Proving that \sqrt{p} is irrational

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In summary, the conversation is about a proof that shows the irrationality of the square root of a prime number. The proof involves assuming the contrary and then reaching a contradiction, proving the irrationality of the square root. The validity of the proof is confirmed by the conversation participants.
  • #1
Mandelbroth
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I'm aware of the standard proof.

What I'm wondering is why we can't just do the following. Given, I haven't slept well and I'm currently out of caffeine, so this one might be trivial for you guys.

Suppose, by way of contradiction, that ##\sqrt{p}=\frac{m}{n}##, for ##m,n\in\mathbb{Z}## coprime. Then, ##p=\frac{m^2}{n^2}##. Because ##p## is an integer, ##\frac{m^2}{n^2}## must be as well. However, because ##m## and ##n## are coprime, so are ##m^2## and ##n^2##. Thus, ##n^2=1## is necessary for ##\frac{m^2}{n^2}## to be an integer. But that means ##m^2=p##, and ##p## is prime. Thus, a contradiction is met and we see that ##\sqrt{p}## is irrational.

Is this valid? I'll probably figure out my mistake (if I made one) by the time I get back with adequate caffeination, but until then, I'd like to make sure I figure it out.

Thank you.

Edit: Gee, I guess it might be important to mention that ##p## is prime. :facepalm:
 
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  • #2
In my opinion, it's valid.
 
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  • #3
It looks OK to me.
 
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  • #4
Completely valid. The last step basically assumes that a prime cannot be a perfect square, which is true and fairly obvious. But if you really want to make the proof completely obvious, you can state: ##m^2 = p##. Hence ##m|p## (m divides p), which is a contradiction.
 
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  • #5
Mandelbroth said:
Edit: Gee, I guess it might be important to mention that ##p## is prime. :facepalm:

Actually, it doesn't matter that p is prime. It's valid for any +ve integer: the square root is either an integer or irrational; it cannot be a proper rational.
 
  • #6
Thank you, all.
 

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

A number is considered irrational if it cannot be expressed as a ratio of two integers. In other words, it cannot be written as a fraction with a finite number of digits. Examples of irrational numbers include pi and the square root of 2.

2. How do you prove that the square root of a prime number is irrational?

To prove that the square root of a prime number (p) is irrational, we assume the opposite - that it is rational. This means that there exist two integers, a and b, such that √p = a/b. Then, we square both sides to get p = a^2/b^2. Since p is a prime number, it can only be divided by 1 and itself. Therefore, the only possible values for a^2 and b^2 are 1 and p. However, this contradicts our initial assumption that a and b are integers, as p is not a perfect square. Thus, our assumption was wrong and the square root of a prime number is irrational.

3. Why is it important to prove that the square root of a prime number is irrational?

Proving that the square root of a prime number is irrational is important because it helps us understand the nature of numbers and their relationships. It also has applications in various fields, such as cryptography and number theory.

4. Can the proof for the irrationality of the square root of a prime number be applied to other numbers?

Yes, the proof for the irrationality of the square root of a prime number can be applied to other numbers. In general, the proof can be extended to any number that is not a perfect square. This includes non-prime numbers such as 6 and 10.

5. Is there a simpler way to prove that the square root of a prime number is irrational?

There are other proofs for the irrationality of the square root of a prime number, but they all involve similar principles and assumptions. Some may be more concise or easier to understand, but they ultimately lead to the same conclusion - the square root of a prime number is irrational.

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