Title: Default for 1.1 Mathematical Notation and Symbols
Course: HELM Workbooks 2023-2024

Exercise

Fully simplify without using a calculator

1. \(-[value] \times 3 =\)
2. \([value] \times -3 =\)
3. \(-[value] \times -3 =\)
4. \(\dfrac{[value]}{-3} =\)
5. \(\dfrac{[value]}{7} =\)
6. \(\dfrac{[value]}{-12} =\)
Feedback: 1. The product of a negative and a positive is a negative \(\implies -[value] \times 3 = [value]\) 2. The product of a positive and a negative is a negative \(\implies [value] \times -3 = [value]\) 3. The product of a negative and a negative is a negative \(\implies -[value] \times -3 = [value]\) 4. Dividing a positive number by a negative number gives a negative number \(\implies \dfrac{[value]}{-3} = [value]\) 5. Dividing a negative number by a positive number gives a negative number \(\implies \dfrac{[value]}{7} = [value]\) 6. Dividing a negative number by a negative number gives a positive number \(\implies \dfrac{[value]}{-12} = [value]\)

Exercise

Evaluate

1. \([value] + 2 \times 6=\)
2. \([value] − 2 − 6 =\)
3. \([value] + 2 − 6 =\)
4. \([value] − 3 \times 2 =\)
5. \([value] \times 3 − 2 =\)
6. \(([value] \div 3) + 2 =\)
7. \([value] \div 3 + 2 =\)
8. \([value] + 4 − 11 − 2 =\)
9. \([value] \times 4 + 11 \times 2 =\)
10. \(−(−[value]) =\)
11. \(7 − (−[value]) =\)
12. \(−19 − (−[value]) =\)
13. \(−19 + (−[value]) =\)
Feedback: We use the BODMAS rules to know in which order to perform the calculations 1. We perform the multiplication first and then the addition \(\implies [value] + 2 \times 6 = [value] + 12 = [value]\) 2. The equation contains only subtraction so we evaluate by working from left to right \(\implies [value] − 2 − 6 = [value] - 6 = [value]\) 3. The equation contains only addition and subtraction so we evaluate by working from left to right \(\implies [value] + 2 − 6 = [value] - 6 = [value]\) 4. We first perform the multiplication and then subtract the two remaining terms from one another \(\implies [value] − 3 × 2 = [value] - 6 = [value]\) 5. We first perform the multiplication and then subtract the two remaining terms from one another \(\implies [value]× 3 − 2 = [value] - 2 = [value]\) 6. We first perform the operation within the brackets and then add the remaining terms \(\implies ([value] ÷ 3) + 2 = [value] + 2 = [value]\) 7. We first perform division between the two adjacent terms and then add the remaining two terms together \(\implies [value] ÷ 3 + 2 = [value] + 2 = [value]\) 8. The equation contains only addition and subtraction so we evaluate by working from left to right \(\implies [value] + 4 − 11 − 2 = [value] - 11 - 2 = [value] - 2 = [value]\) 9. Firstly perform multiplication between the two terms furthest to the left, then perform the remaining multiplication and finally add the two remaining terms together\(\implies [value] \times 4 + 11 \times 2 = [value] + 11 \times 2 = [value] + 22 = [value]\) 10. Taking the negative of a negative number returns the positive of that number \(\implies −(−[value]) = [value]\) 11. Subtracting a negative number is equivalent to adding a positive number \(\implies 7 − (−[value]) = 7 + [value] = [value]\) 12. Subtracting a negative number is equivalent to adding a positive number \(\implies −19 − (−[value]) = -19 + [value] = [value]\) 13. Adding a negative number is equivalent to subtracting a positive number \(\implies −19 + (−[value]) = -19 - [value] = [value]\)

Exercise

Evaluate

1. \(|[value]| =\)
2. \(|[value]| =\)
3. \(|-[value]| =\)
4. \(|[value]| =\)
5. \(|0.01-0.001| =\)
6. \([value]! =\)
7. \([value]! - 4! =\)
8. \(\dfrac{[value]!}{[value]!} =\)
Feedback: We evaluate these terms using the fact that the modulus of a number is the size of that number regardless of its sign and that the factorial of the number is given by the equation: \(n! = n × (n − 1) \times (n − 2) . . . 5 \times 4 \times 3 \times 2 \times 1\) 1. \(|[value]| = [value]\) 2. \(|[value]| = [value]\) 3. \(|-[value]| = [value]\) 4. \(|[value]| = [value]\) 5. \(|0.01-0.001| = |0.009| = [value]\) 6. \([value]! = [value]\) 7. \([value]! - 4! = [value] - [value] = [value]\) 8. \(\dfrac{[value]!}{[value]!} = \dfrac{[value] \times [value] \times [value] \times ... \times 2 \times 1}{[value] \times [value] \times ... \times 2 \times 1} = [value] \times \dfrac{[value] \times [value] \times ... \times 2 \times 1}{[value] \times [value] \times ... \times 2 \times 1} = [value] \times 1 = [value]\)

Exercise

Evaluate

1. \([value]+(-9) =\)
2. \([value]-(-8) =\)
3. \(-[value]+(-2) =\)
4. \([value]-(-3) =\)
Feedback: 1. \([value]+(-9) = [value]-9 = [value]\) 2. \([value]-(-8) = [value]+8 = [value]\) 3. \(-[value]+(-2) = -[value]-2 = [value]\) 4. \([value]-(-3) = [value]+3 = [value]\)

Exercise

State the reciprocal of

1. \([value]\)
2. \(\dfrac{[value]}{16}\)
3. \(\dfrac{[value]}{[value]}\)
4. \(\dfrac{[value]}{[value]}\)
Feedback: The reciprocal of a number is found by inverting it hence: 1. The reciprocal of \([value]\) is \(\dfrac{1}{[value]}\) 2. The reciprocal of \(\dfrac{[value]}{16}\) is \(\dfrac{16}{[value]}\) 3. The reciprocal of \(\dfrac{[value]}{[value]}\) is \(\dfrac{[value]}{[value]}\) 4. The reciprocal of \(\dfrac{[value]}{[value]}\) is \(\dfrac{[value]}{[value]}\)

Exercise

Which of the following statements are true ?

