All rational numbers have the fraction type $$\frac a b,$$ where a and also b space integers($b\neq0$).

You are watching: Is a square root a rational number

My question is: for what $a$ and also $b$ does the fraction have reasonable square root? The straightforward answer would certainly be when both room perfect squares, yet if 2 perfect squares are multiplied through a usual integer $n$, the an outcome may not be two perfect squares. Like:$$\frac49 \to \frac 8 18$$

And intuitively, there is no factoring, $a=8$ and $b=18$ should qualify by some traditional to have a reasonable square root.

Once this is solved, can this be expanded to any degree that roots? favor for what $a$ and also $b$ walk the portion have reasonable $n$th root?

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edited Aug 25 "15 in ~ 12:19

Bart Michels
request Jul 20 "13 in ~ 16:44

user2386986user2386986
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A nice generalization of the basic theorem the arithmetic is the every reasonable number is uniquely represented as a product the primes raised to essence powers. For example:

$$\frac49 = 2^2*3^-2$$

This is the natural generalization the factoring integers to reasonable numbers. Optimistic powers are part of the numerator, negative powers part of the denominator (since $a^-b = \frac1a^b$).

When you take it the $n$th root, you division each power by $n$:

$$\sqrt2^p_2*3^p_3*5^p_5... = 2^p_2/n*3^p_3/n*5^p_5/n...$$

For example:

$$\sqrt\frac49 = 2^2/2*3^-2/2 = \frac23$$

In order for the powers to proceed being integers when we division (and hence the result a reasonable number), they should be multiples that $n$. In the instance where $n$ is $2$, that way the numerator and denominator, in their decreased form, space squares. (And for $n=3$, cubes, and also so on...)

In your example, when you main point the numerator and denominator by the exact same number, they proceed to it is in the very same rational number, just represented differently.

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$$\frac2*42*9 = 2^2+1-1*3^-2 = 2^2*3^-2$$

You appropriately recognize the crucial of factoring, despite you don"t really want to use it in her answer. However the most natural way to check if the fraction produced by separating $a$ by $b$ has a rational $n$th root, is to element $a/b$ and look in ~ the powers. Or, equivalently, reduce the fraction and identify if the numerator and also denominator room integers elevated to the power of $n$.