Question

How do you compare \[\dfrac{16}{17}\] and \[\sqrt{\dfrac{9}{16}}\]?

Answer 1

Hint: From the question we are asked to compare the two fractions \[\dfrac{16}{17}\] and \[\sqrt{\dfrac{9}{16}}\]. So, we will simplify the second fraction and use the property that it divides the two fractions and check whether the resultant is greater than or less than 1. By finding whether the resultant is greater than or less than one, we can compare the two given fractions.

Complete step by step solution:
Firstly, for the second fraction we will simplify it by removing the square root by making it as a whole square of some other fraction.
We will make the second fraction which is \[\sqrt{\dfrac{9}{16}}\] as square root of \[{{\left( \dfrac{3}{4} \right)}^{2}}\]. So, after doing the simplification we convert the second fraction as follows.
\[\Rightarrow \sqrt{\dfrac{9}{16}}\]
\[\Rightarrow \sqrt{{{\left( \dfrac{3}{4} \right)}^{2}}}\]
So, we can also write the second fraction as follows.
\[\Rightarrow \dfrac{3}{4}\]
Now, we will use the basic mathematical operation that is division to the first fraction and the simplified second fraction and check whether the resultant is greater than or less than 1.
So, we get,
\[\Rightarrow \dfrac{\dfrac{16}{17}}{\dfrac{3}{4}}\]
\[\Rightarrow \dfrac{16}{17}\times \dfrac{4}{3}\]
\[\Rightarrow \dfrac{64}{51}>1\]
Here we got that result after dividing both the fractions as greater than 1. So, we know that when we divide two integers or fractions if the resultant is greater than 1 then the numerator will be greater than 1.
Therefore, from this we can conclude that \[\dfrac{16}{17}>\sqrt{\dfrac{9}{16}}\].

Note: We must be very careful in doing the calculations. We must know basic mathematical operations like division and its properties. We must not do mistake in calculation like for example in this step \[\dfrac{\dfrac{16}{17}}{\dfrac{3}{4}}\] if we write the next step as \[\dfrac{16}{17}\times \dfrac{3}{4}\] instead of \[ \dfrac{16}{17}\times \dfrac{4}{3}\] our solution will be wrong.