The role of 0 and 1 — lesson. Mathematics State Board, Class 8.

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Zero

$(0)$ is a rational number, and the sum of any rational number with zero

$(0)$ results in the same rational number.

Thus,

$(\frac{a}{b} + 0) = (0 + \frac{a}{b}) = \frac{a}{b}$ , for every rational number

$\frac{a}{b}$ .

$0$ is called the additive identity for rational numbers.

Example:

(i) Consider the rational number

$\frac{3}{5}$ .

Then, we have

$(\frac{3}{5} + 0) = (0 + \frac{3}{5}) = \frac{3}{5}$ .

$(\frac{3}{5} + 0) = (\frac{3}{5} + \frac{0}{5}) = (\frac{3 + 0}{5}) = \frac{3}{5}$

Therefore,

$(\frac{3}{5} + 0) = (0 + \frac{3}{5}) = \frac{3}{5}$ .

(ii) Consider the rational number

$\frac{- 2}{3}$ .

Then, we have

$(\frac{- 2}{3} + 0) = (0 + \frac{- 2}{3}) = \frac{- 2}{3}$ .

$(\frac{- 2}{3} + 0) = (\frac{- 2}{3} + \frac{0}{3}) = (\frac{- 2 + 0}{3}) = \frac{- 2}{3}$

Therefore,

$(\frac{- 2}{3} + 0) = (0 + \frac{- 2}{3}) = \frac{- 2}{3}$ .

One

$(1)$ is a rational number and the product of any rational number with one

$(1)$ result in the same rational number.

For any rational number

$\frac{a}{b}$ , we have

$(\frac{a}{b} \times 1) = (1 \times \frac{a}{b}) = \frac{a}{b}$ .

$1$ is called the multiplicative identity for rational numbers.

Example:

(i) Consider the rational number

$\frac{3}{4}$ .

Then, we have

$(\frac{3}{4} \times 1) = (1 \times \frac{3}{4}) = \frac{3}{4}$ .

$(\frac{3}{4} \times 1) = (\frac{3}{4} \times \frac{1}{4}) = (\frac{3 \times 1}{4}) = \frac{3}{4}$ and

$(1 \times \frac{3}{4}) = (\frac{1}{4} \times \frac{3}{4}) = (\frac{1 \times 3}{4}) = \frac{3}{4}$ .

Therefore,

$(\frac{3}{4} \times 1) = (1 \times \frac{3}{4}) = \frac{3}{4}$ .

(ii) Consider the rational

$\frac{- 9}{13}$ .

Then, we have

$(\frac{- 9}{13} \times 1) = (1 \times \frac{- 9}{13}) = \frac{- 9}{13}$

$(\frac{- 9}{13} \times 1) = (\frac{- 9}{13} \times \frac{1}{13}) = (\frac{- 9 \times 1}{13}) = \frac{- 9}{13}$ and

$(1 \times \frac{- 9}{13}) = (\frac{1}{13} \times \frac{- 9}{13}) = (\frac{1 \times (- 9)}{13}) = \frac{- 9}{13}$ .

Therefore,

$(\frac{- 9}{13} \times 1) = (1 \times \frac{- 9}{13}) = \frac{- 9}{13}$ .

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5. The role of 0 and 1

Theory: