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Section 34.1 Numbers That Are Annoyingly Big or Small

In science courses, there are important values that are sometimes very large or very small. For example, in a course such as astronomy, you might learn that the average distance from the earth to the sun is approximately 149600000000 meters. In chemistry, you will run into a unit called a mole, which represents approximately 60221407600000000000000 molecules. You might also have very small numbers, such as atoms being approximately 0.0000000001 meters in size.

One of the challenges with numbers like these is that it can be very difficult to compare them. For example, determine which of the two numbers is the larger number: 10000000000000000000 or 5000000000000000000. It can be an annoying and frustrating process to have to count out all of the digits in those numbers, and it's easy to make mistakes. Some people will use commas or small spaces every three digits (starting from the decimal point) in order to help, but this is still a lot of writing and a lot of opportunities for errors.

This is addressed in multiple ways. We are going to focus on scientific notation in this section. The idea of scientific notation is to take these long numbers and write them in a more compact form that makes them easier to think about and easier to work with.

Definition 34.1.1. Scientific Notation.

Scientific notation is a representation of numbers (that are typically very large or very small) using the form \(a \times 10^n\text{,}\) where \(a\) is a decimal number between 1 and 10 (including 1 and excluding 10) and \(n\) is some integer.

The value of scientific notation is that it replaces the process of counting out zeros and displays that value explicitly. It comes out of the fact that multiplying or dividing a number by a power of corresponds to "moving the decimal point" a certain number of places.

Activity 34.1.1. Converting from Scientific Notation.

The idea of "moving the decimal point" is extremely common when performing these arithmetic calculations. This idea is just a reflection of the fact that we work with a system for writing down numbers. When multiplying by a positive power of the multiplication makes the final result bigger.

And when multiplying by a negative power of the multiplication makes the final result smaller.

Try it!

Convert the numbers \(1.234 \times 10^5\) and \(9.87 \times 10^{-6}\) into standard form.

Solution.
\begin{equation*} 1.234 \times 10^5 = 123400 \qquad 9.87 \times 10^{-6} = 0.00000987 \end{equation*}

Activity 34.1.2. Converting to Scientific Notation.

Putting a number into scientific notation follows the same idea. However, in this case you must decide the number of positions that the decimal point must move. Some people say that the signs are "switched" when putting the number into scientific notation, but that tends to cause confusion because students often forget which way the correct way is and so when they "switch" they still get it wrong. The best way to avoid errors is simply to take a moment and think about the final answer. If you started with a big number, then your scientific notation should represent a big number. And if you started with a small number, your scientific notation should represent a small number.

The other important factor to remember is that your value should be a number between and but not itself. Another way to think about it is that you only want one digit to the left of the decimal point (and that digit should not be zero) when you're done.

Try it!

Convert the numbers and to scientific notation.

Solution.
\begin{equation*} 3480000000 = 3.48 \times 10^{9} \qquad 0.0000736 = 7.36 \times 10^{-5} \end{equation*}

Another reason for scientific notation is that it allows us to more easily perform arithmetic. The product \(2000000000 \cdot 400000000000\) is definitely an 8 followed by some number of zeros, but counting those zeros runs into the same risks of errors as we saw above.

Activity 34.1.3. Multiplication in Scientific Notation.

To perform arithmetic with scientific notation correctly, it is helpful to think about some basic algebra ideas. Instead of thinking about \(a \times 10^n\text{,}\) it can be more helpful to think about \(a \times x^n\) where \(x = 10\text{.}\) The benefit to this is that it allows you to use your past experience and intuition.

We'll start with multiplication. What is \(2x^7 \cdot 3x^9\text{?}\) You might recall that we multiply the number parts and then multiply the variable parts, and that the variable parts satisfy \(x^n \cdot x^m = x^{n+m}\text{,}\) which gives you as the final result.

Try it!

Calculate \((2 \times 10^7) \cdot (3 \times 10^9)\text{.}\)

Solution.
\begin{equation*} (2 \times 10^7) \cdot (3 \times 10^9) = 6 \times 10^{16} \end{equation*}

Activity 34.1.4. Adding and Subtracting with Scientific Notation.

When adding and subtracting using scientific notation, you need to be a bit more careful. We need to have like terms in order for us to combine them. For example, \(6x^8 + 2x^7\) cannot be simplified. In order to get around this, we need to rewrite the numbers so that they have the same power of 10.

In order to do that, we need to think about other representations of numbers. Here is a collection of expressions that are all equivalent to

\begin{equation*} 5000 = 5000 \times 10^0 = 500 \times 10^1 = 50 \times 10^2 = 5 \times 10^3 = 0.5 \times 10^4 = 0.05 \times 10^5 = \cdots \end{equation*}

The important feature to notice is that as the power of increases, the other number decreases. This balancing act makes sense if you think about the overall result. If the overall result is not going to change, then if one number of the product gets larger, the other must get smaller.

Once that pattern is recognized, the calculations are simply a matter of execution. Here are two common approaches to the same calculation.

\begin{equation*} \begin{aligned} (6 \times 10^8) + (2 \times 10^7) \amp = (6 \times 10^8) + (0.2 \times 10^8) \\ \amp = 6.2 \times 10^8 \end{aligned} \end{equation*}

And here is the second:

\begin{equation*} \begin{aligned} (6 \times 10^8) + (2 \times 10^7) \amp = (6 \times 10^8) + (0.2 \times 10^8) \\ \amp = 6.2 \times 10^8 \end{aligned} \end{equation*}

Some prefer this first way because it avoids decimals. Some prefer the second way because it's shorter. It doesn't matter which way you do it as long as your steps are valid.

Try it!

Calculate \((8 \times 10^6) + (6 \times 10^8)\text{.}\) Give your final answer in scientific notation.

Solution.
\begin{equation*} \begin{aligned} (8 \times 10^6) + (6 \times 10^8) \amp = (8 \times 10^6) + (600 \times 10^6) \\ \amp = 608 \times 10^6 \\ \amp = 6.08 \times 10^8 \end{aligned} \end{equation*}

Activity 34.1.5. Adding and Subtracting with Negative Powers and Scientific Notation.

The same idea can be applied when the exponent is negative, but you need to be careful. As the exponent gets more negative, the power of becomes smaller and so the other number must get bigger.

\begin{equation*} 0.007 = 0.007 \times 10^0 = 0.07 \times 10^{-1} = 0.7 \times 10^{-2} = 7 \times 10^{-3} = 70 \times 10^{-4} = \cdots \end{equation*}

Try it!

Calculate \((5 \times 10^{-6}) + (4 \times 10^{-5})\text{.}\) Give your final answer in scientific notation.

Solution.
\begin{equation*} \begin{aligned} (5 \times 10^{-6}) + (4 \times 10^{-5}) \amp = (5 \times 10^{-6}) + (40 \times 10^{-6}) \\ \amp = 45 \times 10^{-6} \\ \amp = 4.5 \times 10^{-5} \end{aligned} \end{equation*}