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Number and Algebra

In Year 7 and Year 8 we introduced powers with whole number indices and developed the index laws for these powers.

We now extend our study to rational (fractional), including negative, indices.

We live in a world of very large and very small numbers. Population sizes, government spending, intergalactic distances and the size of computer memories are examples of very large numbers. Thickness of materials, circuit diagrams and subatomic particles are examples of very small numbers. Some examples of small and large numbers are:

\begin{itemize}
  \item the time taken by light to travel one metre is approximately 0.000000003 seconds
  \item the radius of a hydrogen atom is approximately 0.000000000025 metres
  \item the current Big Bang model of astronomy suggests that the Universe is about 13.7 billion years old.
\end{itemize}

This chapter introduces scientific notation, which is a convenient way of writing such numbers. Significant figures are also discussed.

\section*{The index laws}
\section*{Index notation}
We recall the following from ICE-EM Mathematics Year 8.

\begin{itemize}
  \item A power is the product of a certain number of factors, all of which are the same.
\end{itemize}

For example,

\[
2^{4}=2 \times 2 \times 2 \times 2 \text { is the fourth power of } 2 .
\]

\begin{itemize}
  \item For any number \(b, b^{1}=b\).
  \item In general, \(b^{n}=\underbrace{b \times b \times b \times \ldots \times b}_{n}\), where there are \(n\) common factors in the product.
\end{itemize}

Here \(b\) is the base and \(n\) the index.


The prime decomposition of these numbers can be found by a process of repeated division (see ICE-EM Mathematics Year 8). This process is particularly useful when the decomposition contains more than one type of prime number.

\section*{Example 3}
Express as a product of powers of prime numbers.
a 9000
b 66150

\section*{Solution}
By repeated division:

a \(9000=2^{3} \times 3^{2} \times 5^{3}\)

b \(66150=1323 \times 50\)

\[
=2 \times 3^{3} \times 5^{2} \times 7^{2}
\]

The following laws for indices were discussed in Chapter 3 of ICE-EM Mathematics Year 8 for powers with whole number indices.

In the following, \(a\) and \(b\) are integers and \(m\) and \(n\) are non-zero whole numbers.


\section*{Example 4}

\section*{Solution}

\section*{Example 5}
Simplify in power form, and hence evaluate where appropriate.

\section*{Solution}

\section*{Example 6}
Simplify, expressing the answer in index form.

\section*{Solution}


\section*{Example 7}
Simplify by expanding the brackets.

Solution


\section*{A number raised to the power zero}
Clearly \(\frac{4^{3}}{4^{3}}=1\). If the index laws are to apply then \(4^{3-3}=4^{0}=1\).

Hence we define \(a^{0}=1\), for all non-zero numbers \(a\).

Note: \(0^{0}\) is not defined. If \(x\) is a pronumeral, \(x^{0}=1(x \neq 0)\).

\section*{Example 8}
Simplify:

\section*{Solution}

\section*{Example 9}
Simplify:

\section*{The zero power}
For all non-zero numbers \(a\), we define \(a^{0}=1\).

\section*{Exercise 8A}



15 Simplify:
c \(\left(b^{4}\right)^{2} \div\left(b^{3}\right)^{2}\)
g \(\frac{4\left(x^{3}\right)^{2} y^{4}}{3 x^{4} y^{3}} \times \frac{3 x^{3}\left(y^{2}\right)^{2}}{8 x y^{5}}\)
h \(\frac{8 a^{2}\left(b^{3}\right)^{2}}{3 a b^{2}} \div \frac{16 a^{5} b^{3}}{9\left(a^{3}\right)^{2}}\)



\section*{Negative indices}
In the last section, we defined \(a^{n}\) to be the product of \(n\) factors of \(a\), where \(a\) is any number and \(n\) is a positive integer. We also defined \(a^{0}=1\) for all non-zero numbers \(a\).

