$(\frac{- 2}{9}) - (\frac{7}{9}) =___________.$

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Solution

Rational number
A rational number, in Mathematics, can be defined as any number which can be represented in the form of $\frac{p}{q}$ where $q \neq 0$ . When the rational number is divided, the result will be in decimal form, which may be either terminating decimal or repeating decimal. The set of rational numerals includes positive, negative numbers, and zero.
Given that, $(\frac{- 2}{9}) - (\frac{7}{9})$
Taking LCM we get
$\begin{array}{rcl} (\frac{- 2}{9}) - (\frac{7}{9}) & = & \frac{(- 2 - 7)}{9} \\ = & \frac{- 9}{9} \\ = & - 1 \\ ∴ (\frac{- 2}{9}) - (\frac{7}{9}) & = & - 1 \end{array}$

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Similar questions

Using the given pattern, find the missing numbers.
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + . . .^{2} = 21^{2}$
$5^{2} + . . .^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + . . . .^{2} = . . . .^{2}$ .

Using the given pattern, find the missing numbers.

1² + 2² + 2² = 3²

2² + 3² + 6² = 7²

3² + 4² + 12² = 13²

4² + 5² + _ ² = 21²

5² + _ ² + 30² = 31²

6² + 7² + _ ² = __²

Question 6
Using the given pattern, find the missing numbers:
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + {- --- -}^{2} = 21^{2}$
$5^{2} + {- --- -}^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + - --- - 2 = - --- - 2$