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

• Oct 5th 2013, 08:52 PM
farmeruser1
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!
• Oct 6th 2013, 05:37 AM
Plato
Re: proving √n is irration (n=square free + integer)
Quote:

Originally Posted by farmeruser1
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).

Suppose that $\sqrt{n}\notin\mathbb{Z}^+$ but it is rational, then $\exists p\in\mathbb{Z}^+[p\sqrt{n}\in\mathbb{Z}^+]$, WHY?

By the well order of the positive integers, that is a least positive integer with at at property. Call it $k$.

The greatest integer in $\sqrt{n}$ is $\left\lfloor {\sqrt n } \right\rfloor$ and we know that $0 < \sqrt n - \left\lfloor {\sqrt n } \right\rfloor < 1$.

Can you explain why $\left( {\sqrt n - \left\lfloor {\sqrt n } \right\rfloor } \right)k\in\mathbb{Z}^+~?$

Can you explain why $\left( {\sqrt n - \left\lfloor {\sqrt n } \right\rfloor } \right)k

Can you explain why $\left( {\sqrt n - \left\lfloor {\sqrt n } \right\rfloor } \right)k\sqrt n\in\mathbb{Z}^+~?$

So we have found a smaller positive integer with the property of $k$ that is a contradiction.
• Oct 6th 2013, 07:14 AM
farmeruser1
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?
• Oct 6th 2013, 08:17 AM
Plato
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, $18=2\cdot 3^2$ is not square-free. But we have shown that $\sqrt{18}$ is not rational.