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

\(98765 =\) $9.8765 \cdot 10^{4}$	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. $\begin{array}{l} 1. 2345600.0 \\ \leftarrow \end{array}$

\(0.012345 =\) $1. 2345 \cdot 10^{- 2}$	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. $\begin{array}{l} 0 . 0007. 89 \\ \to \end{array}$

8. The normal form of a number

Theory: