# 1.3: Handling Large and Small Numbers


Example 3 from Measurements, Quantities and Unity Factors illustrates a common occurrence in science—results often involve very large numbers or very small fractions. The United States used 66 500 000 000 000 000 000 J (joules) of energy in 1971, and the mass of a water molecule is 0.000 000 000 000 000 000 000 029 9 g. Such numbers are inconvenient to write and hard to read correctly. (We have divided the digits into groups of three to make it easier to locate the decimal point. Spaces are used instead of commas because many countries use a comma to indicate the decimal.)

There are two ways of handling this problem. We can express a quantity in larger or smaller units, as in Example 4 from Measurements, Quantities and Unity Factors, or we can use a better way to write small and large numbers. The latter approach involves what is called scientific notation or exponential notation. The position of the decimal point is indicated by a power (or exponent) of 10. For example,

$138= 13.8\cdot 10= 1.38\cdot 10\cdot 10= 1.38\cdot 10^2$

$0.004\ 83= \frac{4.83}{10\cdot10\cdot10} = 4.83\cdot\frac{1}{10^3} = 4.83\cdot10^{-3}$

A number with a negative exponent is simply the reciprocal of (one divided by) the same number with the equivalent positive exponent. Therefore decimal fractions (numbers between zero and one) may be expressed using negative powers of 10. Numbers between 1 and 10 require no exponential part, and those larger than 10 involve positive exponents. By convention the power of 10 is chosen so that there is one digit to the left of the decimal point in the ordinary number. That is, we would usually write 5280 as 5.28 × 103 not as 0.528 × 104 or 52.8 × 102.

To convert a number from ordinary to scientific notation, count how many places the decimal point must be shifted to arrive at a number between 1 and 10. If these shifts are to the left, the number was large to begin with and we multiply by a large (that is, positive) power of 10. If the shift is to the right, a reciprocal (negative) power of 10 must be used.

Example $$\PageIndex{1}$$ : Scientific Notation

Express the following numbers in scientific notation: (a) 7563; (b) 0.0156.

Solution:

a) In this case the decimal point must be shifted left three places:

Therefore we use an exponent of +3:

$7563= 7.563\cdot10^3$

b) Shifting the decimal point two places to the right yields a number between 1 and 10:

Therefore the exponent is –2:

$0.0156= 1.56\cdot10^{-2}$

When working with exponential notation, it is often necessary to add, subtract, multiply, or divide numbers. When multiplying and dividing, you must remember that multiplication corresponds to addition of exponents, and division to their subtraction.

Multiplication: $$\text{10}^a\cdot\text{10}^b = \text{10}^{\small(a + b\small)}$$

Division: $$\frac{10^a}{10^b} = 10^{(a - b)}$$

Hence $$(3.0\cdot10^5)\cdot(5.0\cdot10^3) = 15.0\cdot10^{(5+3)}= 15.0\cdot10^8= 1.50\cdot10^9$$ and $$\frac{3.0\cdot10^5}{5.0\cdot10^3} =0.6\cdot10^{(5- 3)} = 0.6\cdot10^2= 6.0\cdot10$$

Example $$\PageIndex{2}$$ : Exponential Notation

1. $$(3.89 \cdot 10^5) \cdot (1.09 \cdot 10^{-3})$$
2. $$(6.41\cdot10^{-5}) \cdot(2.72 \cdot10^{-2})$$
3. $$\frac{(5.0\cdot10^6)}{(3.98\cdot10^8)}$$
4. $$\frac{(7.53\cdot10^{-3})}{(8.57\cdot10^{-5})}$$

Solution:

1. $$(3.89\cdot10^5)\cdot(1.09 \cdot10^{-3}) = 3.89\cdot1.09\cdot10^{5 + (-3)}= 4.24\cdot10^2$$
2. $$(6.41\cdot10^{-5})\cdot(2.72 \cdot10^{-2})$$= $$6.41\cdot2.72\cdot10^{-5 + (-2)}$$ = $$17.43 \cdot10^{-7}$$ = $$1.743 \cdot10^{-6}$$
3. $$\frac{(5.0\cdot10^6)}{(3.98\cdot10^8)}= \frac{5.0}{3.98} \cdot10^{6-8}= 1.26\cdot10^{-2}$$
4. $$\frac{(7.53\cdot10^{-3})}{(8.57\cdot10^{-5})} = \frac{7.53}{8.57}\cdot10^{-3 - (-5)} = 0.879\cdot10^2 = 8.79\cdot10^1$$

Addition and subtraction require that all numbers be converted to the same power of 10. (This corresponds to lining up the decimal points.)

Example $$\PageIndex{3}$$ : Scientific Notation

a)$$(6.32\cdot10^2) – (1.83 \cdot10^\cdot{-1})$$

b)$$(3.72 \cdot10^4) + (1.63\cdot10^5) – (1.7 10^3)$$

Solution:

a) First convert to the same power of 10; then add the ordinary numbers.

\begin{align}&&6.32\cdot10^2&=& 632 \\&&–1.83\cdot10^{-1}&=& – 0.183 \\632 – 0.183&=& 631.817&=& 6.318 17\cdot10^2 \end{align}

b) Convert all powers of 10 to 104.

\begin{align}3.72\cdot10^4 &=& 3.72 \cdot10^4 &=& 3.72 × 10^4 \\1.63\cdot10^5&=& 1.63\cdot10 \cdot10^4&=& 16.3 \cdot10^4 \\–1.7 \cdot10^3&=& –1.7 \cdot10^{-1}\cdot10^4&=& – 0.17\cdot10^4 \\ (3.72\cdot10^4) + (16.3\cdot10^4) - (0.17 \cdot10^4)&=& 19.85 \cdot10^4&=& 1.985\cdot10^5\end{align}

Scientific notation is becoming more common every day. Many electronic pocket calculators use it to express numbers which otherwise would not fit into their displays. For example, an eight-digit calculator could not display the number 6 800 000 000.The decimal point would remain fixed on the right, and the 6 and the 8 would “overflow” to the left side. Such a number is often displayed as 6.8 09, which means 6.8 × 109. If you use a calculator which does not have scientific notation, we recommend that you express all numbers as powers of 10 before doing any arithmetic. Follow the rules in the last two examples, using your calculator to do arithmetic on the ordinary numbers. You should be able to add or subtract the powers of 10 in your head.

Computers also are prone to print results in scientific notation, and they use yet another minor modification. The printed number 2.3074 E-07 means 2.3074 × 10–7 for example. In this case the E indicates that the number following is an exponent of 10.

Figure $$\PageIndex{1}$$ The level of a liquid in a graduated cylinder. The saucer-shaped surface of a liquid in a tube is called a meniscus.