1. \(-[value] \le [value]\)
2. \(-[value] \le -[value]\)
3. \(-[value] \le |[value]|\)
4. \(|-[value]| < [value]\)
5. \(-[value] \le [value]\)
6. \([value]! \le [value]!\)
7. \([value]! \le [value]!\)

Numbers, Operations and Common Notations

A knowledge of the properties of numbers is fundamental to the study of engineering mathematics. Students who possess this knowledge will be well-prepared for the study of algebra. Much of the terminology used throughout the rest of this Section can be most easily illustrated by applying it to numbers. For this reason we strongly recommend that you work through this Section even if the material is familiar.

The number line

A useful way of picturing numbers is to use a number line. Figure 1 shows part of this line. Positive numbers are represented on the right-hand side of this line, negative numbers on the left-hand side. Any whole or fractional number can be represented by a point on this line which is also called the real number line, or simply the real line. Study Figure 1 and note that a minus sign is always used to indicate that a number is negative, whereas the use of a plus sign is optional when describing positive numbers.

The line extends indefinitely both to the left and to the right. Mathematically we say that the line extends from minus infinity to plus infinity. The symbol for infinity is \(\infty\).

[Image: Numbers can be represented on a number line]
Numbers can be represented on a number line

The symbol \(>\) means ‘greater than’; for example \(6>4\). Given any number, all numbers to the right of it on the number line are greater than the given number. The symbol \(<\) means ‘less than’; for example \(-3<19\). We also use the symbols \(\ge\) meaning ‘greater than or equal to’ and \(\le\) meaning ‘less than or equal to’. For example, \(7\le 10\) and \(7\le 7\) are both true statements.

Sometimes we are interested in only a small section, or interval, of the real line. We write \([1,3]\) to denote all the real numbers between 1 and 3 inclusive, that is 1 and 3 are included in the interval. Therefore the interval \([1,3]\) consists of all real numbers \(x\), such that \(1\le x\le 3\). The square brackets, \([,]\) mean that the end-points are included in the interval and such an interval is said to be closed. We write \((1,3)\) to represent all real numbers between 1 and 3, but not including the end-points. Thus \((1,3)\) means all real numbers \(x\) such that \(1<x<3\), and such an interval is said to be open. An interval may be closed at one end and open at the other. For example, \((1,3]\) consists of all numbers \(x\) such that \(1<x\le 3\). Intervals can be represented on a number line. A closed end-point is denoted by \(\bullet\); an open end-point is denoted by \(\circ\). The intervals \((-6,-4)\), \([-1,2]\) and \((3,4]\) are illustrated in Figure 2.

[Image: The intervals (−6, −4), [−1, 2] and (3, 4] depicted on the real line]
The intervals \((-6,-4)\), \([-1,2]\) and \((3,4]\) depicted on the real line

Calculation with Numbers

To perform calculations with numbers we use the operations, \(+\), \(-\), \(\times\) and \(\div\).

Addition (\(+\))

We say that \(4+5\) is the sum of 4 and 5. Note that \(4+5\) is equal to \(5+4\) so that the order in which we write down the numbers does not matter when we are adding them. Because the order does not matter, addition is said to be commutative. This first property is called commutativity. When more than two numbers are to be added, as in \(4+8+9\), it makes no difference whether we add the 4 and 8 first to get \(12+9\), or whether we add the 8 and 9 first to get \(4+17\). Whichever way we work we will obtain the same result, 21. Addition is said to be associative. This second property is called associativity.

Subtraction (\(-\))

We say that \(8-3\) is the difference of \(8\) and \(3\). Note that \(8-3\) is not the same as \(3-8\) and so the order in which we write down the numbers is important when we are subtracting them i.e. subtraction is not commutative. Subtracting a negative number is equivalent to adding a positive number, thus \(7-(-3)=7+3=10\).

The plus or minus sign (\(\pm\))

In engineering calculations we often use the notation plus or minus, \(\pm\). For example, we write \(12\pm 8\) as shorthand for the two numbers \(12+8\) and \(12-8\), that is 20 and 4. If we say a number lies in the range \(12\pm 8\) we mean that the number can lie between 4 and 20 inclusive.

Multiplication (\(\times\))

The instruction to multiply, or obtain the product of, the numbers 6 and 7 is written \(6\times 7\). Sometimes the multiplication sign is missed out altogether and we write \((6)(7)\). Note that \((6)(7)\) is the same as \((7)(6)\) so multiplication of numbers is commutative. If we are multiplying three numbers, as in \(2\times 3\times 4\), we obtain the same result whether we multiply the 2 and 3 first to obtain \(6\times 4\), or whether we multiply the 3 and 4 first to obtain \(2\times 12\). Either way the result is 24. Multiplication of numbers is associative. Recall that when multiplying positive and negative numbers the sign of the result is given by the rules given in the Key Point below.

