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 argument 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 argument I suggested. In fancy mathematical jargon, there's an isomorphism between our arguements.

 

[micromass]

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 argument 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!

 

[Bacle2]

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