Exercise 1.4 Practice
Operations on Real Numbers
Q1: Classify Numbers
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Classify the following numbers as rational or irrational:
(i) $2 - \sqrt{5}$
(ii) $(3 + \sqrt{23}) - \sqrt{23}$
(iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
(iv) $\frac{1}{\sqrt{2}}$
(v) $2\pi$
(i) $2 - \sqrt{5}$
(ii) $(3 + \sqrt{23}) - \sqrt{23}$
(iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
(iv) $\frac{1}{\sqrt{2}}$
(v) $2\pi$
(i) $2 - \sqrt{5}$: Difference of a rational and an irrational number is irrational.
(ii) $(3 + \sqrt{23}) - \sqrt{23}$: $3 + \sqrt{23} - \sqrt{23} = 3$. This is rational.
(iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$: $\frac{2}{7}$. This is rational.
(iv) $\frac{1}{\sqrt{2}}$: Quotient of a non-zero rational and an irrational number is irrational.
(v) $2\pi$: Product of a non-zero rational and an irrational number is irrational.
(i) Irrational, (ii) Rational, (iii) Rational, (iv) Irrational, (v) Irrational
Q2: Simplify Expressions
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Simplify each of the following expressions:
(i) $(3 + \sqrt{3})(2 + \sqrt{2})$
(ii) $(3 + \sqrt{3})(3 - \sqrt{3})$
(iii) $(\sqrt{5} + \sqrt{2})^2$
(iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
(i) $(3 + \sqrt{3})(2 + \sqrt{2})$
(ii) $(3 + \sqrt{3})(3 - \sqrt{3})$
(iii) $(\sqrt{5} + \sqrt{2})^2$
(iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
(i): $3(2) + 3(\sqrt{2}) + \sqrt{3}(2) + \sqrt{3}(\sqrt{2}) = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$.
(ii): Identity $(a+b)(a-b) = a^2 - b^2$.
$3^2 - (\sqrt{3})^2 = 9 - 3 = 6$.
$3^2 - (\sqrt{3})^2 = 9 - 3 = 6$.
(iii): Identity $(a+b)^2 = a^2 + 2ab + b^2$.
$(\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2 = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}$.
$(\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2 = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}$.
(iv): Identity $a^2 - b^2$.
$(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$.
$(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$.
(i) $6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$, (ii) 6, (iii) $7 + 2\sqrt{10}$, (iv) 3
Q3: Conceptual
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Recall, $\pi$ is defined as the ratio of the circumference (say $c$) of a circle to its diameter (say $d$). That is, $\pi = \frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?
There is no contradiction.
When we measure a length with a scale or any other device, we only obtain an approximate rational value.
Therefore, we may not realize that either $c$ or $d$ is irrational. The ratio of an irrational number to a rational number (or vice versa) is irrational.
Measurement is approximate; $c$ or $d$ is irrational.
Q4: Construction
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Represent $\sqrt{9.3}$ on the number line.
Step 1: Draw a line segment AB = 9.3 units.
Step 2: Extend AB to C such that BC = 1 unit. Total length AC = 10.3 units.
Step 3: Find the midpoint O of AC. Draw a semicircle with center O and radius OA.
Step 4: Draw a perpendicular at B intersecting the semicircle at D. The length BD is $\sqrt{9.3}$.
Step 5: Taking B as center and BD as radius, draw an arc intersecting the number line at E. Point E represents $\sqrt{9.3}$.
Construct semicircle on 10.3 units; perpendicular at 9.3 gives height $\sqrt{9.3}$.
Q5: Rationalisation
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Rationalise the denominators of the following:
(i) $\frac{1}{\sqrt{7}}$
(ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
(iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
(iv) $\frac{1}{\sqrt{7} - 2}$
(i) $\frac{1}{\sqrt{7}}$
(ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
(iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
(iv) $\frac{1}{\sqrt{7} - 2}$
(i) $\frac{1}{\sqrt{7}}$: Multiply by $\frac{\sqrt{7}}{\sqrt{7}}$.
= $\frac{\sqrt{7}}{7}$.
= $\frac{\sqrt{7}}{7}$.
(ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$: Multiply by $\frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}}$.
= $\frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}$.
= $\frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}$.
(iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$: Multiply by $\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$.
= $\frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}$.
= $\frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}$.
(iv) $\frac{1}{\sqrt{7} - 2}$: Multiply by $\frac{\sqrt{7} + 2}{\sqrt{7} + 2}$.
= $\frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}$.
= $\frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}$.
(i) $\frac{\sqrt{7}}{7}$, (ii) $\sqrt{7}+\sqrt{6}$, (iii) $\frac{\sqrt{5}-\sqrt{2}}{3}$, (iv) $\frac{\sqrt{7}+2}{3}$