We now give meaning to negative integer indices. For example, we want to give a meaning to \(2^{-4}, 3^{-100}\) and so on. To work towards a useful definition, look at the following pattern:

\[
\begin{array}{lll}
2^{5}=32, & 2^{4}=16, & 2^{3}=8, \\
2^{2}=4, & 2^{1}=2, & 2^{0}=1
\end{array}
\]

Each index is one less than the preceding one, and each number to the right of an equal sign is half the number in the previous expression.

This can be continued purely as a pattern:

\[
2^{-1}=\frac{1}{2}, \quad 2^{-2}=\frac{1}{4}=\frac{1}{2^{2}}, \quad 2^{-3}=\frac{1}{8}=\frac{1}{2^{3}} \quad \text { and so on. }
\]

This suggests we define \(2^{-n}=\frac{1}{2^{n}}\), for any positive integer \(n\).

\section*{Negative indices}
Define

\[
a^{-n}=\frac{1}{a^{n}}
\]

where \(a\) is a non-zero number and \(n\) is a positive integer.

Note that this tells us that \(a^{-n} \times a^{n}=1\). So \(a^{n}=\frac{1}{a^{-n}}\)

For example,

\[
2^{10}=\frac{1}{2^{-10}}
\]

So we can say that for all \(n\), whether \(n\) is positive or negative, \(a^{-n}=\frac{1}{a^{n}}\)

\section*{Example 10}
\section*{Evaluate:}

\section*{Fractions and negative indices}
The reciprocal of a fraction such as \(\frac{4}{3}\) is \(\frac{3}{4}\). The index -1 means 'the reciprocal of', so

\[
\left(\frac{3}{4}\right)^{-1}=\frac{4}{3}
\]

When raising a fraction to other negative indices take the reciprocal first.

\[
\left(\frac{2}{3}\right)^{-3}=\left(\frac{3}{2}\right)^{3}=\frac{27}{8}
\]

\section*{Example 11}
Evaluate:

The index laws we revised in Section 8A are also valid when using negative integer indices.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
\multicolumn{2}{|c|}{For all integers \(m\) and \(n\) and non-zero numbers \(a\) and \(b\) the following are true.} &  \\
\hline
Zero index & \(a^{0}=1\) &  \\
\hline
Negative index &  & \(a^{-n}=\frac{1}{a^{n}}\) \\
\hline
Index law 1 & Product of powers & \(a^{m} a^{n}=a^{m+n}\) \\
\hline
Index law 2 & Quotient of powers & \(a^{m} \div a^{n}=\frac{a^{m}}{a^{n}}=a^{m-n}\) \\
\hline
Index law 3 & Power of a power & \(\left(a^{m}\right)^{n}=a^{m n}\) \\
\hline
Index law 4 & Power of a product & \((a b)^{n}=a^{n} b^{n}\) \\
\hline
Index law 5 & Power of a quotient & \(\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}\) \\
\hline
\end{tabular}
\end{center}

The results for negative indices \(m\) and \(n\) can be proved using the index laws for positive integer indices. An indication of the proof appears as a question in the Challenge exercise at the end of the chapter.

\section*{Example 12}
Write as a single power and then evaluate.

\section*{Solution}



\section*{Example 13}
Write as a single power and then evaluate.

\section*{Solution}

\section*{Example 14}
Simplify, expressing the answers with positive indices.

\section*{Solution}

\section*{Exercise 8B}

\section*{Fractional indices}
\section*{Fractional indices with numerator one}
We are now going to extend our study of indices by looking at fractional indices. We begin by considering what we mean by powers such as \(3^{\frac{1}{2}}, 2^{\frac{1}{3}}\) and \(7^{\frac{1}{10}}\), in which the index is the reciprocal of a positive integer.

If \(a\) is a positive number, then \(\sqrt{a}\) is the positive square root and \((\sqrt{a})^{2}=a=a^{1}\).

