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If the radicand of a square root is a non-square (making the root an irrational), and if the non-square is either a prime number, or a composite number that does not have a square divisor (other than 1), does this mean that the square root is not divisible by an integral divisor (that it does not have an integral factor)?

For example, $\sqrt{200} = \sqrt{100\times2} = \sqrt{100}\times\sqrt{2}=10\sqrt{2}$, so $\sqrt{200}$ has an integral factor of 10 (is divisible by 10). However, $\sqrt{6} = \sqrt{3\times2} = \sqrt{3}\times\sqrt{2}$

I'm mainly wondering for the purpose of reducing fractions that contain radicals and integers.

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up vote 2 down vote accepted

The phrase "does not have an integral factor" is a little imprecise, since anything has an integral factor: we have $x=17\left(\frac{x}{17}\right)$.

So we rephrase the question. Suppose that $n$ is a square-free integer. Do there exist integers $d$ and $k$, with $k\gt 1$, such that $$\sqrt{n}=k\sqrt{d}?\tag{1}$$ Once we express the question like that, the answer is quick. Suppose to the contrary that there are integers $k,d$, with $k\gt 1$, such that (1) holds. Squaring both sides, we get $n=k^2d$, so $n$ is divisible by a square greater than $1$.

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Thanks for your answer but I only follow you up until statement (1). You say that $n$ is divisible by a square, but I'm wondering if $\sqrt{n}$ is divisible by k. Could you please elaborate? – yroc Sep 18 '13 at 18:10
By $\sqrt{n}$ divisible by $k\gt 1$, I assume you mean that there is an integer $d$ such that $\sqrt{n}=k\sqrt{d}$. Then the solution above says this cannot happen if $n$ is square-free. If you allow things like $\sqrt{6}=2\sqrt{3/2}$, then $\sqrt{n}$ is always divisible by any positive integer $k$. – André Nicolas Sep 18 '13 at 18:14
So you're saying that if $n$ is a square-free integer and $d$ is an integer, then integer $k$ does not exist? – yroc Sep 18 '13 at 18:22
There is then no integer $k\gt 1$ such that $\sqrt{n}=k\sqrt{d}$. Proof given in the post. – André Nicolas Sep 18 '13 at 18:23
You are welcome. It is a good thing to want clarity about the logic of an argument. – André Nicolas Sep 18 '13 at 18:47

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