# Which of these numbers are rational: sqrt(1), sqrt(2), sqrt(65), sqrt(196), sqrt(225)?

Sep 20, 2017

$\sqrt{1}$, $\sqrt{196}$ and $\sqrt{225}$.

#### Explanation:

The question is, which number does not have a radical sign after you simplify it.

So... the square root of $1$ is $1$, so $\sqrt{1}$ is rational.
The square root of $2$ cannot be simplified further, because $2$ is not a perfect square. $\sqrt{2}$ is not rational.
$\sqrt{65} = \sqrt{5 \cdot 13}$. This still has a radical sign and we cannot simplify it further, so this isn't rational.
$\sqrt{196} = \sqrt{4 \cdot 49} = \sqrt{{2}^{2} \cdot {7}^{2}} = 14$
$\sqrt{196}$ is rational, because we get a whole number without a radical${.}^{1}$
$\sqrt{225} = \sqrt{25 \cdot 9} = \sqrt{{5}^{2} \cdot {3}^{2}} = 15$

$\sqrt{225}$ is rational, because we get a whole number without a radical.

So, the rational radicals are: $\sqrt{1}$, $\sqrt{196}$ and $\sqrt{225}$.

Footnote $1$: Not all rational numbers have to be whole ones. For example, $0. \overline{11}$ is rational, because it can simplify into a fraction. All rational numbers are by definition, a number that can simplify into a fraction. So, whole numbers are rational, but not all rational numbers are whole.