How can I prove that the square root of a prime is irrational?

Question

Preferably through contradiction. So far, I have:

Assume the square root of some prime p is rational

Therefore, square root of p can be represented by the division of two non-canceling integers, a and b.

If we square both sides, we get p is equal to a squared over b squared.

....

Where do I go from here?

Answer 1

First, assume that there exists some prime number 'p' whose square root is a rational numer 'r'.

√p = r

Now square both sides:

p = r²

Since 'r' is a rational number, it can be represented as a fraction 'a/b', in which 'a' and 'b' are both positive, prime integers:

p = (a/b)² = (a/b)*(a/b)

*** You can clearly see that (a/b)² = (a/b)*(a/b) = p. But this means that 'p' isn't prime, because it has 'a/b' (an integer) as a divisor, so we have a contradiction of the given fact that 'p' is prime.

The girl below me literally copied her proof from this website:
http://www.everything2.com/index.pl?node_id=1458559

Answer 2

Same idea but a little different nonetheless

Assum sqrt(p) is rational for some prime p. Then we can write sqrt(p) = a/b for relatively prime positive integers a and b. Squaring both sides yields

p = (a^2)/(b^2)
p(b^2) = a^2.

Clearly the right hand side is a perfect square. So the LHS should be a perfect square as well. For this to happen p must be of the form p = c^2 for positive integral c. This contradicts the fact that p is prime. As a result, our assumption was false, and the square root of a prime is irrational.

Answer 3

Let p be any prime number.

Assume √p is a rational number. √p can therefore be written as a fraction, a/b where a and b are coprime integers. (That it can be written as a fraction comes from the definition of rational but to choose a and b to be coprime we require the fundamental theorem of arithmetic. If you don't know a proof of this I suggest you read the node for the sake of thoroughness.)

√p= a/b

b×√p=a

Ok next stage (this isn't related to the first bit.) Take the highest integer lower than √p and call this number c.(e.g. √5 is approximately 2.236 so c would be 2.)

b×(√p-c)=b×√p-b×c

Now b×√p is an integer and b×c is an integer so the result, let's call it d, must also be an integer. The next step is to multiply this result by √p

b×√p-b×c=d

d×√p=b×p- b×c×√p

Now b×p is an integer and c×(b×√p) is also an integer. Therefore d×√p is an integer.

d is less than b. {d=b(√p-c} but in choosing and a and b to be comprime we ensured that b was the smallest interger which when multiplied by √p gave an interger.

Voila the contradiction!

This proof not only covers primes but extendeds to all intergers with nonintergal square roots. (If p is a perfect square then (√p - c) is zero and the rest of the proof goes down the tube, unsurprisingly.)

Answer 4

I have not previously seen the proof that cocooo posted. A neat thing about it is that you can see at a glance why the argument fails for perfect squares. Very nice.