Key Point

Multiplication

- positive \(\times\) positive = positive
- positive \(\times\) negative = negative
- negative \(\times\) positive = negative
- negative \(\times\) negative = positive

For example, \((-4)\times 5 = -20\), and \((-3)\times (-6)=18\).
When dealing with fractions we sometimes use the word ‘of’ as in ‘find \(\dfrac{1}{2}\) of 36’. In this context ‘of’ is equivalent to multiply, that is
\(\qquad \dfrac{1}{2}\) of 36 is equivalent to \(\dfrac{1}{2}\times 36 = 18\)
Division (\(\div\)) or (\(/\))
The quantity \(8\div 4\) means \(8\) divided by \(4\). This is also written as \(8/4\) or \(\dfrac{8}{4}\) and is known as the quotient of \(8\) and \(4\). In the fraction \(\dfrac{8}{4}\) the top line is called the numerator and the bottom line is called the denominator. Note that \(8/4\) is not the same as \(4/8\) and so the order in which we write down the numbers is important. Division is not commutative.
When dividing positive and negative numbers, recall the rules in the Key Point below for determining the sign of the result:
Key Point
Division
  - \(\dfrac{\text{positive}}{\text{positive}} = \text{positive}\)
  - \(\dfrac{\text{positive}}{\text{negative}} = \text{negative}\)
  - \(\dfrac{\text{negative}}{\text{positive}} = \text{negative}\)
  - \(\dfrac{\text{negative}}{\text{negative}} = \text{positive}\)

The reciprocal of a number
The reciprocal of a number is found by inverting it. If the number \(\dfrac{2}{3}\) is inverted we get \(\dfrac{3}{2}\). So the reciprocal of \(\dfrac{2}{3}\) is \(\dfrac{3}{2}\). Because we can write 4 as \(\dfrac{4}{1}\), the reciprocal of 4 is \(\dfrac{1}{4}\).

The modulus notation (\(| |\))

The modulus of a number is the size of that number regardless of its sign. For example \(|4|\) is equal to 4, and \(|-3|\) is equal to 3. The modulus of a number is thus never negative.

Exercise

State the reciprocal of

1. The reciprocal of \(\dfrac{[value]}{11}\) is
2. The reciprocal of \(\dfrac{1}{[value]}\) is
3. The reciprocal of [value] is
Feedback: The reciprocal of a number is found by inverting it hence: 1. The reciprocal of \(\dfrac{[value]}{11}\) is \(\dfrac{11}{[value]}.\) 2. The reciprocal of \(\dfrac{1}{[value]}\) is \([value].\) 3. The reciprocal of [value] is \(-\dfrac{1}{[value]}.\)

The factorial symbol (\(!\))

Another commonly used notation is the factorial, denoted by the exclamation mark ‘!’. The number \(5!\), read ‘five factorial’, or ‘factorial five’, is a shorthand notation for the expression \(5\times 4 \times 3 \times 2 \times 1\), and the number \(7!\) is shorthand for \(7\times 6\times 5\times 4 \times 3 \times 2 \times 1\). Note that \(1!\) equals 1, and by convention \(0!\) is defined as 1 also. Your scientific calculator is probably able to evaluate factorials of small integers. It is important to note that factorials only apply to positive integers.

Key Point

Factorial notation

If n is a positve integer then \(\qquad n! = n\times (n-1)\times (n-2)\times ... \times 4 \times 3 \times 2 \times 1\)

Example

1. Evaluate 4! and 5! without using a calculator.
2. Use your calculator to find \(10\)!.

Solution

1. \(4! = 4 \times 3 \times 2 \times 1 = 24\) Similarly, 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 Note that 5! = 5 \times 4! = 5 \times 24 = 120
2. \(10! = 3, 628, 800\)

Exercise

State the modulus of

1. The modulus of \(-\dfrac{[value]}{7}\) is
2. The modulus of \(\dfrac{1}{[value]}\) is
3. The modulus of [value] is
Feedback: The modulus of a number is the size of that number regardless of its sign hence: 1. \(|-\dfrac{[value]}{7}| = \dfrac{[value]}{7}.\) 2. \(|\dfrac{1}{[value]}| = \dfrac{1}{[value]}.\) 3. \(|[value]| = [value].\)

Exercise

Find the factorial button on your calculator and hence compute \([value]!\)

\([value]!\) =
Feedback: \([value]! = [value] \times [value] \times [value] \times ... \times 2 \times 1 = [value].\)

Rounding to n Decimal Places

In general, a calculator or computer is unable to store every decimal place of a real number. Real numbers are rounded. To round a number to \(n\) decimal places we look at the \((n+1)^{th}\) digit in the decimal expansion of the number.

- If the \((n+1)^{th}\) digit is 0, 1, 2, 3 or 4 then we : that is, we simply chop to \(n\) places. (In other words we neglect the \((n+1)^{th}\) digit and any digits to its right.)
- If the \((n+1)^{th}\) digit is 5, 6, 7, 8 or 9 then we : we add \(1\) to the \(n^{th}\) decimal place and then chop to \(n\) places.

For example

\(\qquad\qquad\qquad\begin{aligned} \dfrac{1}{3} &= 0.3333 & \qquad & \mbox{rounded to 4 decimal places} \\ \pi &= 3.14159 & \qquad & \mbox{rounded to 5 decimal places}\\ \dfrac{8}{3} &= 2.667 & \qquad & \mbox{rounded to 3 decimal places}\\ 2136 &= 2000.000 & \qquad & \mbox{rounded to 3 decimal places}\\ 36.8 &= 37 & \qquad & \mbox{rounded to 1 decimal place}\\ 6.2399 &= 6.240 & \qquad & \mbox{rounded to 3 decimal places} \end{aligned}\)

Sometimes the phrase ‘decimal places’ is abbreviated to ‘d.p.’ or ‘dec.pl.’.

