# Thread: proof of infinite n's s.t. sqrt(n) = irrational

1. ## proof of infinite n's s.t. sqrt(n) = irrational

Hey; I need help with the following problem:

"Prove that there are infinitely many positive integers 'n' such that sqrt(n) is irrational."

I think I'm supposed to do this with a proof by contradiction, but I'm even struggling with that distinction... which part do I negate? the infinite part or the irrational part?

I'm really struggling with my proofs involving irrationality... if anyone could give general tips concerning irrationals in proofs, I would also greatly appreciate it.

Thanks,

Martin

2. The negation would be "there exist a finite number of integers whose squareroot is irrational".

Just saying "there exist an infinite number of integers whose squareroot is rational" won't help because that is true! There exist an infinite number of integers whose squareroot is rational and an infinite number of integers whose squareroot is irrational.

But I don't know if a proof by contradiction is the best way to go. It would be sufficient to show that the squareroot of any prime is irrational.

3. Originally Posted by mart1n
Hey; I need help with the following problem:

"Prove that there are infinitely many positive integers 'n' such that sqrt(n) is irrational."

I think I'm supposed to do this with a proof by contradiction, but I'm even struggling with that distinction... which part do I negate? the infinite part or the irrational part?

I'm really struggling with my proofs involving irrationality... if anyone could give general tips concerning irrationals in proofs, I would also greatly appreciate it.

Thanks,

Martin
As hallsoflvy already mentioned, it suffices to prove that Sqrt(p) is irrational for any prime p. The following are highlights of the proof:

== Supose Sqrt(p) = a/b, with a, b coprime integers (i.e., the fraction a/b is reduced)

== It follows that p*b^2 = a^2, and using the decomposition of naturals in product of primes and the definition of prime, deduce p divides a^2 and thus a ==> a = p^n*x, with (p,x) = 1 (i.e., n is the maximal power of p that divides a). But then

p = a^2/b^2 = p^(2n)*x^2/b^2 ==> b^2 = p^(2n-1)*x^2

== Deduce now that p divides both a, b contradicting that a/b is reduced.

As you can see, the above is basically the same argument used to prove that Sqrt(2) is not rational.

Tonio

4. Originally Posted by mart1n
... if anyone could give general tips concerning irrationals in proofs, I would also greatly appreciate it.
To prove a number is irrational, it often helps to start with trying to prove that it can not be expressed in the form $\displaystyle m/n$ where m, n are both integers.

It's usual to suppose that m and n have no common factors greater than 1, i.e. that the fraction so expressed is in its simplest form.

What is also common is to assume that such a number *can* be so expressed, and then try and prove a contradiction.

The proof that $\displaystyle \sqrt 2$ is irrational is a classic. It's worth learning, like good poetry is worth learning.

5. Thanks a bunch, dudemasters.

6. sqrt(p) is irrational for every prime p (because the only possible rational roots of polynomail x^2 - p are 1,-1,p,-p by rational roots theorem, thus sqrt(p) which is root of that polynomial must be irrational) ... and since there are infinite primes p, there are infinite sqrt(p) irrational numbers

7. Originally Posted by HallsofIvy
But I don't know if a proof by contradiction is the best way to go. It would be sufficient to show that the squareroot of any prime is irrational.
Another approach: If $\displaystyle \sqrt{2}$ is irrational then for all $\displaystyle k\in\mathbb{N}$, $\displaystyle k\sqrt{2}=\sqrt{2k^2}$ is irrational too. Thus it is sufficient to show that $\displaystyle \sqrt{2}\not\in\mathbb{Q}$.

8. Oh, that's nice!

9. Originally Posted by flyingsquirrel
Another approach: If $\displaystyle \sqrt{2}$ is irrational then for all $\displaystyle k\in\mathbb{N}$, $\displaystyle k\sqrt{2}=\sqrt{2k^2}$ is irrational too. Thus it is sufficient to show that $\displaystyle \sqrt{2}\not\in\mathbb{Q}$.
Sweet, neat, petite and oh so elegant.