Proof for irrationality of roots of prime number !

A fellow living near my house asked me to prove
that roots of all prime numbers are always irrational.

I don't even have the idea of where to start !
Can I use Mathematical Induction ? (still no clear idea)..

Start at the biggest number.
I'm sure u find it anywhere in the net.

It's actually a contradiction, not an induction (there might be an induction but the standard way is an induction)

It's actually a proof that the square root of any whole number is either a whole number or an irrational number. A prime number cannot be a perfect square otherwise it's square root would be a factor of itself.

Say we have our number P which is not a pefect square (ie there exists no whole number p s.t. p^2 = P)

If its square root is rational then it can be expressed as A/B where

(A^2) / (B^2) = P

=> A^2 = P(B^2)

as P is prime and (you have to do something a little different for non primes, but I won't bother)

P|A^2

=> P|A

(the | symbol means "divides" so X|Y means there is a whole number n s.t. nX = Y)

P|A

so define
a = A / P (a is a whole number)

(Pa)^2 = P(B^2)

P(a^2) = B^2


so P|(B^2)

=> P|B

so as P|A and P|B A/B cannot be a fraction in it's lowest terms. So the square root of P cannot be expressed as a fraction in it's lowest terms and hence is an irrational number.

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It's actually a proof that the squate of any whole number is either a whole number or an irrational number

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I don't Get this...How can the Square of a Whole number be irrational ?

and then Is this condition valid for the any nth root
of a Prime number ?

Also if 1 is considered a Unique Prime number the root
of it is rational. So this is only valid for Prime numbers
greater than 1 right ?


also I find it difficult to understand
P|A^2 => P|A

But i agree that if P|A then P|A^2

Hope you would clarify this. :)

Sorry, my bad, I meant to say.

The square root of any whole number is either a whole number or is irrational.

So if a whole number is not a perfect square then it's square root is irrational.


Looking through it I've made loads of typos and stupid errors. I'll go back and correct them.

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and then Is this condition valid for the any nth root
of a Prime number ?

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yup, just go through the proof and replace ^2 with ^n

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Also if 1 is considered a Unique Prime number the root
of it is rational. So this is only valid for Prime numbers
greater than 1 right ?

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1 isn't a prime number. The definition of a prime number states that all prime numbers are greater than 1.

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P|A^2 => P|A

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I'll just do this for prime numbers, as it's all you're interested in.
First we split A into it's prime factors.

A = p(1) * p(2) * p(3) * ... (p(1), p(2), ... are all prime numbers)

so A^2 = p(1)^2 * p(2)^2 * p(3)^2 * ...

So if P|A^2 P must be one of the prime numbers {p(1), p(2), ...} As no other prime numbers divide A^2

So P|A

I hope that helps.

It did ! Thankyou very Much ! :)