Question

How do you estimate $\sqrt{35}$ to the nearest hundredth? Is the estimated value irrational or rational? Why?

Answer 1

Hint: In this question we will first find the value of the square root of $35$. Then we will put the value of it to the nearest hundredth value which means we will round it up to $2$ decimal places. And then we will check whether the number is a rational number by checking whether the number can be expressed in the form of two integers $a$ and $b$, as $\dfrac{a}{b}$ where $b\ne 0$, if the number cannot be expressed, then it is an irrational number.

Complete step by step solution:
We have the number given to us as:
$\Rightarrow \sqrt{35}$
Now since there is no integer root of the number, we will use a calculator to get the value of the term.
On using the calculator, we get:
$\Rightarrow \sqrt{35}=5.16079$
Now in the question we have been told to estimate the number to the nearest hundredth. Now since the third decimal in the term is $0$ we don’t have to round the hundredths place; we can write the value as:
$\Rightarrow \sqrt{35}=5.16$
Now we will check whether this number is a rational or an irrational number.
We know that a number is rational if it can be expressed in the form of two integers $a$ and $b$ as $\dfrac{a}{b}$ where $b\ne 0$.
We have the number as:
$\Rightarrow 5.16$
Now since there are two decimal places in the number, to eliminate it we will multiply it with $100$ and to not change the value we will also divide it with $100$. We get:
$\Rightarrow \dfrac{516}{100}$
Now the number can be written as:
$\Rightarrow \dfrac{129\times 4}{25\times 4}$
On simplifying, we get:
$\Rightarrow \dfrac{129}{25}$
Now $129$ and $25$ are both integers and $b\ne 0$ therefore, it is a rational number.

Note:
It is to be remembered when a number cannot be expressed in the form of the quotient of two integers $a$ and $b$ as $\dfrac{a}{b}$ such that $b\ne 0$ then it is an irrational number. An irrational number cannot be expressed on the timeline. Common examples of irrational numbers are $\pi $ and $e$.