Rational Numbers — NCERT Solutions, AI Tutor & Practice

The commutative property states $a + b = b + a$.

LHS: $\frac{-3}{5} + \frac{2}{7}$. LCM of $5$ and $7$ is $35$. $\frac{-21}{35} + \frac{10}{35} = \frac{-11}{35}$

RHS: $\frac{2}{7} + \frac{-3}{5} = \frac{10}{35} + \frac{-21}{35} = \frac{-11}{35}$

Since LHS $=$ RHS $= \frac{-11}{35}$, the commutative property of addition is verified.

Commutativity would require $a - b = b - a$.

$a - b$: $\frac{4}{9} - \left(\frac{-1}{3}\right) = \frac{4}{9} + \frac{1}{3} = \frac{4}{9} + \frac{3}{9} = \frac{7}{9}$

$b - a$: $\frac{-1}{3} - \frac{4}{9} = \frac{-3}{9} - \frac{4}{9} = \frac{-7}{9}$

Since $\frac{7}{9} \ne \frac{-7}{9}$, subtraction is not commutative for rational numbers.

Additive inverse: the number that gives $0$ on addition. $\frac{-7}{12} + x = 0 \Rightarrow x = \frac{7}{12}$ So the additive inverse is $\frac{7}{12}$.

Multiplicative inverse (reciprocal): the number that gives $1$ on multiplication. $\frac{-7}{12} \times y = 1 \Rightarrow y = \frac{12}{-7} = \frac{-12}{7}$ So the reciprocal is $\frac{-12}{7}$.

The distributive property: $a \times (b + c) = a\times b + a \times c$.

Method 1 (add inside bracket first): $\frac{3}{7} + \frac{-1}{7} = \frac{2}{7}$ $\frac{2}{5} \times \frac{2}{7} = \frac{4}{35}$

Method 2 (distribute): $\frac{2}{5}\times\frac{3}{7} + \frac{2}{5}\times\frac{-1}{7} = \frac{6}{35} + \frac{-2}{35} = \frac{4}{35}$

Both give $\frac{4}{35}$.

(a) Multiplying by $1$ leaves the number unchanged — this is the multiplicative identity property ($1$ is the multiplicative identity).

(b) Adding $0$ leaves the number unchanged — this is the additive identity property ($0$ is the additive identity).

(c) The order of multiplication is swapped without changing the result — this is the commutative property of multiplication.

A rational number between two numbers is their average. $\text{Mean} = \frac{1}{2}\left(\frac{1}{4} + \frac{1}{2}\right)$ $= \frac{1}{2}\left(\frac{1}{4} + \frac{2}{4}\right) = \frac{1}{2}\times\frac{3}{4} = \frac{3}{8}$

So $\frac{3}{8}$ lies between $\frac{1}{4}$ and $\frac{1}{2}$, since $\frac{1}{4} = \frac{2}{8} < \frac{3}{8} < \frac{4}{8} = \frac{1}{2}$.

Take the LCM of $3$ and $5$, which is $15$: $\frac{2}{3} = \frac{10}{15}, \qquad \frac{4}{5} = \frac{12}{15}$ Only one number ($\frac{11}{15}$) sits directly between, so multiply numerator and denominator by a larger factor, say $10$: $\frac{100}{150} \quad \text{and} \quad \frac{120}{150}$ Now five rational numbers between them are: $\frac{101}{150},\ \frac{103}{150},\ \frac{105}{150},\ \frac{108}{150},\ \frac{115}{150}$ (Any five fractions with numerators between $100$ and $120$ are acceptable.)

LCM of $6, 3, 2$ is $6$. $\frac{-5}{6} + \frac{2}{3} - \frac{1}{2} = \frac{-5}{6} + \frac{4}{6} - \frac{3}{6}$ $= \frac{-5 + 4 - 3}{6} = \frac{-4}{6} = \frac{-2}{3}$

The answer is $\frac{-2}{3}$.

Associativity requires $(a\times b)\times c = a\times(b\times c)$.

LHS: $\left(\frac{-1}{2}\times\frac{3}{4}\right)\times\frac{-2}{5} = \frac{-3}{8}\times\frac{-2}{5} = \frac{6}{40} = \frac{3}{20}$

RHS: $\frac{-1}{2}\times\left(\frac{3}{4}\times\frac{-2}{5}\right) = \frac{-1}{2}\times\frac{-6}{20} = \frac{6}{40} = \frac{3}{20}$

Since LHS $=$ RHS $= \frac{3}{20}$, the associative property of multiplication is verified.

$\frac{-5}{4}$ is negative, so it lies to the left of $0$.

Since the denominator is $4$, divide each unit length between consecutive integers into $4$ equal parts.

Now $\frac{-5}{4} = -1\frac{1}{4}$, meaning it is $1$ full unit plus $1$ more of the four small parts to the left of $0$.

So, starting at $0$ and moving left, mark the point at the $5$th small division (which lies between $-1$ and $-2$, one small part beyond $-1$).

$-2 \quad \bullet\!\!\underset{\frac{-5}{4}}{} \quad -1 \quad 0$ The marked point between $-1$ and $-2$ (closer to $-1$) represents $\frac{-5}{4}$.

Let the other number be $x$. $\frac{-4}{3}\times x = \frac{-16}{9}$ So, $x = \frac{-16}{9} \div \frac{-4}{3} = \frac{-16}{9}\times\frac{3}{-4}$ $= \frac{-16 \times 3}{9 \times -4} = \frac{-48}{-36} = \frac{4}{3}$

The other rational number is $\frac{4}{3}$.

Check: $\frac{-4}{3}\times\frac{4}{3} = \frac{-16}{9}$ ✓

Each term has the common factor $\frac{7}{9}$, so take it out (distributive property): $\frac{7}{9}\left(\frac{3}{5} + \frac{2}{5} - 1\right)$ Inside the bracket: $\frac{3}{5} + \frac{2}{5} - 1 = \frac{5}{5} - 1 = 1 - 1 = 0$ Therefore: $\frac{7}{9}\times 0 = 0$

The value of the expression is $0$.