# Prove Square Root of 15 is Irrational

luke8ball

## Homework Statement

Prove Square Root of 15 is Irrational

## The Attempt at a Solution

Here's what I have. I believe it's valid, but I want confirmation.

As usual, for contradiction, assume 15.5=p/q, where p,q are coprime integers and q is non-zero.

Thus, 15q2 = 5*3*q2 = p2

Since 5 and 3 are prime, they must divide p. However, since the lcm(5,3) = 15, it must be the case that 15 divides p. Thus, p=15k for some k.

Then 15q2 = 15*15*k2, so q2=15k2. By the same argument, this implies 15 divides q.

However, we have reached a contradiction, since we assumed that p and q were coprime.

Is this a valid argument?

Reptillian
If I understand the arguement correctly, it's saying...the square root of any prime number is necessarily irrational by definition of prime. Since sqrt(x*y)=sqrt(x)*sqrt(y), and since 5 and 3 are both prime, the square root of their product must be irrational. Makes sense to me.

luke8ball
Hmm, well one of my main questions is if it's valid to say that 15 divides p and q, with the lcm argument. Does that park make sense?

Reptillian
I think what you're saying is analogous to the arguement I suggested. In fancy mathematical jargon, there's an isomorphism between our arguements.

Staff Emeritus
Homework Helper

## Homework Statement

Prove Square Root of 15 is Irrational

## The Attempt at a Solution

Here's what I have. I believe it's valid, but I want confirmation.

As usual, for contradiction, assume 15.5=p/q, where p,q are coprime integers and q is non-zero.

Thus, 15q2 = 5*3*q2 = p2

Since 5 and 3 are prime, they must divide p. However, since the lcm(5,3) = 15, it must be the case that 15 divides p. Thus, p=15k for some k.

Then 15q2 = 15*15*k2, so q2=15k2. By the same argument, this implies 15 divides q.

However, we have reached a contradiction, since we assumed that p and q were coprime.

Is this a valid argument?

Yes, this is valid.

If I understand the arguement correctly, it's saying...the square root of any prime number is necessarily irrational by definition of prime. Since sqrt(x*y)=sqrt(x)*sqrt(y), and since 5 and 3 are both prime, the square root of their product must be irrational. Makes sense to me.

So you're saying that the product of irrational numbers is irrational?? Then what about $\sqrt{2}\sqrt{2}$?

luke8ball
Hahaha, thanks for your help. The only reason I was confirming is that my professor essentially said what you said, and we're supposed to give another type of argument..

I just wanted to make sure that if 3 divides p and 5 divides p, 15 divides p. Lol, making sure I'm not making things up!

Hmm, well one of my main questions is if it's valid to say that 15 divides p and q, with the lcm argument. Does that park make sense?

Yes. Try to do it separately for 3 and for 5, to show that 3|p and 5|p . Then

p=3t , p=5s , so 5s=3t . But, as you said , since gcd(3,5)=1, we must have 3|s, so

s=3q , so p=5(3q).

A more general result is that the square root of a rational number x is rational iff x is a perfect square.

luke8ball
Thanks again!

Reptillian
So you're saying that the product of irrational numbers is irrational?? Then what about $\sqrt{2}\sqrt{2}$?

Lol, good point! :D