Number Systems

NCERT Solutions • Class 9 Maths • Chapter 1
Exercise 1.4
1. Classify the following numbers as rational or irrational:
Solution:
  1. (i) $2 – \sqrt{5}$ Difference between a rational and an irrational number is always irrational.
    Classification: Irrational
  2. (ii) $(3 + \sqrt{23}) – \sqrt{23}$ Simplify: $3 + \sqrt{23} – \sqrt{23} = 3$
    $3$ can be written as $\frac{3}{1}$.
    Classification: Rational
  3. (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$ Simplify: $\frac{2}{7}$ (since $\sqrt{7}$ cancels out).
    Classification: Rational
  4. (iv) $\frac{1}{\sqrt{2}}$ Quotient of a non-zero rational number and an irrational number is irrational.
    Classification: Irrational
  5. (v) $2\pi$ Product of a non-zero rational number and an irrational number is irrational.
    Classification: Irrational
2. Simplify each of the following expressions:
Solution:
  1. (i) $(3 + \sqrt{3})(2 + \sqrt{2})$ Apply distributive property: $$ = 3(2 + \sqrt{2}) + \sqrt{3}(2 + \sqrt{2}) $$ $$ = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6} $$
  2. (ii) $(3 + \sqrt{3})(3 – \sqrt{3})$ Using identity $(a+b)(a-b) = a^2 – b^2$: $$ = (3)^2 – (\sqrt{3})^2 $$ $$ = 9 – 3 = \mathbf{6} $$
  3. (iii) $(\sqrt{5} + \sqrt{2})^2$ Using identity $(a+b)^2 = a^2 + b^2 + 2ab$: $$ = (\sqrt{5})^2 + (\sqrt{2})^2 + 2(\sqrt{5})(\sqrt{2}) $$ $$ = 5 + 2 + 2\sqrt{10} $$ $$ = \mathbf{7 + 2\sqrt{10}} $$
  4. (iv) $(\sqrt{5} – \sqrt{2})(\sqrt{5} + \sqrt{2})$ Using identity $(a-b)(a+b) = a^2 – b^2$: $$ = (\sqrt{5})^2 – (\sqrt{2})^2 $$ $$ = 5 – 2 = \mathbf{3} $$
3. Recall, $\pi$ is defined as the ratio of the circumference ($c$) of a circle to its diameter ($d$). That is, $\pi = \frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?
Solution: There is no contradiction.
Reason: When we measure a length (circumference $c$ or diameter $d$) with a scale or any other device, we only get an approximate rational value. We never obtain the exact mathematical value.
Therefore, we may not realize that either $c$ or $d$ is irrational. Since at least one of them is irrational, the ratio $\frac{c}{d}$ becomes irrational ($\pi$).
4. Represent $\sqrt{9.3}$ on the number line.
Solution: We represent this using geometric construction.

Steps of Construction:
  • Draw a line segment $AB = 9.3$ units.
  • Extend the line from point $B$ to $C$ such that $BC = 1$ unit. So, total length $AC = 10.3$ units.
  • Find the midpoint of $AC$ and name it $O$. Draw a semicircle with center $O$ and radius $OA$.
  • Draw a line perpendicular to $AC$ at point $B$, intersecting the semicircle at point $D$.
  • The length $BD$ represents $\sqrt{9.3}$.
  • Treating the line as a number line with $B$ as zero (0), place a compass at $B$, measure $BD$, and draw an arc to intersect the number line at point $E$.
  • Point $E$ represents $\sqrt{9.3}$ on the number line.
5. Rationalise the denominators of the following:
Solution:
  1. (i) $\frac{1}{\sqrt{7}}$ Multiply numerator and denominator by $\sqrt{7}$: $$ \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \mathbf{\frac{\sqrt{7}}{7}} $$
  2. (ii) $\frac{1}{\sqrt{7} – \sqrt{6}}$ Multiply by the conjugate $(\sqrt{7} + \sqrt{6})$: $$ \frac{1(\sqrt{7} + \sqrt{6})}{(\sqrt{7} – \sqrt{6})(\sqrt{7} + \sqrt{6})} $$ $$ = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 – (\sqrt{6})^2} $$ $$ = \frac{\sqrt{7} + \sqrt{6}}{7 – 6} = \mathbf{\sqrt{7} + \sqrt{6}} $$
  3. (iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$ Multiply by the conjugate $(\sqrt{5} – \sqrt{2})$: $$ \frac{1(\sqrt{5} – \sqrt{2})}{(\sqrt{5} + \sqrt{2})(\sqrt{5} – \sqrt{2})} $$ $$ = \frac{\sqrt{5} – \sqrt{2}}{5 – 2} = \mathbf{\frac{\sqrt{5} – \sqrt{2}}{3}} $$
  4. (iv) $\frac{1}{\sqrt{7} – 2}$ Multiply by the conjugate $(\sqrt{7} + 2)$: $$ \frac{1(\sqrt{7} + 2)}{(\sqrt{7} – 2)(\sqrt{7} + 2)} $$ $$ = \frac{\sqrt{7} + 2}{(\sqrt{7})^2 – (2)^2} $$ $$ = \frac{\sqrt{7} + 2}{7 – 4} = \mathbf{\frac{\sqrt{7} + 2}{3}} $$
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