts the fact that p is a prime number, and its only divisors are 1 and itself.
In both cases, we have arrived at a contradiction. Therefore, the assumption that √p is rational is false, and we can conclude that √p is irrational for any prime number p.
This proof can be generalized to square roots of other prime numbers. To find more proofs regarding the irrationality of square roots, including √2, you can refer to the link you provided - http://www.cut-the-knot.org/proofs/sq_root.shtml.