We now introduce the alternative notation \(a^{\frac{1}{2}}\) for \(\sqrt{a}\). We do this because the third index law then continues to hold, that is:

\[
\left(a^{\frac{1}{2}}\right)^{2}=a^{2 \times \frac{1}{2}}=a^{1}
\]

Keep in mind that \(a^{\frac{1}{2}}\) is nothing more than an alternative notation for \(\sqrt{a}\).

For example, \(49^{\frac{1}{2}}=7,64^{\frac{1}{2}}=8,100^{\frac{1}{2}}=10\), and so on.

Every positive number \(a\) has a cube root \(\sqrt[3]{a}\). It is the positive number whose cube is \(a\).

For example:

\[
\sqrt[3]{8}=2 \text { because } 2^{3}=8 \quad \sqrt[3]{27}=3 \text { because } 3^{3}=27
\]

We define \(a^{\frac{1}{3}}\) to be \(\sqrt[3]{a}\). The third index laws continues to hold.

\[
\left(a^{\frac{1}{3}}\right)^{3}=a^{3 \times \frac{1}{3}}=a^{1}
\]

Similarly we define

\[
a^{\frac{1}{4}}=\sqrt[4]{a}, a^{\frac{1}{5}}=\sqrt[5]{a} \text {, and so on. }
\]

\section*{Fractional indices}
Let \(a\) be positive or zero and let \(n\) be a positive integer.

Define \(a^{\frac{1}{n}}\) to be the \(n^{\text {th }}\) root of \(a\). That is, \(a^{\frac{1}{n}}=\sqrt[n]{a}\).

For example, \(a^{\frac{1}{2}}=\sqrt{a}\) and \(a^{\frac{1}{3}}=\sqrt[3]{a}\)

Square roots, cube roots, fourth roots and so on are usually irrational numbers, and a calculator can be used to obtain approximations. Using a calculator, we obtain:

\[
10^{\frac{1}{2}}=\sqrt{10} \approx 3.1623, \quad 10^{\frac{1}{3}}=\sqrt[3]{10} \approx 2.1544, \quad 10^{\frac{1}{4}}=\sqrt[4]{10} \approx 1.7783, \ldots
\]

Using our new notation, here are some other numerical approximations, all recorded correct to 4 decimal places. You can use your calculator to check these.

\[
2^{\frac{1}{5}} \approx 1.1487, \quad 10^{\frac{1}{8}} \approx 1.3335, \quad 0.2^{\frac{1}{4}} \approx 0.6687, \quad 3.2^{\frac{1}{6}} \approx 1.2139
\]

\section*{Example 15}
Evaluate:
a \(\sqrt[3]{27}\)
b \(\sqrt[4]{16}\)
c \(\sqrt[5]{243}\)
d \(\sqrt[3]{125}\)

\section*{Solution}
a \(\sqrt[3]{27}\) is the number that when cubed is \(27 . \quad\) b \(\sqrt[4]{16}=2\) because \(2^{4}=16\).

Thus \(3^{3}=27\), because \(\sqrt[3]{27}=3\)
c \(\sqrt[5]{243}=\sqrt[5]{3^{5}}\)
d \(\sqrt[3]{125}=\sqrt[3]{5^{3}}\)
\(=3\)
\(=5\)

\section*{Example 16}
Evaluate:
a \(9^{\frac{1}{2}}\)
b \(27^{\frac{1}{3}}\)
c \(16^{\frac{1}{4}}\)
d \(1000000^{\overline{6}}\)

\section*{Solution}
a \(9^{\frac{1}{2}}=3\)
(since \(\left.3^{2}=9\right)\)
b \(27^{\frac{1}{3}}=3\)
(since \(\left.3^{3}=27\right)\)
c \(16^{\frac{1}{4}}=2 \quad\left(\right.\) since \(\left.2^{4}=16\right)\)
d \(1000000^{\frac{1}{6}}=10\)
(since \(\left.10^{6}=1000000\right)\)

\section*{Positive fractional indices}
In the previous section, we defined \(8^{\frac{1}{3}}=\sqrt[3]{8}\). For consistency with the index laws we now define \(8^{\frac{2}{3}}=\left(8^{\frac{1}{3}}\right)^{2}\).

