Prove, if p and q are distinct prime numbers

[ayusuf]

Prove, if p and q are distinct prime numbers, then sqrt(p/q) is irrational. I know how to prove that if p is a distinct prime number, then sqrt(p) is irrational. From there let sqrt(p) = q/r and then prove but for this I'm stuck. Do we let sqrt(p/q) = (a/b)/(c/d).Thanks.

 

[VeeEight]

Assume not and that sqrt(p/q) is equal to a/b, for integers a, b and in lowest terms..
Then you can say a²q = b²p. This implies that p divides a²q, but q is prime, so p divides a². Since p is prime and divides a²=(a)(a), we know it divides a. So write a as a multiple of p and write out your equation and rearrange to isolate b² (or b). Can you find a contradiction here?

 

[ayusuf]

So from a²q = b²p which implies that p divides a². I understand up to there but when you say write a as a multiple of p I get confused. Could you elaborate a little bit more on that? Thanks.

 

[VeeEight]

If p divides a² then p divides a (why? There is the famous fact that if p is prime and p divides ab then p divides a or p divides b).
So since p divides a, we can write out a=pc for some integer c.
Now, back to the original equation a²q = b²p
a=pc implies that p²c²q = b²p
Now, isolate b² in this equation. Can you say anything about p dividing b here?

 

[ayusuf]

Okay now I understand what you meant by multiple but where are you getting that famous fact from? Okay so now if we isolate b² in this equation we will have
b² = pc²q and by taking the square root of both sides we'll have
b = sqrt(pc²q) = csqrt(pq).

Why would p divide b?

 

[VeeEight]

You have correctly isolated b² but you do not need to reduce it.
b² = pc²q means that b² is a multiple of p, which means that p divides b².
This implies that p divides b.
Now p divides a AND p divides b, contradicting that a/b was in lowest terms.

The famous fact (that prime p divides ab implies p divides a or p divides b) will be in the first few chapters of any number theory book. Here is a link: http://en.wikipedia.org/wiki/Euclid's_lemma. This fact is very important in number theory and even in group theory & ring theory so I would advise that you understand it.

 

[ayusuf]

The reason your saying b² is a multiple of p is because c/d = k and if we isolate c we get c = dk. So in this c is b², d is p and k is c²q right?

also how do go from p dividing b² to p dividing b? Thanks.

 

[VeeEight]

The reason your saying b² is a multiple of p is because c/d = k and if we isolate c we get c = dk. So in this c is b², d is p and k is c²q right?

You yourself found the equation b² = pc²q, which is the same thing as b² = p(c²q), which means that b² is equal to p times something. It doesn't get much clearer than that.

also how do go from p dividing b² to p dividing b? Thanks.

b² = bb and using the famous fact we see that p divides b.

 

[ayusuf]

Okay for the first part kind of what I was saying that c = dk. It's like saying 12 = 6*k where k = 2 right? I know it's pretty clear but I just want to make sure I fully digest this stuff.

Okay now for p divides b² because according to Euclid's lemma it must divide one of the factors right? Thanks. I really appreciate your help.

 

[VeeEight]

Okay for the first part kind of what I was saying that c = dk. It's like saying 12 = 6*k where k = 2 right? I know it's pretty clear but I just want to make sure I fully digest this stuff.

Yup.

Okay now for p divides b² because according to Euclid's lemma it must divide one of the factors right? Thanks. I really appreciate your help.

Yes. p is prime and divides bb so by the lemma it divides b.