Prove Square Root of 15 is Irrational

In summary, the conversation discusses a proof for the irrationality of the square root of 15. The argument involves assuming the opposite and reaching a contradiction, as well as using the definition of prime numbers. It is also mentioned that the product of irrational numbers is not necessarily irrational.
  • #1
luke8ball
22
0

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?
 
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  • #2
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.
 
  • #3
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
I think what you're saying is analogous to the argument I suggested. In fancy mathematical jargon, there's an isomorphism between our arguements.
 
  • #5
luke8ball said:

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.

Reptillian said:
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 [itex]\sqrt{2}\sqrt{2}[/itex]?
 
  • #6
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
luke8ball said:
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
Thanks again!
 
  • #9
micromass said:
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
 

1. What does it mean for a number to be irrational?

A number is considered irrational if it cannot be written as a ratio of two integers, and therefore has an infinite number of non-repeating decimal places.

2. How do you prove that the square root of 15 is irrational?

To prove that the square root of 15 is irrational, we must assume the opposite, which is that the square root of 15 is rational. From there, we can use a proof by contradiction to show that this assumption leads to a contradiction, thus proving that the square root of 15 is indeed irrational.

3. Can you give an example of a proof by contradiction for the square root of 15 being irrational?

One example of a proof by contradiction for the square root of 15 being irrational is as follows: Assume that the square root of 15 is rational, and can be written as a fraction, a/b, where a and b are integers with no common factors. Then, a/b = √15, which can be rewritten as a^2 = 15b^2. This means that a^2 is a multiple of 15, and therefore a must be a multiple of 15. Thus, a = 15k, where k is an integer. Substituting this back into our equation, we get 15k^2 = 15b^2, which simplifies to k^2 = b^2. This means that b is also a multiple of 15, which contradicts our initial assumption that a and b have no common factors. Therefore, our assumption that the square root of 15 is rational must be false, and thus it is irrational.

4. Are there any other methods to prove the irrationality of the square root of 15?

Yes, there are other methods to prove the irrationality of the square root of 15. One method is to use the prime factorization of 15, which is 3*5. We can then show that the square root of 15 cannot be written as a ratio of two integers by showing that one of the factors, either 3 or 5, will have an odd exponent when we take the square root. Another method is to use the decimal expansion of the square root of 15 and show that it is a non-repeating and non-terminating decimal, which is a characteristic of irrational numbers.

5. How is the proof for the square root of 15 being irrational applicable to other square roots?

The proof for the square root of 15 being irrational is applicable to other square roots in a similar fashion. By assuming that the square root of any number is rational and using a proof by contradiction, we can show that the square root is actually irrational. However, the specific steps and methods used may vary depending on the number in question.

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