1.2 Mathematical skills
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1.2 Mathematical skills (ESAB)
You should be comfortable with scientific notation and how to write scientific notation. You should also be able to easily convert between different units and change the subject of a formula. In addition, concepts such as rate, direct and indirect proportion, fractions and ratios and the use of constants in equations are important.
Rounding off (ESAC)
Certain numbers may take an infinite amount of paper and ink to write out. Not only is that impossible, but writing numbers out to a high precision (many decimal places) is very inconvenient and rarely gives better answers. For this reason we often estimate the number to a certain number of decimal places.
Rounding off a decimal number to a given number of decimal places is the quickest way to approximate a number. For example, if you wanted to roundoff \(\text{2,6525272}\) to three decimal places then you would first count three places after the decimal. Next you mark this point with a \(\): \(\text{2,652}5272\). All numbers to the right of \(\) are ignored after you determine whether the number in the third decimal place must be rounded up or rounded down. You round up the final digit (make the digit one more) if the first digit after the \(\) is greater than or equal to \(\text{5}\) and round down (leave the digit alone) otherwise. So, since the first digit after the \(\) is a \(\text{5}\), we must round up the digit in the third decimal place to a \(\text{3}\) and the final answer of \(\text{2,6525272}\) rounded to three decimal places is \(\text{2,653}\).
In a calculation that has many steps, it is best to leave the rounding off right until the end. This ensures that your answer is more accurate.
Scientific notation (ESAD)
In science one often needs to work with very large or very small numbers. These can be written more easily (and more compactly) in scientific notation, in the general form:
\[N\times {10}^{n}\]where \(N\) is a decimal number between \(\text{0}\) and \(\text{10}\) that is rounded off to a few decimal places. \(n\) is known as the exponent and is an integer. If \(n>0\) it represents how many times the decimal place in \(N\) should be moved to the right. If \(n<0\), then it represents how many times the decimal place in \(N\) should be moved to the left. For example, \(\text{3,24} \times \text{10}^{\text{3}}\) represents \(\text{3 240}\) (the decimal moved three places to the right) and \(\text{3,24} \times \text{10}^{\text{3}}\) represents \(\text{0,00324}\) (the decimal moved three places to the left).
If a number must be converted into scientific notation, we need to work out how many times the number must be multiplied or divided by \(\text{10}\) to make it into a number between \(\text{1}\) and \(\text{10}\) (i.e. the value of \(n\)\(\text{1}\) and \(\text{10}\) is (the value of \(N\)). We do this by counting the number of decimal places the decimal comma must move.
For example, write the speed of light (\(\text{299 792 458}\) \(\text{m·s$^{1}$}\)) in scientific notation, to two decimal places. First, we find where the decimal comma must go for two decimal places (to find \(N\)) and then count how many places there are after the decimal comma to determine \(n\).
In this example, the decimal comma must go after the first \(\text{2}\), but since the number after the \(\text{9}\) is \(\text{7}\), \(N=3,00\). \(n=8\) because there are \(\text{8}\) digits left after the decimal comma. So the speed of light in scientific notation, to two decimal places is \(\text{3,00} \times \text{10}^{\text{8}}\) \(\text{m·s$^{1}$}\).
We can also perform addition, subtraction, multiplication and division with scientific notation. The following two worked examples show how to do this:
Worked example 1: Addition and subtraction with scientific notation
\(\text{1,99} \times \text{10}^{\text{26}} + \text{1,67} \times \text{10}^{\text{27}}  \text{2,79} \times \text{10}^{\text{25}} = ?\)
Make all the exponents the same
To add or subtract numbers in scientific notation we must make all the exponents the same:
\begin{align*} \text{1,99} \times \text{10}^{\text{26}} & = \text{0,199} \times \text{10}^{\text{25}} \\ \text{and } \text{1,67} \times \text{10}^{\text{27}} & = \text{0,0167} \times \text{10}^{\text{25}} \end{align*}Carry out the addition and subtraction
Now that the exponents are the same we can simply add or subtract the \(N\) part of each number:
\[\text{0,199} + \text{0,0167}  \text{2,79} = \text{2,5743}\]Write the final answer
To get the final answer we put the common exponent back:
\(\text{2,5743} \times \text{10}^{\text{25}}\)
Note that we follow the same process if the exponents are positive. For example \(\text{5,1} \times \text{10}^{\text{3}} + \text{4,2} \times \text{10}^{\text{4}} = \text{4,71} \times \text{10}^{\text{4}}\)
Worked example 2: Multiplication and division with scientific notation
\(\text{1,6} \times \text{10}^{\text{19}} \times \text{3,2} \times \text{10}^{\text{19}} \div \text{5} \times \text{10}^{\text{21}}\)
Carry out the multiplication
For multiplication and division the exponents do not need to be the same. For multiplication we add the exponents and multiply the \(N\) terms:
\[\text{1,6} \times \text{10}^{\text{19}} \times \text{3,2} \times \text{10}^{\text{19}} = \left(\text{1,6} \times \text{3,2}\right) \times {10}^{19+\left(19\right)} = \text{5,12} \times \text{10}^{\text{38}}\]Carry out the division
For division we subtract the exponents and divide the \(N\) terms. Using our result from the previous step we get:
\[\text{5,12} \times \text{10}^{\text{38}} \div \text{5} \times \text{10}^{\text{21}} = \left(\text{5,12} \div 5\right)\times {10}^{38\left(21\right)} = \text{1,024} \times \text{10}^{\text{17}}\]Write the final answer
The answer is: \(\text{1,024} \times \text{10}^{\text{17}}\)
Note that we follow the same process if the exponents are positive. For example: \(\text{5,1} \times \text{10}^{\text{3}} \times \text{4,2} \times \text{10}^{\text{4}} = \text{21,42} \times \text{10}^{\text{7}} = \text{2,142} \times \text{10}^{\text{8}}\)
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