Problems on Irrational Numbers

Till here we have learnt many concepts regarding irrational numbers. Under this topic we will be solving some problems related to irrational numbers. It will contain problems from all topics of irrational numbers.

Before moving to problems, one should look at the basic concepts regarding the comparison of irrational numbers.

For comparing them, we should always keep in mind that if square or cube roots of two numbers (‘a’ and ‘b’) are to be compared, such that ‘a’ is greater than ‘b’, then a$^{2}$ will be greater than b$^{2}$ and a$^{3}$ will be greater than b$^{2}$ and so on, i.e., n$^{th}$ power of ‘a’ will be greater than n$^{th}$ power of ‘b’.

The same concept is to be applied for the comparison between rational and irrational numbers.

So, now let’s have look at some problems given below:

1. Compare √11 and √21.

Solution:

Since the given numbers are not the perfect square roots so the numbers are irrational numbers. To compare them let us first compare them into rational numbers. So,

(√11)$^{2}$ = √11 × √11 = 11.

(√21)$^{2}$ = √21 × √21 = 21.

Now it is easier to compare 11 and 21.

Since, 21 > 11. So, √21 > √11.

2. Compare √39 and √19.

Solution:

Since the given numbers are not the perfect square roots of any number, so they are irrational numbers. To compare them, we will first compare them into rational numbers and then perform the comparison. So,

(√39)$^{2}$ = √39 × √39 = 39.

(√19)$^{2}$ = √19 × √19 = 19

Now it is easier to compare 39 and 19. Since, 39 > 19.

So,√39 > √19.

3. Compare $\sqrt[3]{15}$ and $\sqrt[3]{11}$.

Solution:

Since the given numbers are not the perfect cube roots. So, to make comparison between them e first need to convert them into rational numbers and then perform the comparison. So,

$(\sqrt[3]{15})^{3}$ = $\sqrt[3]{15}$ × $\sqrt[3]{15}$ × $\sqrt[3]{15}$ = 15.

$(\sqrt[3]{11})^{3}$ = $\sqrt[3]{11}$ × $\sqrt[3]{11}$ × $\sqrt[3]{11}$ = 11.

Since, 15 > 11. So, $\sqrt[3]{15}$ > $\sqrt[3]{11}$.

4. Compare 5 and √17.

Solution:

Among the numbers given, one of them is rational while other one is irrational. So, to make comparison between them, we will raise both of them to them to the same power such that the irrational one becomes rational. So,

(5)$^{2}$ = 5 × 5 = 25.

(√17)$^{2}$ = √17 x × √17 = 17.

Since, 25 > 17. So, 5 > √17.

5. Compare 4 and $\sqrt[3]{32}$.

Solution:

Among the given numbers to make comparison, one of them is rational while other one is irrational. So, to make comparison both numbers will be raised to the same power such that the irrational one becomes rational. So,

4$^{3}$= 4 × 4 × 4 = 64.

$(\sqrt[3]{32})^{3}$ = $\sqrt[3]{32}$ × $\sqrt[3]{32}$ × $\sqrt[3]{32}$ = 32.

Since, 64 > 32. So, 4 > $\sqrt[3]{32}$.

6. Rationalize $\frac{1}{4 + \sqrt{2}}$.

Solution:

Since the given fraction contains irrational denominator, so we need to convert it into a rational denominator so that calculations may become easier and simplified ones. To do so we will multiply both numerator and denominator by the conjugate of the denominator. So,

$\frac{1}{4 + \sqrt{2}} \times (\frac{4 - \sqrt{2}}{4 - \sqrt{2}})$

⟹ $\frac{4 - \sqrt{2}}{4^{2} - \sqrt{2^{2}}}$

⟹ $\frac{4 - \sqrt{2}}{16 - 2}$

⟹ $\frac{4 - \sqrt{2}}{14}$

So the rationalized fraction is: $\frac{4 - \sqrt{2}}{14}$.

7. Rationalize $\frac{2}{14 - \sqrt{26}}$.

Solution:

$\frac{2}{14 - \sqrt{26}} \times \frac{14 + \sqrt{26}}{14 + \sqrt{26}}$

⟹ $\frac{2(14 - \sqrt{26})}{14^{2} - \sqrt{26^{2}}}$

⟹ $\frac{2(14 - \sqrt{26})}{196 - 26}$

⟹ $\frac{2(14 - \sqrt{26})}{170}$

So, the rationalized fraction is: $\frac{2(14 - \sqrt{26})}{170}$.