Question

Arrange these numbers in ascending order $2\sqrt{3},4,\sqrt{15},3\sqrt{2}$.

Answer 1

Hint: Rewrite all the given terms in proper square root and then compare each of them. Arrange them in ascending order, i.e., from the smallest to the greatest one.

Complete step-by-step answer:
We have to arrange the given numbers $2\sqrt{3},4,\sqrt{15},3\sqrt{2}$ in ascending order.
We will rewrite each of the numbers in proper square root form and then compare them.

We will begin by writing $2\sqrt{3}$ in proper square root form. We know that $2=\sqrt{4}$.
Thus, we can rewrite $2\sqrt{3}$ as $2\sqrt{3}=\sqrt{4}\times \sqrt{3}=\sqrt{4\times 3}=\sqrt{12}$.

Now, we will write 4 in proper square root form. We know that $4=\sqrt{16}$.

Thus, we can rewrite 4 as $4=\sqrt{16}$.

We observe that $\sqrt{15}$ is already in proper square root form.

We will now write $3\sqrt{2}$ in proper square root form. We know that$3=\sqrt{9}$ .

Thus, we can rewrite $3\sqrt{2}$ as $3\sqrt{2}=\sqrt{9}\times \sqrt{2}=\sqrt{9\times 2}=\sqrt{18}$.

So, we will now compare all the terms $2\sqrt{3}=\sqrt{12},4=\sqrt{16},\sqrt{15},3\sqrt{2}=\sqrt{18}$.

We know that $a>b$ if any only if $\sqrt{a}>\sqrt{b}$.

Thus, we will rearrange the numbers $\sqrt{12},\sqrt{16},\sqrt{15},\sqrt{18}$ in ascending order.

Arranging the terms in ascending order, we have $\sqrt{12}<\sqrt{15}<\sqrt{16}<\sqrt{18}$.
Thus, we have $2\sqrt{3}<\sqrt{15}<4<3\sqrt{2}$.

Hence, by arranging the terms in ascending order, we have $2\sqrt{3}<\sqrt{15}<4<3\sqrt{2}$.

Note: We can also solve this question by calculating the exact value of each of the numbers in decimal form and then arranging them in ascending order. Numbers are said to be in ascending order when the smallest number is placed at the first and the largest number is placed at the last. However, numbers are said to be in descending order when they are arranged from the largest number to the smallest one. One must be careful about using greater than or less than sign while comparing the numbers.