Example

Write down each of these numbers rounded to 4 decimal places:

1. \(0.12345\)
2. \(-0.44444\)
3. \(0.5555555\)
4. \(0.000127351\)
5. \(0.000005\)
6. \(123.456789\)

Solution

1. \(0.12345\) rounded to 4 decimal place becomes \(0.1235.\)
2. \(-0.44444\) rounded to 4 decimal place becomes \(-0.4444.\)
3. \(0.5555555\) rounded to 4 decimal place becomes \(0.5556.\)
4. \(0.000127351\) rounded to 4 decimal place becomes \(0.0001.\)
5. \(0.000005\) rounded to 4 decimal place becomes \(0.0000.\)
6. \(123.456789\) rounded to 4 decimal place becomes \(123.4568.\)

Rounding to n Significant Figures

This process is similar to rounding to decimal places but there are some subtle differences.

To round a number to \(n\) significant figures we look at the \((n+1)^{th}\) digit in the decimal expansion of the number.

- If the \((n+1)^{th}\) digit is 0, 1, 2, 3 or 4 then we : that is, we simply chop to \(n\) places, inserting zeros if necessary before the decimal point. (In other words we neglect the \((n+1)^{th}\) digit and any digits to its right.)
- If the \((n+1)^{th}\) digit is 5, 6, 7, 8 or 9 then we : we add \(1\) to the \(n^{th}\) decimal place and then chop to \(n\) places, inserting zeros if necessary before the decimal point.

For example

\(\qquad\qquad\qquad\begin{aligned} \dfrac{1}{3} &= 0.3333 & \qquad & \mbox{rounded to 4 significant figures} \\ \pi &= 3.14159 & \qquad & \mbox{rounded to 6 significant figures}\\ \dfrac{8}{3} &= 2.667 & \qquad & \mbox{rounded to 4 significant figures}\\ 2136 &= 2000 & \qquad & \mbox{rounded to 1 significant figure}\\ 36.78 &= 37 & \qquad & \mbox{rounded to 2 significant figures}\\ 6.2399 &= 6.240 & \qquad & \mbox{rounded to 4 significant figures} \end{aligned}\)

Sometimes the phrase “significant figures" is abbreviated as “s.f." or “sig.fig."

Example

Write down each of these numbers rounded to 4 significant figures:

1. \(0.12345\)
2. \(-0.44444\)
3. \(0.5555555\)
4. \(0.000127351\)
5. \(25679\)
6. \(123.456789\)
7. \(3456543\)

Solution

1. \(0.12345\) rounded to 4 significant figures becomes \(0.1235.\)
2. \(-0.44444\) rounded to 4 significant figures becomes \(-0.4444.\)
3. \(0.5555555\) rounded to 4 significant figures becomes \(0.5556.\)
4. \(0.000127351\) rounded to 4 significant figures becomes \(0.0001274.\)
5. \(25679\) rounded to 4 significant figures becomes \(25680.\)
6. \(123.456789\) rounded to 4 significant figures becomes \(123.5.\)
7. \(3456543\) rounded to 4 significant figures becomes \(3457000.\)

Arithmetical expressions

A quantity made up of numbers and one or more of the operations \(+\), \(-\), \(\times\) and \(/\) is called an arithmetical expression. Frequent use is also made of brackets, or parentheses, \((\ )\), to separate different parts of an expression. When evaluating an expression it is conventional to evaluate quantities within brackets first. Often a division line implies bracketed quantities. For example in the expression \(\dfrac{3+4}{7+9}\) there is implied bracketing of the numerator and denominator i.e. the expression is \(\dfrac{(3+4)}{(7+9)}\) and the bracketed quantities would be evaluated first resulting in the number \(\dfrac{7}{16}\).

The BODMAS rule

When several arithmetical operations are combined in one expression we need to know in which order to perform the calculation. This order is found by applying rules known as precedence rules which specify which operation has priority. The convention is that bracketed expressions are evaluated first. Any multiplications and divisions are then performed, and finally any additions and subtractions. For short, this is called the BODMAS rule.

Key Point

The BODMAS rule

\(\begin{aligned} &\text{Brackets, ( )} & \qquad & \text{First priority: evaluate terms within brackets} \\ \\ &\text{Of} & \qquad & \\ &\text{Division, } \div & \qquad & \text{Second priority: carry out all multiplications and divisions}\\ &\text{Multiplication, } \times & \qquad & \\ \\ &\text{Addition, } + & \qquad & \text{Third priority: carry out all additions and subtractions}\\ &\text{Subtraction, } - & \qquad & \end{aligned}\)

If an expression contains only multiplication and division we evaluate by working from left to right. Similarly, if an expression contains only addition and subtraction we evaluate by working from left to right. In Section 1.2 we will meet another operation called exponentiation, or raising to a power. We shall see that, in the simplest case, this operation is repeated multiplication and it is usually carried out once any brackets have been evaluated.

Example

Evaluate \(4 − 3 + 7 \times 2\)

Solution

The BODMAS rule tells us to perform the multiplication before the addition and subtraction. Thus

\(\qquad 4 − 3 + 7 \times 2 = 4 − 3 + 14\)

Finally, because the resulting expression contains just addition and subtraction we work from the left to the right, that is

\(\qquad 4 − 3 + 14 = 1 + 14 = 15\)

Example

Evaluate \(8 \div 2 − (4 − 5)\)

Solution

The bracketed expression is evaluated first:

\(\qquad 8 \div 2 − (4 − 5) = 8 \div 2 − (−1)\)

Division has higher priority than subtraction and so this is carried out next giving

\(\qquad 8 \div 2 − (−1) = 4 − (−1)\)

Subtracting a negative number is equivalent to adding a positive number. Thus

\(\qquad 4 − (−1) = 4 + 1 = 5\)