\section*{Positive fractional indices}
If \(a\) is a positive number or zero and, \(p\) and \(q\) are positive integers, define:

\[
a^{\frac{p}{q}}=\left(a^{\frac{1}{q}}\right)^{p}=(\sqrt[q]{a})^{p}
\]

\section*{Example 17}
Evaluate:
a \(4^{\frac{3}{2}}\)
b \(8^{\frac{2}{3}}\)
c \(81^{\frac{3}{4}}\)

\section*{Solution}
a \(4^{\frac{3}{2}}=\left(4^{\frac{1}{2}}\right)^{3}\)
b \(8^{\frac{2}{3}}=\left(8^{\frac{1}{3}}\right)^{2}\)
\(=3^{3} \quad\left(\right.\) since \(\left.2^{2}=4\right)\)
\(=2^{2}\)
(since \(\left.2^{3}=8\right)\)
\(=8\)
\(=4\)

c \(81^{\frac{3}{4}}=\left(81^{\frac{1}{4}}\right)^{3}\)

\[
\begin{aligned}
& =3^{3} \\
& =27
\end{aligned}
\]

The index laws also hold for fractional indices. A proof of these more general results is given as a question in the Challenge exercise.

\section*{Example 18}
Simplify:
a \(\left(d^{21}\right)^{\frac{1}{7}}\)
b \(p^{\frac{2}{3}} \times p^{\frac{4}{5}}\)
c \(q^{\frac{4}{5}} \div q^{\frac{2}{3}}\)

\section*{Solution}
a \(\left(d^{21}\right)^{\frac{1}{7}}=d^{21 \times \frac{1}{7}}\)
b \(p^{\frac{2}{3}} \times p^{\frac{4}{5}}=p^{\frac{2}{3}+\frac{4}{5}}\)
c \(q^{\frac{4}{5}} \div q^{\frac{2}{3}}=q^{\frac{4}{5}-\frac{2}{3}}\)
\(=d^{3}\)
\(=p^{\frac{10}{15}+\frac{12}{15}}\)
\(=q^{\frac{12-10}{15}}\)
\(=p^{\frac{22}{15}}\)
\(=q^{\frac{2}{15}}\)

\section*{Negative fractional indices}
Since \(4^{-1}=\frac{1}{4}\), we define \(4^{-\frac{3}{2}}=\frac{1}{4^{\frac{3}{2}}}=\left(\frac{1}{4}\right)^{\frac{3}{2}}\).

We can now combine the definitions of negative indices and fractional indices.

\section*{Negative fractional indices}
Let \(a\) be a positive number or zero, and let \(p\) and \(q\) be positive integers. Define

\[
a^{-\frac{p}{q}}=\frac{1}{a^{\frac{p}{q}}}=\left(\frac{1}{a}\right)^{\frac{p}{q}}
\]

Note: If \(a>0\) and \(b>0,\left(\frac{a}{b}\right)^{-\frac{p}{q}}=\left(\frac{b}{a}\right)^{\frac{p}{q}}\).

\section*{Example 19}
Evaluate:
a \(4^{-\frac{3}{2}}\)
b \(\left(\frac{8}{27}\right)^{-\frac{1}{3}}\)

\section*{Solution}
a \(4^{-\frac{3}{2}}=\frac{1}{4^{\frac{3}{2}}}\)
b \(\left(\frac{8}{27}\right)^{-\frac{1}{3}}=\left(\frac{27}{8}\right)^{\frac{1}{3}}\)
\(=\frac{1}{8}\)
\(=\frac{27^{\frac{1}{3}}}{8^{\frac{1}{3}}}\)
\(=\frac{3}{2}\)

\section*{Example 20}
Simplify, writing each answer with positive indices.
a \(p^{-\frac{1}{4}} \times p^{\frac{2}{3}}\)
b \(x^{\frac{2}{3}} \div x^{-\frac{1}{2}}\)
c \(\left(125 n^{-6}\right)^{\frac{1}{3}}\)
d \(x^{\frac{1}{5}} \div x^{\frac{1}{3}}\)

\section*{Solution}
a \(p^{-\frac{1}{4}} \times p^{\frac{2}{3}}=p^{-\frac{1}{4}+\frac{2}{3}} \quad\) b \(x^{\frac{2}{3}} \div x^{-\frac{1}{2}}=x^{\frac{2}{3}-\left(-\frac{1}{2}\right)}\)

\[
=p^{-\frac{3}{12}+\frac{8}{12}}
\]

\[
\begin{array}{ll}
=p^{-\frac{3}{12}+\frac{8}{12}} & =x^{\frac{4}{6}+\frac{3}{6}} \\
=p^{\frac{5}{12}} & =x^{\frac{7}{6}}
\end{array}
\]

c \(\left(125 n^{-6}\right)^{\frac{1}{3}}=125^{\frac{1}{3}} \times n^{-6 \times \frac{1}{3}}\)

d \(x^{\frac{1}{5}} \div x^{\frac{1}{3}}=x^{\frac{3}{15}-\frac{5}{15}}\)

\[
\begin{aligned}
& =5 \times n^{-2} \\
& =\frac{5}{n^{2}}
\end{aligned}
\]

\[
\begin{aligned}
& =x^{-\frac{2}{15}} \\
& =\frac{1}{x^{\frac{2}{15}}}
\end{aligned}
\]

\section*{Exercise 8C}
1 Evaluate:
a \(\sqrt[3]{8}\)
b \(\sqrt[5]{32}\)
c \(\sqrt[3]{216}\)
d \(\sqrt[4]{81}\)
e \(\sqrt[3]{64}\)
f \(\sqrt[5]{2^{10}}\)

2 Write using fractional indices. Evaluate, correct to 4 decimal places.
a \(\sqrt{14}\)
b \(\sqrt[4]{64}\)
c \(\sqrt[5]{7}\)
d \(\sqrt[7]{11}\)
e \(\sqrt[3]{2^{7}}\)

Example 16 3 Evaluate:
a \(4^{\frac{1}{2}}\)
b \(27^{\frac{1}{3}}\)
c \(243^{\frac{1}{5}}\)
d \(81^{\frac{1}{4}}\)
e \(64^{\frac{1}{2}}\)
f \(25^{\frac{1}{2}}\)
g \(125^{\frac{1}{3}}\)
h \(64^{\frac{1}{3}}\)
i \(32^{\frac{1}{5}}\)
j \(625^{\frac{1}{4}}\)
k \(216^{\frac{1}{3}}\)
l \(49^{\frac{1}{2}}\)

4 Evaluate:
a \(4^{\frac{5}{2}}\)
b \(25^{\frac{3}{2}}\)
c \(125^{\frac{2}{3}}\)
d \(64 \frac{5}{6}\)
e \(32^{\frac{2}{5}}\)
f \(81^{\frac{3}{4}}\)
g \(216^{\frac{2}{3}}\)
h \(243^{\overline{5}}\)
i \((\sqrt[4]{16})^{3}\)
j \((\sqrt[3]{27})^{2}\)
k \(\sqrt[5]{32^{4}}\)
l \(\sqrt[3]{2^{6}}\)

5 Simplify:
a \(\left(a^{\frac{1}{2}}\right)^{2}\)
b \(\left(b^{\frac{1}{3}}\right)^{6}\)
c \(\left(c^{12}\right)^{\frac{1}{4}}\)
d \(\left(c^{10}\right)^{\frac{1}{5}}\)

e \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\)

f \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\)

g \(p^{\frac{3}{4}} \times p^{\frac{2}{5}}\)

h \(q^{\frac{3}{2}} \times q^{\frac{2}{3}}\)

i \(x^{\frac{3}{2}} \div x^{\frac{1}{2}}\)

j \(y^{\frac{2}{3}} \div y^{\frac{1}{3}}\)

k \(p^{\frac{3}{4}} \div p^{\frac{2}{5}}\)

l \(q^{\frac{3}{2}} \div q^{\frac{2}{3}}\)

\(\mathbf{m}\left(4 m^{6}\right)^{\frac{1}{2}}\)

n \(\left(27 n^{12}\right)^{\frac{1}{3}}\)

o \(\left(2 x^{\frac{2}{3}}\right)^{3}\)

p \(\left(3 y^{\frac{1}{2}}\right)^{4}\)

6 Evaluate:
a \(4^{-\frac{1}{2}}\)
b \(25^{-\frac{1}{2}}\)
c \(\left(\frac{8}{125}\right)^{-\frac{1}{3}}\)
d \(\left(\frac{64}{27}\right)^{-\frac{1}{3}}\)
e \(32^{-\frac{2}{5}}\)
f \(\left(\frac{1}{81}\right)^{-\frac{1}{4}}\)
g \(81^{-\frac{1}{4}}\)
h \(\left(\frac{1}{25}\right)^{-\frac{1}{2}}\)
i \(\left(\frac{16}{81}\right)^{-\frac{1}{4}}\)
j \(\left(\frac{32}{243}\right)^{-\frac{1}{5}}\)

7 Simplify, expressing the answer with positive indices.
a \(\left(a^{\frac{1}{2}}\right)^{-2}\)
b \(\left(b^{-\frac{2}{3}}\right)^{6}\)
c \(\left(2 x^{\frac{2}{3}}\right)^{-3}\)
d \(\left(3 y^{\frac{1}{2}}\right)^{-4}\)
e \(x^{\frac{1}{2}} \times x^{-\frac{3}{2}}\)
f \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\)
g \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\)
h \(q^{\frac{3}{2}} \times q^{-\frac{2}{3}}\)
i \(x^{\frac{3}{2}} \div x^{-\frac{1}{2}}\)
j \(y^{\frac{2}{3}} \div y^{-\frac{1}{3}}\)
k \(p^{\frac{3}{4}} \div p^{-\frac{2}{5}}\)
l \(q^{\frac{3}{2}} \div q^{-\frac{2}{3}}\)
\(\mathbf{m}\left(4 m^{-6}\right)^{\frac{1}{2}}\)
n \(\left(27 n^{-12}\right)^{\frac{1}{3}}\)
o \(\left(2 x^{-\frac{2}{5}}\right)^{5}\)

\section*{Scientific notation}
Scientific notation, or standard form, is a convenient way to represent very large or very small numbers. It allows such numbers to be easily read.

Here is an example with a very large number. The star Sirius A is approximately \(81362000000000 \mathrm{~km}\) from the Sun. This is about 81 trillion kilometres. This distance can be written neatly in scientific notation as \(8.1362 \times 10^{13} \mathrm{~km}\). You can verify that if we move the decimal point 13 places to the right, inserting the necessary zeros, we arrive back at the number we started with.

We can also use this notation for very small numbers. For example, an angstrom \((\AA)\) is a unit of length equal to \(0.0000000001 \mathrm{~m}\), which is the approximate diameter of a small atom. In scientific notation this is written as \(1.0 \times 10^{-10} \mathrm{~m}\) or \(1 \times 10^{-10} \mathrm{~m}\). When we move the decimal point 10 places to the left, inserting the zeros, we arrive back at the original number. To further this example, the approximate diameter of a uranium atom is \(0.00000000038 \mathrm{~m}\), or \(3.8 \times 10^{-10} \mathrm{~m}\), or \(3.8 \AA\).

By definition, a positive number is in scientific notation (or standard form) if it is written as \(a \times 10^{b}\), where \(1 \leq a<10\) and \(b\) is an integer.

We convert a number into scientific notation by placing a decimal point after the first non-zero digit and multiplying by the appropriate power of 10 .

Note: If the number to be written in scientific notation is greater than 1 , then the index is positive or zero. If the number is positive and less than 1 , then the index is negative.

\section*{Example 21}
Write in scientific notation.

\section*{Solution}
a \(610=6.1 \times 100\)
b \(21000=2.1 \times 10000\)
\(=6.1 \times 10^{2}\)
\(=2.1 \times 10^{4}\)
c \(46000000=4.6 \times 10000000\)
d \(81=8.1 \times 10\)
\(=4.6 \times 10^{7}\)
\(=8.1 \times 10^{1}\)
e \(0.0067=6.7 \div 1000\)
f \(0.00002=2 \div 100000\)
\(=6.7 \times \frac{1}{1000}\)
\(=2 \times \frac{1}{100000}\)
\(=6.7 \times 10^{-3}\)
\(=2 \times 10^{-5}\)
g \(0.07=7 \div 100\)
h \(8.17=8.17 \times 10^{0}\)
\(=7 \times 10^{-2}\)

\section*{Example 22}
Write in decimal form.

\section*{Solution}
a \(2.1 \times 10^{3}=2.1 \times 1000\)
b \(6.3 \times 10^{5}=6.3 \times 100000\)
\(=2100\)
\(=630000\)
c \(5 \times 10^{-4}=5 \times \frac{1}{10000}\)
d \(8.12 \times 10^{-2}=8.12 \times \frac{1}{100}\)
\(=5 \div 10000\)
\(=8.12 \div 100\)
\(=0.0005\)
\(=0.0812\)

Since numbers written in scientific notation involve powers, when these numbers are multiplied, divided or raised to a power, the index laws come into play.

\section*{Example 23}

\section*{Scientific notation}
\begin{itemize}
  \item Scientific notation, or standard form, is a convenient way to represent very large and very small numbers.
  \item To represent a number in scientific notation, insert a decimal point after the first non-zero digit and multiply by an appropriate power of 10.
\end{itemize}

For example:

\(81362000000000=8.1362 \times 10^{13}\) and \(0.00000000038=3.8 \times 10^{-10}\)

Some of the exercises in the rest of this chapter are best done using a calculator.

\section*{Exercise 8D}
1 Write as a power of 10.
a 10
b 100
c 1000
d 10000
e 1000000
f 1000000000

g a googol - which is 1 followed by 100 zeros.

Note: \(10^{6}\) is a million, \(10^{9}\) is a billion and \(10^{12}\) is a trillion.

2 Write as a power of 10 .



5 The mass of the Earth is approximately \(6000000000000000000000000 \mathrm{~kg}\). Write this value in scientific notation.






\section*{Significant figures}
Suppose that we are using a ruler marked in centimetres and millimetres to measure the length of a sheet of paper. We find that the length of the edge of the paper falls between \(15.2 \mathrm{~cm}\) and \(15.3 \mathrm{~cm}\) but is closer to \(15.3 \mathrm{~cm}\). The normal procedure is to use rounding and write the measurement as \(15.3 \mathrm{~cm}\).

In scientific notation, our measurement is \(1.53 \times 10^{1} \mathrm{~cm}\). We say that the length is \(1.53 \times 10^{1} \mathrm{~cm}\) correct to 3 significant figures. This means that the length is between \(1.525 \mathrm{~cm}\) and \(1.535 \mathrm{~cm}\).

