Ok proof by contradiction here we go.

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, letâ€™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.)