Exercise

1. \([value] \pm 3\) represents the two numbers and
2. \([value] \pm 7\) represents the two numbers and
3. \(-[value] \pm 0.5\) represents the two numbers and
4. \(-[value] \pm 0.05\) represents the two numbers and
5. \(|-[value]| \pm 13\) represents the two numbers and
6. \(|-2| \pm [value]\) represents the two numbers and
Feedback: \(a \pm b\) is short hand for the two numbers \(a - b\) and \(a + b\) 1. \([value] \pm 3 \implies [value] - 3\) and \([value] + 3 \implies [value]\) and \([value]\) 2. \([value] \pm 7 \implies [value] - 7\) and \([value] + 7 \implies [value]\) and \([value]\) 3. \(-[value] \pm 0.5 \implies -[value] - 0.5\) and \(-[value] + 0.5 \implies [value]\) and \([value]\) 4. \(-[value] \pm 0.05 \implies -[value] - 0.05\) and \(-[value] + 0.05 \implies [value]\) and \([value]\) 5. \(|-[value]| \pm 13 \implies |-[value]| - 13\) and \(|-[value]| + 13 \implies [value] - 13\) and \([value] + 13 \implies [value]\) and \([value]\) 6. \(|-2| \pm [value] \implies |-2| - [value]\) and \(|-2| + [value] \implies 2 - [value]\) and \(2 + [value] \implies [value]\) and \([value]\)

Exercise

Write down each of these numbers rounded to 3 decimal places:

1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \(−0.5555,\)
5. \(45.6789\)
6. \([value]\)

Solution
Feedback: 1. \([value]\) rounded to 3 decimal places becomes \( [value].\) 2. \([value]\) rounded to 3 decimal places becomes \( [value].\) 3. \([value]\) rounded to 3 decimal places becomes \( [value].\) 4. \(−0.5555,\) rounded to 3 decimal places becomes \( [value].\) 5. \(45.6789\) rounded to 3 decimal places becomes \( [value].\) 6. \([value]\) rounded to 3 decimal places becomes \( [value].\)

Exercise

Write down each of these numbers rounded to 3 significant figures:

1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \(−0.5555,\)
5. \(45.6789\)
6. \([value]\)

Solution
Feedback: 1. \([value]\) rounded to 3 significant figurs becomes \( [value].\) 2. \([value]\) rounded to 3 significant figurs becomes \( [value].\) 3. \([value]\) rounded to 3 significant figurs becomes \( [value].\) 4. \(−0.5555,\) rounded to 3 significant figurs becomes \( [value].\) 5. \(45.6789\) rounded to 3 significant figurs becomes \( [value].\) 6. \([value]\) rounded to 3 significant figurs becomes \( [value].\)

Exercise

Evaluate the expression below

\([value] + [value] \times [value] =\)
Feedback: The BODMAS rule tells us to perform the multiplication before the addition. Thus \(\qquad [value] + [value] \times [value] = [value] + [value]\) We then perform addition to give the final answer \(\qquad [value] + [value] = [value]\)

Exercise

Evaluate the expression below

\(([value] - [value]) \times [value] =\)
Feedback: The BODMAS rule tells us the bracketed quantity must be evaluated first. Thus \(\qquad ([value] - [value]) \times [value] = [value] \times [value]\) We then perform multiplication to give the final answer \(\qquad [value] \times [value] = [value]\)

Exercise

Evaluate the expression below

\(\dfrac{[value]-[value]}{9-[value]} =\)
Feedback: The dividing line implies that brackets are present around the numerator and around the denominator. Thus \(\qquad \dfrac{[value]-[value]}{9-[value]} = \dfrac{([value]-[value])}{(9-[value])}\) We then perform the subtraction within the brackets to give the final answer \(\qquad \dfrac{([value]-[value])}{(9-[value])} = \dfrac{[value]}{[value]}\)

Using Symbols

Mathematics provides a very rich language for the communication of engineering concepts and ideas, and a set of powerful tools for the solution of engineering problems. In order to use this language it is essential to appreciate how symbols are used to represent physical quantities, and to understand the rules and conventions which have been developed to manipulate these symbols.

The choice of which letters or other symbols to use is largely up to the user although it is helpful to choose letters which have some meaning in any particular context. For instance if we wish to choose a symbol to represent the temperature in a room we might use the capital letter \(T\). Similarly the lower case letter \(t\) is often used to represent time. Because both time and temperature can vary we refer to \(T\) and \(t\) as variables.

In a particular calculation some symbols represent fixed and unchanging quantities and we call these constants. Often we reserve the letters \(x\), \(y\) and \(z\) to stand for variables and use the earlier letters of the alphabet, such as \(a\), \(b\) and \(c\), to represent constants. The Greek letter pi, written \(\pi\), is used to represent the constant 3.14159.... which appears for example in the formula for the area of a circle. Other Greek letters are frequently used as symbols, and for reference, the Greek alphabet is given in the Table below.

\(\begin{aligned} \qquad & A \quad \alpha \quad \text{alpha} && \qquad I \quad \iota \quad \text{iota} && \qquad R \quad \rho \quad \text{rho} \\ \qquad & B \quad \beta \quad \text{beta} && \qquad \Lambda \quad \lambda \quad \text{lambda} && \qquad T \quad \tau \quad \text{tau} \\ \qquad & \Gamma \quad \gamma \quad \text{gamma} && \qquad K \quad \kappa \quad \text{kappa} && \qquad \Sigma \quad \sigma \quad \text{sigma} \\ \qquad & \Delta \quad \delta \quad \text{delta} && \qquad M \quad \mu \quad \text{mu} && \qquad \Upsilon \quad \upsilon \quad \text{upsilon} \\ \qquad & E \quad \epsilon \quad \text{epsilon} && \qquad N \quad \nu \quad \text{nu} && \qquad \Phi \quad \phi \quad \text{phi} \\ \qquad & Z \quad \zeta \quad \text{zeta} && \qquad Xi \quad \xi \quad \text{xi} && \qquad X \quad \chi \quad \text{chi} \\ \qquad & H \quad \eta \quad \text{eta} && \qquad O \quad \omicron \quad \text{omicron} && \qquad \Psi \quad \psi \quad \text{psi} \\ \qquad & \Theta \quad \theta \quad \text{theta} && \qquad \Pi \quad \pi \quad \text{pi} && \qquad \Omega \quad \omega \quad \text{omega} \end{aligned}\)

\(\qquad\qquad\qquad\qquad\qquad \)Table: The Greek alphabet

Mathematics is a very precise language and care must be taken to note the exact position of any symbol in relation to any other. If \(x\) and \(y\) are two symbols, then the quantities \(xy\), \(x^y\), \(x_y\) can all mean different things. In the expression \(x^y\) you will note that the symbol \(y\) is placed to the right of and slightly higher than the symbol \(x\). In this context \(y\) is called a superscript. In the expression \(x_y\), \(y\) is placed lower than and to the right of \(x\), and is called a subscript.

