Number Systems — NCERT Solutions, AI Tutor & Practice

Definition: A number is rational if it can be written as $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$.

Checking zero: $0 = \frac{0}{1} = \frac{0}{2} = \frac{0}{5} = \dots$

Here $p = 0$ and $q$ can be any non-zero integer. Both are integers and the denominator is not zero.

Conclusion: Yes, zero is a rational number. It can be expressed in infinitely many ways as $\frac{p}{q}$.

Idea: To find rationals between two fractions, give them a larger common denominator so there is room in between.

Multiply numerator and denominator by $10$: $\frac{2}{5} = \frac{20}{50}, \qquad \frac{3}{5} = \frac{30}{50}$

Now choose numbers between $\frac{20}{50}$ and $\frac{30}{50}$: $\frac{21}{50}, \quad \frac{25}{50}, \quad \frac{27}{50}$

Conclusion: Three rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$ are $\frac{21}{50}, \frac{25}{50}, \frac{27}{50}$ (many other answers are possible).

Long division of $3 \div 8$:

$3.000 \div 8 = 0.375$

Steps: $30 \div 8 = 3$ remainder $6$; $60 \div 8 = 7$ remainder $4$; $40 \div 8 = 5$ remainder $0$.

Since the remainder becomes $0$, the expansion ends: $\frac{3}{8} = 0.375$

Check with prime factors: $8 = 2^3$. A fraction (in lowest terms) with denominator of the form $2^m 5^n$ has a terminating decimal.

Conclusion: $\frac{3}{8} = 0.375$, which is a terminating decimal.

Let $x = 0.6666\dots$

Since one digit repeats, multiply by $10$: $10x = 6.6666\dots$

Subtract the first equation from the second: $10x - x = 6.6666\dots - 0.6666\dots$ $9x = 6$ $x = \frac{6}{9} = \frac{2}{3}$

Conclusion: $0.\overline{6} = \frac{2}{3}$.

(i) $\sqrt{25} = 5$, a whole number. Rational.

(ii) $7$ is not a perfect square, so $\sqrt{7}$ has a non-terminating, non-recurring decimal expansion. Irrational.

(iii) $0.3796$ terminates. Rational.

(iv) $7.478478\dots = 7.\overline{478}$ is non-terminating but recurring. Rational.

Conclusion: Only $\sqrt{7}$ is irrational; the rest are rational.

(i) Use $(a+b)(a-b) = a^2 - b^2$: $(3+\sqrt{3})(3-\sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6$ This is a whole number — rational.

(ii) Use $(a+b)^2 = a^2 + 2ab + b^2$: $(\sqrt{5}+\sqrt{2})^2 = (\sqrt{5})^2 + 2\sqrt{5}\sqrt{2} + (\sqrt{2})^2$ $= 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}$ Since $\sqrt{10}$ is irrational, $7 + 2\sqrt{10}$ is irrational.

Conclusion: (i) is rational ($6$); (ii) is irrational ($7 + 2\sqrt{10}$).

Convert to decimals: $\frac{5}{7} = 0.714285\dots, \qquad \frac{9}{11} = 0.8181\dots$

So we need irrational numbers between about $0.7142\dots$ and $0.8181\dots$. Construct decimals that are non-terminating and non-recurring (no repeating block):

$0.7401001000100001\dots$ $0.7502002000200002\dots$ $0.8080080008\dots$

Each lies between the two given numbers and has a non-repeating pattern, so each is irrational.

Conclusion: Three such irrational numbers are given above (many other answers possible).

(i) $a^m \cdot a^n = a^{m+n}$: $2^{2/3} \cdot 2^{1/3} = 2^{\frac{2}{3}+\frac{1}{3}} = 2^{1} = 2$

(ii) $(a^m)^n = a^{mn}$: $\left(3^{1/2}\right)^4 = 3^{\frac{1}{2}\times 4} = 3^{2} = 9$

(iii) $a^m \div a^n = a^{m-n}$: $7^{1/3} \div 7^{1/4} = 7^{\frac{1}{3}-\frac{1}{4}} = 7^{\frac{4-3}{12}} = 7^{1/12}$

Conclusion: (i) $2$, (ii) $9$, (iii) $7^{1/12}$.

Conjugate method: Multiply numerator and denominator by the conjugate $(\sqrt{7} + \sqrt{6})$:

$\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

Denominator (using $(a-b)(a+b)=a^2-b^2$): $(\sqrt{7})^2 - (\sqrt{6})^2 = 7 - 6 = 1$

Result: $= \frac{\sqrt{7}+\sqrt{6}}{1} = \sqrt{7} + \sqrt{6}$

Conclusion: $\dfrac{1}{\sqrt{7}-\sqrt{6}} = \sqrt{7} + \sqrt{6}$.

Multiply numerator and denominator by $\sqrt{3}$:

$\frac{4}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{2 \times 3} = \frac{4\sqrt{3}}{6}$

Simplify the fraction: $= \frac{2\sqrt{3}}{3}$

Conclusion: $\dfrac{4}{2\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$.

Idea: Use the Pythagoras theorem, since $\sqrt{5} = \sqrt{2^2 + 1^2}$.

Construction steps:

1. Draw a number line and mark point $O$ at $0$ and point $A$ at $2$ (so $OA = 2$ units).
2. At $A$, draw a line segment $AB$ perpendicular to the number line with $AB = 1$ unit.
3. Join $OB$. By Pythagoras: $OB = \sqrt{OA^2 + AB^2} = \sqrt{2^2 + 1^2} = \sqrt{4+1} = \sqrt{5}$
4. With $O$ as centre and radius $OB = \sqrt{5}$, draw an arc cutting the number line at point $P$.

Conclusion: Point $P$ represents $\sqrt{5}$ on the number line.

Building the spiral:

1. Draw $OA = 1$ unit. At $A$ draw $AB \perp OA$ with $AB = 1$. Then $OB = \sqrt{1^2 + 1^2} = \sqrt{2}.$
2. At $B$, draw $BC \perp OB$ with $BC = 1$ unit. By Pythagoras in triangle $OBC$: $OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}.$

Conclusion: The segment $OC$ has length $\sqrt{3}$. Continuing this process (adding unit perpendiculars) generates $\sqrt{4}, \sqrt{5}, \dots$, forming the square-root spiral.