Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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 $1$.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.