Example

The temperature in a room is measured at four points as shown in the Figure below

[Image: The temperature is measured at four points]

Rather than use different letters to represent the four measurements we can use one symbol, \(T\), together with four subscripts to represent the temperature. Thus the four measurements are denoted by \(T_1\), \(T_2\), \(T_3\) and \(T_4\).

Combining Numbers Together Using \(+,-,\times,\div\)

Addition (\(+\))

If the letters \(x\) and \(y\) represent two numbers, then their sum is written as \(x+y\). Note that \(x+y\) is the same as \(y+x\) just as \(4+7\) is equal to \(7+4\).

Subtraction (\(-\))

Subtracting \(y\) from \(x\) yields \(x-y\). Note that \(x-y\) is not the same as \(y-x\) just as \(11-7\) is not the same as \(7-11\), however in both cases the difference is said to be 4.

Multiplication (\(\times\))

The instruction to multiply \(x\) and \(y\) together is written as \(x\times y\). Usually the multiplication sign is omitted and we write simply \(xy\). An alternative notation is to use a dot to represent multiplication and so we could write \(x.y\) The quantity \(xy\) is called the product of \(x\) and \(y\). As discussed earlier multiplication is both commutative and associative:

\(\qquad \text{i.e.}\qquad x\times y=y\times x\qquad \text{and}\qquad (x\times y)\times z = x\times (y\times z)\)

This last expression can thus be written \(x\times y\times z\) without ambiguity. When mixing numbers and symbols it is usual to write the numbers first. Thus \(3\times x\times y\times 4 = 3\times 4\times x\times y = 12xy\).

Example

Simplify

1. \(9(2y)\)
2. \(-3(5z)\)
3. \(4(2a)\)
4. \(2x \times (2y)\)

Solution

1. Note that \(9(2y)\) means \(9\times(2\times y)\). Because of the associativity of multiplication \(9\times (2\times y)\) means the same as \((9 \times 2) \times y\), that is \(18y\).
2. \(−3(5z)\) means \(−3 \times (5 \times z)\). Because of associativity this is the same as \((−3 \times 5) \times z\), that is \(−15z\).
3. \(4(2a)\) means \(4 \times (2 \times a)\). We can write this as \((4 \times 2) \times a\), that is \(8a\).
4. Because of the associativity of multiplication, the brackets are not needed and we can write \(2x \times (2y) = 2x \times 2y\) which equals
\(2 \times x \times 2 \times y = 2 \times 2 \times x \times y = 4xy\).

Example

What is the distinction between \(9(−2y)\) and \(9 − 2y\)?

Solution

The expression \(9(−2y)\) means \(9 \times (−2y)\). Because of associativity of multiplication we can write this as \(9 \times (−2) \times y\) which equals \(-18y\).

On the other hand \(9 − 2y\) means subtract \(2y\) from \(9\). This cannot be simplified.

Division (\(\div\))

The quantity \(x\div y\) means \(x\) divided by \(y\). This is also written as \(x/y\) or \(\dfrac{x}{y}\) and is known as the quotient of \(x\) and \(y\). In the expression \(\dfrac{x}{y}\) the symbol \(x\) is called the numerator and the symbol \(y\) is called the denominator. Note that \(x/y\) is not the same as \(y/x\). Division by 1 leaves a quantity unchanged so that \(\dfrac{x}{1}\) is simply \(x\).

Algebraic Expressions

A quantity made up of symbols and the operations \(+\), \(-\), \(\times\) and \(/\) is called an algebraic expression. One algebraic expression divided by another is called an algebraic fraction. Thus \(\dfrac{x+7}{x-3} \hbox{ and } \dfrac{3x-y}{2x+z}\) are algebraic fractions. The reciprocal of an algebraic fraction is found by inverting it. Thus the reciprocal of \(\dfrac{2}{x}\) is \(\dfrac{x}{2}\). The reciprocal of \(\dfrac{x+7}{x-3}\) is \(\dfrac{x-3}{x+7}\).

Example

State the reciprocal of each of the following expressions:

1. \(\dfrac{y}{z}\)
2. \(\dfrac{x+z}{a-b}\)
3. \(3y\)
4. \(\dfrac{1}{ a+2b}\)
5. \(-\dfrac{1}{y}\)

Solution

1. \(\dfrac{z}{y}\)
2. \(\dfrac{a-b}{x+z}\)
3. \(3y\) is the same as \(\dfrac{3y}{1}\) so the reciprocal of \(3y\) is \(\dfrac{1}{3y}\).
4. The reciprocal of \(\dfrac{1}{a + 2b}\) is \(\dfrac{a + 2b}{1}\) or simply \(a + 2b\).
5. The reciprocal of \(-\dfrac{1}{y}\) is \(-\dfrac{y}{1}\) or simply \(−y\).

Finding the reciprocal of complicated expressions can cause confusion. Study the following Example carefully.

