The normal form of a number — lesson. Mathematics CBSE, Class 7.

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Grade	Solution	The value of the step
$10^{3}$ =	$10 \cdot 10 \cdot 10$	$= 1000$
$10^{2}$ =	$10 \cdot 10$	$= 100$
$10^{1}$ =	any number in the first degree is equal to itself	$= 10$
$10^{0}$ =	any number in the zero degree is equal to 1	$= 1$
$10^{- 1}$ =	$\frac{1}{10^{1}} = \frac{1}{10}$	$= 0.1$
$10^{- 2}$ =	$\frac{1}{10^{2}} = \frac{1}{10 \cdot 10} = \frac{1}{100}$	$= 0.01$
$10^{- 3}$ =	$\frac{1}{10^{3}} = \frac{1}{10 \cdot 10 \cdot 10} = \frac{1}{1000}$	$= 0.001$

$a^{- n} = \frac{1}{a^{n}}$

To record very large or very small numbers, use the normal form.

The normal form of a number is called the multiple of this number:

$a \cdot 10^{n} where 1 \leq a < 10$

Note that greater than

$1$ and less than

$10.$

If the number is greater than or equal to

$10,$ then writes in normal form

$10$ with a positive lever,

for example, blue whale mass is approx

$1.9 \cdot 10^{5}$

$kg$ .

If the number is less than

$1,$ then writes in normal form

$10$ with a negative gain,

for example, the mass of the smallest ants is approx

$0.000001 kg =$

$1 \cdot 10^{- 6}$

$kg$ .

Remember that an integer is after the last digit.

$1 = 1.0\$	$300 = 300.0\$	$50,000 = 50,000.0$
$20 = 20.0$	$4000 = 4000.0\$	$600,000 = 600,$ $000.0$

Important!

When switching from a number to a normal notation (or vice versa), move the period to the right or left in the number and multiply it by

$10$ the appropriate degree.

If the normal format is to write a number greater than

$10,$ then move the period to the left.

Example:

$98765 =$ $9.8765 \cdot 10^{4}$	Period moved $4$ places to the left. $\begin{array}{l} 9. 8765.0 \\ \leftarrow \end{array}$
$12345600 =$ $1.23456 \cdot 10^{7}$	Period moved $7$ places to the left. $\begin{array}{l} 1. 2345600.0 \\ \leftarrow \end{array}$

If the normal form is to write a number smaller than

$1,$ then move the period to the right.

Example:

$0.012345 =$ $1. 2345 \cdot 10^{- 2}$	Period moved $2$ places to the right. $\begin{array}{l} 0.0 1. 2345 \\ \to \end{array}$
$0.00234567 =$ $2. 34567 \cdot 10^{- 3}$	Period moved $3$ places to the right. $\begin{array}{l} 0 . 002. 34567 \\ \to \end{array}$
$0.000789 =$ $7. 89 \cdot 10^{- 4}$	Period moved $4$ places to the right. $\begin{array}{l} 0 . 0007. 89 \\ \to \end{array}$

Reference:

Mathematics for 7th grade / Ilze France, Gunta Lace, Ligita Pickaine, Anita Mikelsone. - Lielvarde: Lielvārde, 2007. - 248 pp. - References: 119-120. p.

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8. The normal form of a number

Theory: