 UPSKILL MATH PLUS

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### Theory:

Zero $$(0)$$ is a rational number, and the sum of any rational number with zero$$(0)$$ results in the same rational number.
Thus, $\left(\frac{a}{b}+0\right)=\left(0+\frac{a}{b}\right)=\frac{a}{b}$, 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 $\left(\frac{3}{5}+0\right)=\left(0+\frac{3}{5}\right)=\frac{3}{5}$.

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

Therefore, $\left(\frac{3}{5}+0\right)=\left(0+\frac{3}{5}\right)=\frac{3}{5}$.

(ii) Consider the rational number $\frac{-2}{3}$.

Then, we have $\left(\frac{-2}{3}+0\right)=\left(0+\frac{-2}{3}\right)=\frac{-2}{3}$.

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

Therefore, $\left(\frac{-2}{3}+0\right)=\left(0+\frac{-2}{3}\right)=\frac{-2}{3}$.
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 $\left(\frac{a}{b}×1\right)=\left(1×\frac{a}{b}\right)=\frac{a}{b}$.

$$1$$ is called the multiplicative identity for rational numbers.
Example:
(i) Consider the rational number $\frac{3}{4}$.

Then, we have $\left(\frac{3}{4}×1\right)=\left(1×\frac{3}{4}\right)=\frac{3}{4}$.

$\left(\frac{3}{4}×1\right)=\left(\frac{3}{4}×\frac{1}{4}\right)=\left(\frac{3×1}{4}\right)=\frac{3}{4}$ and $\left(1×\frac{3}{4}\right)=\left(\frac{1}{4}×\frac{3}{4}\right)=\left(\frac{1×3}{4}\right)=\frac{3}{4}$.

Therefore, $\left(\frac{3}{4}×1\right)=\left(1×\frac{3}{4}\right)=\frac{3}{4}$.

(ii) Consider the rational $\frac{-9}{13}$.

Then, we have $\left(\frac{-9}{13}×1\right)=\left(1×\frac{-9}{13}\right)=\frac{-9}{13}$

$\left(\frac{-9}{13}×1\right)=\left(\frac{-9}{13}×\frac{1}{13}\right)=\left(\frac{-9×1}{13}\right)=\frac{-9}{13}$ and $\left(1×\frac{-9}{13}\right)=\left(\frac{1}{13}×\frac{-9}{13}\right)=\left(\frac{1×\left(-9\right)}{13}\right)=\frac{-9}{13}$.

Therefore, $\left(\frac{-9}{13}×1\right)=\left(1×\frac{-9}{13}\right)=\frac{-9}{13}$.