Proof for irrationality of roots of prime number !

[Active]

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)..

[TB]

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

[Sam Finch]

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.

[Active]

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

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.

[Sam Finch]

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.

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

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

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 ?

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

P|A^2 => P|A

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.

[Active]

It did ! Thankyou very Much !