All measurements involve rounding to one level of accuracy or another. For example, you may read that the mass of an electron is about \(0.00000000000000000000000000091093826 \mathrm{~g}\). This usually means that a measurement was made and the last digit 6 is a rounding digit and is therefore not completely accurate. The other digits are accurate. Writing this in scientific notation, we say that the mass of an electron is \(9.1093826 \times 10^{-28} \mathrm{~g}\) correct to \(\mathbf{8}\) significant figures because there are 8 digits in the factor 9.1093826 before the power of 10 .

The mass of the Earth is about \(5.9736 \times 10^{24} \mathrm{~kg}\). How many significant digits are there in this measurement?

There are 5 digits in 5.9736 , so the measurement is correct to 5 significant figures.

Notice that, to use the idea of significant figures, we must first express the number in scientific notation.

People sometimes apply the same kind of ideas to shorten a given decimal number containing many digits. This need not have anything to do with measurement. It is a way of abbreviating the information you are given. The procedure is called writing the number to a certain number of significant figures.

To write a number to a specified number of significant figures, first write the number in scientific notation and then round to the required number of significant figures.

For example, \(0.00034061=3.4061 \times 10^{-4}\)

\[
\begin{aligned}
& \approx 3 \times 10^{-4} \quad(\text { correct to } 1 \text { significant figure }) \\
& \approx 3.4 \times 10^{-4} \quad(\text { correct to } 2 \text { significant figures }) \\
& \approx 3.41 \times 10^{-4} \quad(\text { correct to } 3 \text { significant figures }) \\
& \approx 3.406 \times 10^{-4} \quad(\text { correct to } 4 \text { significant figures })
\end{aligned}
\]

\section*{Example 24}
Write in scientific notation and then round correct to 3 significant figures.
a 235.674
b 0.00724546

\section*{Solution}
a \(235.674=2.35674 \times 10^{2}\)

\[
\approx 2.36 \times 10^{2}
\]

b \(0.00724546=7.24546 \times 10^{-3}\)

\[
\approx 7.25 \times 10^{-3}
\]

\section*{Example 25}

\section*{Solution}
a \(276000000=2.76 \times 10^{8}\)
b \(0.000000654=6.54 \times 10^{-7}\)
\(\approx 2.8 \times 10^{8}\)
\(\approx 6.5 \times 10^{-7}\)

\section*{Significant figures}
\begin{itemize}
  \item A number may be expressed with different numbers of significant figures.
\end{itemize}

For example: \(\pi\) is 3.1 to 2 significant figures, 3.14 to 3 significant figures, 3.142 to 4 significant figures and so on.

\begin{itemize}
  \item To write a number correct to a specified number of significant figures, first write the number in scientific notation and then round to the required number of significant figures.
\end{itemize}

\section*{Exercise 8E}
1 Write in scientific notation, correct to 3 significant figures.


3 Along each row, write the respective number in scientific notation, correct to the indicated number of significant figures.



e \(\sqrt{4.863 \times 10^{-12}}\)
f \(\frac{177.41 \times 0.048}{16.23}\)
g \(\frac{7.932 \times 10^{12} \times 9.4 \times 10^{-10}}{0.000000000416}\)
h \(\frac{579.2 \times 0.6231}{79.05 \times 115.4}\)
i \(\frac{74510000000}{6.4 \times 10^{-18} \times 4.4 \times 10^{23}}\)
j \(\frac{79.99}{\sqrt{48.92}+11.68^{2}}\)
k \(\frac{15.62^{2}(79.1+111.7)}{12.46+4.48^{3}}\)
l \(56.21 \times 12+\frac{1}{2} \times 9.8 \times 12^{2}\)

6 Estimate each of the following, correct to 1 significant figure, using appropriate units. In each case explain how you obtained your answer.

a The thickness of a sheet of paper

b The volume of your classroom

c The height of a six-storey building

d The area required for a car park for 500 cars

e The total printed area of a 600 -page novel

f The volume of a warehouse that can store 300000 pairs of shoes (still in their boxes)

g The length of a queue if every student in your school is standing in it

Discuss your answers with others in your class and with your teacher.

