proving √n is irration (n=square free + integer)

Hi all!

*I have to show that *√n is irrational, where n is "any square free positive integer". Ive never worked with "Square free" numbers, but I understand the term (no divisor can be a square).

So I assume we proceed by contradiction, so √p=a/b ->p=(a/b)^2, now do we do the same "trick" where we find a common factor between a and b, or is there a different way to proceed because we are dealing with square free numbers instead of prime.

Thanks!

Re: proving √n is irration (n=square free + integer)

Re: proving √n is irration (n=square free + integer)

Pardon my naivety, but I have never worked with Integer square roots before, Let me give this a crack.

To answer your first question, if √n is rational, then √n=q/p, which is p√n=q, q is a positive integer, as wanted.

Now, the next one, if we expand we get k√n-k[√n]. k√n is an integer that we just proved above, and k[√n] is a positive smaller integer by definition.

So we have (√n-[√n])k is a positive integer.

The next part by definition, <0(√n-[√n])<1 so this multiplied by some k is <k.

Now the last part, lets expand, kn-k√n[√n], which by using the last 2 parts we know is a positive integer.

Now because k√n<k, then it cannot be rational, is this correct?

Re: proving √n is irration (n=square free + integer)

Quote:

Originally Posted by

**farmeruser1** Pardon my naivety, but I have never worked with Integer square roots before, Let me give this a crack.

To answer your first question, if √n is rational, then √n=q/p, which is p√n=q, q is a positive integer, as wanted.

Now, the next one, if we expand we get k√n-k[√n]. k√n is an integer that we just proved above, and k[√n] is a positive smaller integer by definition.

So we have (√n-[√n])k is a positive integer.

The next part by definition, <0(√n-[√n])<1 so this multiplied by some k is <k.

Now the last part, lets expand, kn-k√n[√n], which by using the last 2 parts we know is a positive integer.

Now because k√n<k, then it cannot be rational, is this correct?

That is correct. Well done.

If fact, this proves a stronger result than asked for.

For example, is not square-free. But we have shown that is not rational.