# Ex.1.1 Q5 Rational Numbers Solution - NCERT Maths Class 8

## Question

Name the property under multiplication used in each of the following:

(i) \(\begin{align}\;\frac{{ - 4}}{5} \times 1 = 1 \times \frac{{ - 4}}{5} = \frac{{ - 4}}{5}\end{align}\)

(ii) \(\begin{align}\;\frac{{ - 13}}{{17}} \times \frac{{ - 2}}{7} = \frac{{ - 2}}{7} \times \frac{{ - 13}}{{17}}\end{align}\)

(iii) \(\begin{align}\;\frac{{ - 19}}{{29}} \times \frac{{29}}{{ - 19}} = 1\end{align}\)

## Text Solution

**What is known?**

Rational number.

**What is unknown?**

Name of the property.

(i) \(\begin{align}\;\frac{{ - 4}}{5} \times 1 = 1 \times \frac{{ - 4}}{5} = \frac{{ - 4}}{5}\end{align}\)

**Reasoning:**

So, \(1\) is the multiplicative identity.

**Steps:**

\[\begin{align}\;\frac{{ - 4}}{5} \times 1 = 1 \times \frac{{ - 4}}{5} = \frac{{ - 4}}{5}\end{align}\]

\(\therefore 1\) is the multiplicative identity and here, property of multiplicative identity is used.

(ii) \(\begin{align}\;\frac{{ - 13}}{{17}} \times \frac{{ - 2}}{7} = \frac{{ - 2}}{7} \times \frac{{ - 13}}{{17}}\end{align}\)

**Reasoning:**

In general, \(a \times b = b \times a \) for any two rational numbers. This is called commutativity of multiplication.

**Steps:**

\[\begin{align}\frac{{ - 13}}{{17}} \times \frac{{ - 2}}{7} &= \frac{{ - 2}}{7} \times \frac{{ - 13}}{{17}}\\

[a \times b] &= [b \times a]\end{align}\]

Commutativity of multiplication of rational numbers is used here.

(iii) \(\begin{align}\;\frac{{ - 19}}{{29}} \times \frac{{29}}{{ - 19}} = 1\end{align}\)

**Reasoning:**

For a rational number \(\begin{align}\frac{a}{b}\end{align}\) the multiplicative inverse is the reciprocal of that number that is \(\begin{align}\frac{b}{a}\end{align}\) . So that the product of the rational number and its multiplicative inverse is \(1.\)

**Steps:**

\[\begin{align}\frac{{ - 19}}{{29}} \times \frac{{29}}{{ - 19}} &= 1\\\left[ {\left( {\frac{a}{b}} \right) \times \left( {\frac{b}{a}} \right)} \right] &= 1\end{align}\]

Multiplicative Inverse.