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Prove Square Root of 15 is Irrational

  • Thread starter luke8ball
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  • #1
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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?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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?
 
  • #4
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I think what you're saying is analogous to the arguement I suggested. In fancy mathematical jargon, there's an isomorphism between our arguements.
 
  • #5
22,097
3,280

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 [itex]\sqrt{2}\sqrt{2}[/itex]?
 
  • #6
22
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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!
 
  • #7
Bacle2
Science Advisor
1,089
10
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.
 
  • #8
22
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Thanks again!
 
  • #9
66
0
So you're saying that the product of irrational numbers is irrational?? Then what about [itex]\sqrt{2}\sqrt{2}[/itex]?

Lol, good point! :D
 

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