Comparison between Rational and Irrational Numbers

Rational numbers are those which can be written in ‘\(\frac{p}{q}\)’ form where ‘p’ and ‘q’ belong to integers and ‘q’ is not equal to zero. The decimal numbers which are terminating and non- repeating fall under the category of rational numbers. On the other hand, irrational numbers can’t be written in ‘\(\frac{p}{q}\)’ form because they are non- terminating and non- repeating decimals. We can easily make comparison between rational numbers by simply comparing numerators of the rational fractions (in case of like rational fractions), while by taking L.C.M. and then comparing the numerators (in case of unlike rational fractions).

In the previous topic, we have seen how to make comparison between irrational numbers. In this topic we will get to know about the comparison between rational and irrational numbers.

The concept can be understood in a better way by having look at below given solved examples:

1. Compare 2 and \(\sqrt{3}\).

Solution:

 To compare the given numbers, let us first find out the square of both the numbers and then proceed with the comparison. So,

2\(^{2}\)= 2 x 2 = 4.

\((\sqrt{3})^{2}\)  = \(\sqrt{3}\) x \(\sqrt{3}\) = 3.

Since, 4 is greater than 3. 

So, 2 is greater than \(\sqrt{3}\).

2. Compare \(\frac{4}{3}\) and \(\sqrt{5}\)

Solution:

In the given numbers, one of them is rational while other one is irrational. To make the comparison, let us first make the given irrational number into rational number and then carry out the comparison. So, let us square both the given numbers. Hence,

\((\frac{4}{3})^{2}\) = \(\frac{4}{3}\)  x \(\frac{4}{3}\) = \(\frac{16}{9}\).

\((\sqrt{5})^{2}\) = \(\sqrt{5}\) x \(\sqrt{5}\)  = 5.

Now, let us take the L.C.M. of the two rational numbers so formed and compare them. So, we have to compare \(\frac{16}{9}\) and 5. The L.C.M. of 9 and 1 is 9. So, we have to make comparison between \(\frac{16}{9}\) and \(\frac{45}{9}\). Since, \(\frac{16}{9}\) is smaller than \(\frac{45}{9}\). 

So, \(\frac{16}{9}\) will be smaller than 5. 

Hence, \(\frac{4}{3}\) will be smaller than \(\sqrt{5}\).

3. Compare \(\frac{7}{2}\) and \(\sqrt[3]{7}\).

Solution:

In the given numbers for comparison, one of them is rational \(\frac{7}{2}\) while other one is irrational number \(\sqrt[3]{7}\). To make comparison between them, firstly we’ll make both numbers rational numbers and then comparison process will be carried out. So, to make both the numbers rational, let us find the cube of both the numbers. So,

\((\frac{7}{2})^{3}\) = \(\frac{7}{2}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) = \(\frac{343}{8}\).

\[(\sqrt[3]{7})^{3}\] = \(\sqrt[3]{7}\) x \(\sqrt[3]{7}\) x \(\sqrt[3]{7}\) = 7.

Now, L.C.M. of 1 and 8 is 8. So, the two numbers to be compared are \(\frac{343}{8}\) and \(\frac{56}{8}\). Now, the rational fractions have become like rational fractions. So, we just need to compare their numerators. Since, \(\frac{343}{8}\) is greater than \(\frac{56}{8}\). 

So, \(\frac{7}{2}\) is greater than \(\sqrt[3]{7}\).

4. Arrange the following in ascending order:

6, \(\frac{5}{4}\), \(\sqrt[3]{4}\), \(7^\frac{2}{3}\), \(8^\frac{2}{3}\).

Solution:

We have to arrange the given series in ascending order. To do so, let us first of all find the cube of all the elements of the given series. So,

(6)\(^{3}\) = 6 x 6 x 6 = 216.

\((\frac{5}{4})^{3}\) = \(\frac{5}{4}\) x \(\frac{5}{4}\) x \(\frac{5}{4}\) = \(\frac{125}{64}\).

\((\sqrt[3]{4})^{3}\) = \(\sqrt[3]{4}\) x \(\sqrt[3]{4}\) x \(\sqrt[3]{4}\) = 4.

\((7^\frac{2}{3})^{3}\) = \(7^\frac{2}{3}\) x \(7^\frac{2}{3}\) x \(7^\frac{2}{3}\) = 7\(^{2}\)= 49.

\((8^\frac{2}{3})^{3}\) = \(8^\frac{2}{3}\) x \(8^\frac{2}{3}\) x \(8^\frac{2}{3}\) = 8\(^{2}\) = 64.

Now we have to make the comparison between 216, \(\frac{125}{64}\), 4, 49, 64. 

This could be done by converting the series into like fractions and then proceeding. 

So, the series becomes:

\(\frac{13824}{64}\), \(\frac{125}{64}\), \(\frac{256}{64}\), \(\frac{3136}{64}\), \(\frac{4096}{64}\).

Arranging the above series in ascending order we get;

\(\frac{125}{64}\) < \(\frac{256}{64}\) < \(\frac{3136}{64}\) < \(\frac{4096}{64}\) < \(\frac{13824}{64}\).

So, the required series is:

\(\frac{5}{4}\) < \(\sqrt[3]{4}\) < \(7^\frac{2}{3}\) < \(8^\frac{2}{3}\) < 6.

9th Grade Math

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