Question

Fill in the blanks with the correct symbol out of \[ < , > and = \] \[0 \ldots \dfrac{{ - 3}}{{ - 15}}\]

Answer 1

Hint: In this question we have to find the relation between two numbers. For that we are going to change the negative denominator into positive denominator and find the LCM of the denominators. Next, we compare two numbers. And also we are going to multiply in complete step by step solutions.

Complete step-by-step solution:
First we write each of the given numbers with a positive denominator.
$ = \dfrac{{ - 3}}{{ - 15}}$
Next, we multiply the denominator and numerator by -1 and we get,
$ = \dfrac{{ - 3 \times \left( { - 1} \right)}}{{ - 15 \times \left( { - 1} \right)}}$
$ = \dfrac{3}{{15}}$
Finally, we get the positive denominator.
Next, we find the LCM of the denominators 1 and 15. The LCM value is 15 and multiply given two numbers by LCM value 15.
We get,
$\left( {\dfrac{0}{1}} \right) = \left( {\dfrac{{0 \times 15}}{{1 \times 15}}} \right) = \left( {\dfrac{0}{{15}}} \right)$ and
$\left( {\dfrac{3}{{15}}} \right) = \left( {\dfrac{{3 \times 1}}{{15 \times 1}}} \right) = \left( {\dfrac{3}{{15}}} \right)$
The value is $\dfrac{0}{{15}}and\dfrac{3}{{15}}$. Here, the denominator value is the same. Next, we are compare the numerator values and we get,
Now, $0 < 3$
Then, we get $\dfrac{0}{{15}} < \dfrac{3}{{15}}$
Already we know that 0 divided by any value is zero.
Therefore, $\dfrac{0}{{15}} = 0.$
 Then we apply the above step and we get,
$0 < \dfrac{3}{{15}}$
Hence, $0 < \dfrac{{ - 3}}{{ - 15}}$

Therefore, the correct symbol of the given two numbers is less than relation(<).

Note: The given question is very simple to solve only when the mathematical expressions are correct. The students concentrate on writing the mathematical expression from the given information by clearly understanding the question. So, the students should be careful when doing calculations.
There is another little hard way to find the answer.
Our given numbers is $0{\text{ }} and {\text{ }}\dfrac{{ - 3}}{{ - 15}}$
First we take, $\dfrac{{ - 3}}{{ - 15}}$
$\dfrac{{ - 3}}{{ - 15}} = \dfrac{{ - 3 \times \left( { - 1} \right)}}{{ - 15 \times \left( { - 1} \right)}} = \dfrac{3}{{15}}$which is positive number.
Already we know that all positive numbers are greater than 0. This statement is true. So, we use the statement between two numbers $0,\dfrac{3}{{15}}$ and we get,
$0 < \dfrac{3}{{15}}$.
Hence, $0 < \dfrac{{ - 3}}{{ - 15}}$.