# Question a04c1

Sep 4, 2017

$4.56 \cdot {10}^{- 3}$

#### Explanation:

Your goal here is to write your number in normalized scientific notation, which has

$\textcolor{w h i t e}{a a} \textcolor{b l u e}{m} \times {10}^{\textcolor{p u r p \le}{n} \textcolor{w h i t e}{a} \stackrel{\textcolor{w h i t e}{a a a a a a}}{\leftarrow}} \textcolor{w h i t e}{a \textcolor{b l a c k}{\text{the")acolor(purple)("exponent}} a a}$
$\textcolor{w h i t e}{\frac{a}{a} \textcolor{b l a c k}{\uparrow} a a a a}$
$\textcolor{w h i t e}{\textcolor{b l a c k}{\text{the")acolor(blue)("mantissa}} a}$

and

1 <= |color(blue)(m)| < 10" " " "color(darkorange)("(*)")#

As you know, you have

${10}^{0} = 1$

This means that you can write your initial number as

$\textcolor{b l u e}{0.00456} \cdot {10}^{\textcolor{p u r p \le}{0}}$

Now, to start converting the number to scientific notation, multiply it by $1 = \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$ $\to$ keep in mind that we can multiply the number by $1$ because that leaves its value unchanged!

$\textcolor{b l u e}{0.00456} \cdot {10}^{\textcolor{p u r p \le}{0}} \cdot \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$

You can rewrite this as

$\textcolor{b l u e}{0.00456} \cdot \textcolor{b l u e}{10} \cdot {10}^{\textcolor{p u r p \le}{0}} / \textcolor{p u r p \le}{10} = \textcolor{b l u e}{0.0456} \cdot {10}^{\textcolor{p u r p \le}{- 1}}$

At this point, you must check to see if the new value of the mantissa satisfies condition $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$.

Since

$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\le}}} \textcolor{b l u e}{0.0456} \textcolor{red}{\cancel{\textcolor{b l a c k}{<}}} 10$

you must repeat the process again. This time, you have

$\textcolor{b l u e}{0.0456} \cdot {10}^{\textcolor{p u r p \le}{- 1}} \cdot \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$

which is equivalent to

$\textcolor{b l u e}{0.0456} \cdot \textcolor{b l u e}{10} \cdot {10}^{\textcolor{p u r p \le}{- 1}} / \textcolor{p u r p \le}{10} = \textcolor{b l u e}{0.456} \cdot {10}^{\textcolor{p u r p \le}{- 2}}$

Condition $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ is still not satisfied, so you must repeat the process again. This time, you have

$\textcolor{b l u e}{0.456} \cdot {10}^{\textcolor{p u r p \le}{- 2}} \cdot \frac{\textcolor{b l u e}{10}}{\textcolor{p u r p \le}{10}}$

which is equivalent to

$\textcolor{b l u e}{0.456} \cdot \textcolor{b l u e}{10} \cdot {10}^{\textcolor{p u r p \le}{- 2}} / \textcolor{p u r p \le}{10} = \textcolor{b l u e}{4.56} \cdot {10}^{\textcolor{p u r p \le}{- 3}}$

Finally, you have

$1 \le \textcolor{b l u e}{4.56} < 10 \text{ " " } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$

so you can say that

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.00456 = 4.56 \cdot {10}^{- 3}}}}$

Notice that the number written in scientific notation has $3$ sig figs, just like the number written in standard form.