Question

Is \\" alt="\sqrt{\frac{1}{4} } \\" align="absmiddle" class="latex-formula"> a rational number?
Yes or No
is $\sqrt{ \\\\\\\frac{3}{16}$ a rational number?
Yes or no
Explain:

Answer 1

A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. The number 8 is a rational number because it can be written as the fraction 8/1.

Answer 2

Answer:

Yes

No

Step-by-step explanation:

A rational number is a number that can be written in fraction form. Even the infinite decimal, if the infinite decimal can be written in a fraction form, we consider that as a rational number.

$\sqrt{ \frac{1}{4} }$

Any numbers that are inside the square root will be counted as irrational number. Unless if we are able to pull the number inside the square root all out.

From the square root, we are able to pull out 1 and 4.

$\sqrt{ \frac{1 \times 1}{2 \times 2} } \\ \frac{1}{2}$

Therefore, the square root of 1/4 is a rational number.

Next we have the square root of 3/16.

$\sqrt{ \frac{3}{16} }$

We can consider separately, for numerator and denominator.

For numerator which is 3, cannot be pulled out.

For denominator which is 4, can be evaluated as 2.

$\sqrt{ \frac{3}{2 \times 2} } \\ \frac{ \sqrt{3} }{2}$

The reason why I keep the evaluated square root of 2 as a denominator because of the properties.

$\sqrt{ \frac{x}{y} } = \frac{ \sqrt{x} }{ \sqrt{y} }$

So when we evaluate square root of 4 as a denominator, we will get 2 as a denominator.

Because there is still a square root which is square root of 3 and cannot be evaluated by pulling out.

Numbers inside Square Root are an approximately number. Unless can be evaluated by pulling the same two numbers.