Example

Obtain the reciprocal of:

1. \(p+q\)
2. \(\dfrac{1}{R_1} + \dfrac{1}{R_2}\)

Solution

1. Because \(p + q\) can be thought of as \(\dfrac{p + q}{1}\) its reciprocal is \(\dfrac{1}{p + q}\). Note in particular that the reciprocal of \(p + q\) is not \(\dfrac{1}{p} + \dfrac{1}{q}\). This distinction is important and a common cause of error. To avoid an error carefully identify the numerator and denominator in the original expression before inverting.
2. The reciprocal of \(\dfrac{1}{R_1} + \dfrac{1}{R_2}\) is \(\dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}}\). To simplify this further requires knowledge of the addition of algebraic fractions which is dealt with in later workbooks. It is important to note that the reciprocal of \(\dfrac{1}{R_1} + \dfrac{1}{R_2}\) is not \(R_1 + R_2\).

The equals sign (\(=\))

The equals sign, \(=\), is used in several different ways. Firstly, an equals sign is used in equations. The left-hand side and right-hand side of an equation are equal only when the variable involved takes specific values known as solutions of the equation. For example, in the equation \(x-8=0\), the variable is \(x\). The left-hand side and right-hand side are only equal when \(x\) has the value \(8\). If \(x\) has any other value the two sides are not equal.

Secondly, the equals sign is used in formulae. Physical quantities are often related through a formula. For example, the formula for the length, \(C\), of the circumference of a circle expresses the relationship between the circumference of the circle and its radius, \(r\). This formula states \(C=2\pi r\). When used in this way the equals sign expresses the fact that the quantity on the left is found by evaluating the expression on the right.

Thirdly, an equals sign is used in identities. An identity looks just like an equation, but it is true for all values of the variable. We shall see shortly that \((x-1)(x+1)=x^2-1\) for any value of \(x\) whatsoever. This mean that the quantity on the left means exactly the same as that on the right whatever the value of \(x\). To distinguish this usage from other uses of the equals symbol it is more correct to write \((x-1)(x+1)\equiv x^2-1\), where \(\equiv\) means ‘is identically equal to’. However, in practice, the equals sign is often used. We will only use \(\equiv\) where it is particularly important to do so.

The ‘not equals’ sign (\(\ne\))

The sign \(\ne\) means ‘is not equal to’. For example, \(5\ne 6\), \(7\ne -7\).

The notation for the change in a variable (\(\delta\))

The change in the value of a quantity is found by subtracting its initial value from its final value. For example, if the temperature of a mixture is initially \(13^\circ\)C and at a later time is found to be \(17^\circ\)C, the change in temperature is \(17-13=4^\circ\)C. The Greek letter \(\delta\) is often used to indicate such a change. If \(x\) is a variable we write \(\delta x\) to stand for a change in the value of \(x\). We sometimes refer to \(\delta x\) as an increment in \(x\). For example if the value of \(x\) changes from 3 to 3.01 we could write \(\delta x= 3.01-3 = 0.01\). It is important to note that this is not the product of \(\delta\) and \(x\), rather the whole symbol ‘\(\delta x\)’ means ‘the increment in \(x\)’.

Sigma (or summation) notation (\(\sum\))

This provides a concise and convenient way of writing long sums.

The sum

\(\qquad x_1+x_2+x_3+x_4 +\ldots +x_{11}+x_{12}\)

is written using the capital Greek letter sigma, \(\sum\), as

\(\qquad \sum_{k=1}^{12} x_k\)

The symbol \(\sum\) stands for the sum of all the values of \(x_k\) as \(k\) ranges from 1 to 12. Note that the lower-most and upper-most values of \(k\) are written at the bottom and top of the sigma sign respectively.

Example

Write out explicitly what is meant by \(\sum_{k=1}^{5}k^3\)

Solution

We must let k range from 1 to 5:

\(\qquad \sum_{k=1}^{5}k^3 = 1^3+2^3+3^3+4^3+5^3\)

Example

Write out explicitly

1. \(\sum_{k=1}^{3}1\)
2. \(\sum_{k=0}^{4}2\)

Solution

1. Here k does not appear explicitly in the terms to be added. This means add the constant 1, three times.
\(\qquad \sum_{k=1}^{3}1 = 1 + 1 + 1 = 3\)
In general \(\sum_{k=1}^{n}1 = n\)
2. Here \(k\) starts at zero so there are \(n + 1\) terms where \(n = 4\):
\(\qquad \sum_{k=0}^{4}2 = 2 + 2 + 2 + 2 + 2 = 10\)

Exercise

Find the values of \(a\), \(b\) and \(c\) such that \(\sum_{k=a}^{b}c = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}\)

a =

b =

c =
Feedback: The \[\sum_{k=a}^{b}c = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}\] can be writen as: \[\sum_{k=[value]}^[value] [value] \]

Exercise

State the reciprocal of

1. \([value]\)
2. \(\displaystyle[value]\)
3. \([value]\)
4. \(\displaystyle[value]\)
5. \([value]\)
6. \(\displaystyle[value]\)
Feedback: The reciprocal of 1. \([value]\) is \( \displaystyle [value]\) 2. \(\displaystyle [value]\) is \( [value]\) 3. \([value]\) is \(\displaystyle [value]\) 4. \( \displaystyle[value]\) is \( [value]\) 5. \([value]\) is \(\displaystyle[value]\) 6. \(\displaystyle [value]\) is \( \displaystyle [value]\)

Exercise

The pressure \(p\) in a reaction vessel changes from [value] to [value]. Write down the value of \(\delta p\).

