Exercise-1.4, Class 9th, Maths, Chapter 1, NCERT

1. Classify the following numbers as rational or irrational:

(i) $2-\sqrt{5}$
Answer. $\sqrt{5}$ is irrational, and subtracting an irrational from a rational (2) gives an irrational. So $2-\sqrt{5}$ is irrational.

(ii) $(3+\sqrt{23})-\sqrt{23}$
Answer. $(3+\sqrt{23})-\sqrt{23}=3$, which is ${\displaystyle \frac{3}{1}}$, so rational.

(iii) ${\displaystyle \frac{2\sqrt{7}}{7\sqrt{7}}}$
Answer. Simplify:

$$\frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7}\cdot \frac{\sqrt{7}}{\sqrt{7}}=\frac{2}{7}$$

so it is rational.

(iv) ${\displaystyle \frac{1}{\sqrt{2}}}$
Answer. $\sqrt{2}$ is irrational; $1\mathrm{/}\sqrt{2}$ is therefore irrational.

(v) $2\pi$
Answer. $\pi$ is irrational; any nonzero rational multiple of an irrational is irrational. So $2\pi$ is irrational.

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2. Simplify each of the following expressions:

(i) $(3+\sqrt{3})(2+\sqrt{2})$
Answer. Expand:

$$(3+\sqrt{3})(2+\sqrt{2})=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\mathrm{.}$$

(ii) $(3+\sqrt{3})(3-\sqrt{3})$
Answer. Use $a^2-b^2$:

$$(3+\sqrt{3})(3-\sqrt{3})=3^2-(\sqrt{3}{)}^2=9-3=6$$

(iii) $(\sqrt{5}+\sqrt{2}{)}^2$
Answer. Use $(a+b{)}^2$:

$$(\sqrt{5}+\sqrt{2}{)}^2=5+2\sqrt{10}+2=7+2\sqrt{10}\mathrm{.}$$

(iv) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$
Answer. Use $a^2-b^2$:

$$(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=5-2=3$$

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3. Recall, π is defined as the ratio of the circumference (say $c$) of a circle to its diameter (say $d$), i.e. $\pi ={\displaystyle \frac{c}{d}}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?
Answer. There is no contradiction. Defining $\pi$ as the ratio $c\mathrm{/}d$ does not force $c\mathrm{/}d$ to be rational — $c$ and $d$ are real numbers (lengths). A ratio of two real numbers can be irrational (and in fact the ratio of circumference to diameter for a true circle is the particular real number $\pi$, which is irrational). In short: being a ratio of lengths does not imply rationality.

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4. Represent ${\displaystyle \sqrt[3]{9}}$ on the number line.
(Note: the printed exercise shows “Represent 9 3. on the number line.” I interpret this as the cube root $\sqrt[3]{9}$; if you meant something else, tell me and I’ll adjust.)

Answer. $\sqrt[3]{9}$ is the unique positive real $x$ such that $x^3=9$. We know $2^3=8$ and $3^3=27$, so $\sqrt[3]{9}$ lies between 2 and 3. Numerically,

$$\sqrt[3]{9}\approx 2.080083823051904\dots$$

(Computed using Newton–Raphson: iteration $x_{n+1}=\frac{2x_n+9\mathrm{/}{x}_n^2}{3}$Starting with $x_0=2$ gives rapid convergence: $2.0833333$, $2.0800889$, $2.08008382306$, ….)

To place it on the number line: mark 2 and 3, then mark the point about $0.0800838$ units to the right of 2 (or use a compass/scale to position the coordinate $2.080083823$). For a geometric construction one may use higher-level tools (or iterative numerical construction) — the decimal approximation above is sufficient for plotting.

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5. Rationalise the denominators of the following:

(i) ${\displaystyle \frac{1}{\sqrt{7}}}$
Answer.

$$\frac{1}{\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}\mathrm{}$$

(ii) ${\displaystyle \frac{1}{\sqrt{7}-\sqrt{6}}}$
Answer. Multiply by the conjugate $\sqrt{7}+\sqrt{6}$:

$$\frac{1}{\sqrt{7}-\sqrt{6}}\cdot \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}\mathrm{}$$

(iii) ${\displaystyle \frac{1}{\sqrt{5}+2}}$
Answer. Multiply numerator and denominator by $\sqrt{5}-2$:

$$\frac{1}{\sqrt{5}+2}\cdot \frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{(\sqrt{5}{)}^2-2^2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2$$

(iv) ${\displaystyle \frac{1}{\sqrt{7}-2}}$
Answer. Multiply by $\sqrt{7}+2$:

$$\frac{1}{\sqrt{7}-2}\cdot \frac{\sqrt{7}+2}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}\mathrm{.}$$