\(\delta p =\)
Feedback: \(\delta p\) stands for the change in the value of p hence we have: \(\qquad \delta p = [value] - [value] = [value]\)

Exercise

Express as simply as possible

1. \(([value]) \times x \times ([value]) \times y =\)
2. \([value] \times x \times z \times ([value]) =\)
Feedback: 1. Because of the commutativity of multiplication we have: \(\qquad ([value]) \times x \times ([value]) \times y = ([value]) \times ([value]) \times x \times y\) and then because of the associativity of multiplication we have: \(([value]) \times ([value]) \times x \times y = ([value] \times [value]) \times x \times y = ([value]) \times x \times y = [value]xy\) 2. Because of the commutativity of multiplication we have: \(\qquad ([value]) \times x \times z \times ([value]) = ([value]) \times ([value]) \times x \times z\) and then because of the associativity of multiplication we have: \(([value]) \times ([value]) \times x \times z = ([value] \times [value]) \times x \times z = ([value]) \times x \times z = [value]xy\)

Exercise

Simplify

1. \([value]([value]y) =\)
2. \(17x(-2y) =\)
3. \([value]x(8y) =\)
4. \(5x(-8y) =\)
Feedback: 1. Note that \([value]([value]y)\) means \([value] \times ([value] \times y)\) then because of the associativity of multiplication we have: \(\qquad [value] \times ([value] \times y) = ([value] \times [value]) \times y = [value] \times y = [value]\) 2. Note that \(17x(-2y)\) means \(17 \times x \times (-2 \times y)\) then by using the associativity and commutativity of multiplication we have: \(\qquad 17 \times x \times (-2 \times y) = 17 \times (x \times -2) \times y = 17 \times (-2 \times x) \times y = (17 \times -2) \times x \times y = -34 \times x \times y = -34xy\) 3. Note that \([value]x(8y)\) means \([value] \times x \times (8 \times y)\) then by using the associativity and commutativity of multiplication we have: \(\qquad [value] \times x \times (8 \times y) = [value] \times (x \times 8) \times y = 17 \times (8 \times x) \times y = ([value] \times 8) \times x \times y = [value] \times x \times y = [value]xy\) 4. Note that \(5x(-8y)\) means \(5 \times x \times (-8 \times y)\) then by using the associativity and commutativity of multiplication we have: \(\qquad 5 \times x \times (-8 \times y) = 5 \times (x \times -8) \times y = 5 \times (-8 \times x) \times y = (5 \times -8) \times x \times y = -40 \times x \times y = -40xy\)

Exercise

The value of \(x\) is \([value] \pm [value]\). The value of \(y\) is \([value] \pm 5\). Find the maximum and minimum values of the given equations

1. \(\text{max}(x+y) =\)
\(\text{min}(x+y) =\)
2. \(\text{max}(xy) =\)
\(\text{min}(xy) =\)
3. \(\text{max}\left(\dfrac{x}{y}\right) =\)
\(\text{min}\left(\dfrac{x}{y}\right) =\)
4. \(\text{max}\left(\dfrac{y}{x}\right) =\)
\(\text{min}\left(\dfrac{y}{x}\right) =\)
Feedback: We have that \(x\) is given by \([value] \pm [value]\) and hence it can take the values \([value] + [value] = [value]\) or \([value] - [value] = [value]\) Similarly \(y\) is given by \([value] \pm 5\) and hence it can take the values \([value] + 5 = [value]\) or \([value] - 5 = [value]\) 1. A sum of two elements is maximised when the two elements themselves are maximised hence: \(\begin{align} \text{max}(x+y) & = \text{max}(x) + \text{max}(y) \\ &= \text{max}([value],[value]) + \text{max}([value],[value]) \\ &= [value] + [value] = [value] \end{align}\) Similarly the sum of two elements is minimised when the two elements themselves are minimised hence: \(\begin{align} \text{min}(x+y) & = \text{min}(x) + \text{min}(y) \\ &= \text{min}([value],[value]) + \text{min}([value],[value]) \\ &= [value] + [value] = [value] \end{align}\) 2. A procuct of two elements is also maximised when the two elements themselves are maximised hence: \(\begin{align} \text{max}(xy) &= \text{max}(x \times y) = \text{max}(x) \times \text{max}(y) \\ &= \text{max}([value],[value]) \times \text{max}([value],[value]) \\ &= [value] \times [value] = [value] \end{align}\) Similarly the product of two elements is minimised when the two elements themselves are minimised hence: \(\begin{align} \text{min}(xy) &= \text{min}(x \times y) = \text{min}(x) \times \text{min}(y) \\ &= \text{min}([value],[value]) \times \text{min}([value],[value]) \\ &= [value] \times [value] = [value] \end{align}\) 3. A fraction of two elements is maximised when the numerator is maximised and the denominator is minimised: \( \text{max}\left(\dfrac{x}{y}\right) = \dfrac{\text{max}(x)}{\text{min}(y)} = \dfrac{\text{max}([value],[value])}{\text{min}([value],[value])} = \dfrac{[value]}{[value]}\) Similarly a fraction of two elements is minimised when the numerator is minimised and the denominator is maximised: \( \text{min}\left(\dfrac{x}{y}\right) = \dfrac{\text{min}(x)}{\text{max}(y)} = \dfrac{\text{min}([value],[value])}{\text{max}([value],[value])} = \dfrac{[value]}{[value]}\) 4. Just as above the fraction is maximised when the numerator is maximised and the denominator is minimised: \( \text{max}\left(\dfrac{y}{x}\right) = \dfrac{\text{max}(y)}{\text{min}(x)} = \dfrac{\text{max}([value],[value])}{\text{min}([value],[value])} = \dfrac{[value]}{[value]}\) And again the fraction is minimised when the numerator is minimised and the denominator is maximised: \( \text{min}\left(\dfrac{y}{x}\right) = \dfrac{\text{min}(y)}{\text{max}(x)} = \dfrac{\text{min}([value],[value])}{\text{max}([value],[value])} = \dfrac{[value]}{[value]}\)