Introduction

The objective of this worksheet is to use BEDMAS to estimate and work out the solutions to the problems on your calculator.

However, there can be variations in the BEDMAS rules, even between different models of the same calculator.

The general BEDMAS rule is:

  • B - Brackets - all expressions in brackets are evaluated first.
  • E - Exponents - \( x^y \). NOTE: Division is sometimes written using a \( \wedge \) (caret) as in \( x \wedge y \) or a double star \( ** \) as \( x ** y \).
  • D - Division - \( x \div y \). NOTE: Division is sometimes written as \( / \) (slash) as in \( x / y \).
  • M - Multiplication \( x \cdot y \) NOTE: The multiplication symbol \( \times \) is also used, the \( \cdot \) symbol is used to prevent confusion between the common variable name \( x \) and \( \times \). Some notations may use a single star \( * \) as in \( x * y \).
  • A - Addition \( x + y \).
  • S - Subtraction \( x - y \).

This method of assigning priorities works well up to (perhaps) NCEA Level 1. A more sophisticated priority order is required beyond that.

A more universal set of priority rules is as follows. The higher priority operators appear in the list before lower priority operators.

  • Expressions in brackets - these are always evaluated first.
  • Some single-argument or unary operators. These generally include the following common operators: \( \sqrt{x} \), \(\sqrt[3]{x} \), \( x^{-1} \) (or \( \frac{1}{x} \)), \( x^2 \), \(x^3 \), \(\sin x \), \(\cos x \), \( \tan x \), unary \( - \) (negation).
  • Some dual-argument or binary operators. These generally include: exponent \( x^y \), the exponential notation operator \( \mathrm{Exp} \) or \( \times 10^x \), and fractions \( a \frac{b}{c} \).
  • Multiplication (\( \times \) or \( \cdot \) ) and division (\( \div \) or \( / \)). These expressions are read from left to right.
  • Addition (\( + \)) and subtraction (\( - \)). These expressions are read from left to right.

Here are some useful behavioural rules to go by when doing calculations:

  • Operators will bind as few arguments as possible, i.e. operators are not greedy. The expression \( \sqrt{8} + 1 \) will be evaluated as \( \left(\sqrt{8} \right) + 1\), not \( \sqrt{8 + 1} \).
  • Operations will be evaluated from left to right. \( 1 \div 2 \div 3 \) will be read as \( (1 \div 2 ) \div 3\), not \( 1 \div (2 \div 3) \). However, all expressions involving repeated division without brackets should be treated as unclear. Never use multiple \( \div \) operators in a row, or mix \( \div \) with \( \times \) without brackets! Expressions like this are unclear.
  • Subtraction is the same as adding a negative number. \( a - b = a + (-b) \). Expressions like \( a - b - c - d \) are interpreted as \( a + (-b) + (-c) + (-d) \) etc.
  • Division is the same as multiplying by the reciprocal, or inverse (\( x^{-1} \) or \( \frac{1}{x} \)). \( a \div b = a \times b^{-1} = a \times \left ( \frac{1}{b} \right) \). Expressions like \( a \div b \times c \div d \) are interpreted as \( a \times \frac{1}{b} \times c \times \frac{1}{b} \) etc. Never use multiple \( \div \) operators in a row, or mix \( \div \) with \( \times \) without brackets! Expressions like this are unclear.
  • Inverting a division is the same as swapping the order of the operators i.e. \( (a \div b)^{-1} = b \div a \).
  • Multiplying a division: \( a \times (b \div c) = (a \times b) \div c \).
  • Converting a division to a fraction: \( a \div b = \frac{a}{b} \).
  • The fraction \( \frac{a + b}{c + d} = (a + b) \div (c + d) \).
  • Converting a whole number to a fraction: \( a = \frac{a}{1} \).
  • Reciprocal of a fraction: \(\left( \frac{a}{b}\right)^{-1} = \frac{b}{a} \)
  • Fraction multiplication: \( \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \)
  • Fraction division: \( \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \left( \frac{c}{d} \right )^{-1} = \frac{a}{b} \times \frac{d}{c} \).


Assignment Answers and Commentary

Question 1

\( 3 \times 4 + 8 \)

Solution

This is evaluated as following. Brackets are used to make the order of operations more explicit.

\( 3 \times 4 + 8  = (3 \times 4) + 8 = 12 + 8 = 20 \)


Question 2

\( 17 \div 8 + 5 \)


Solution - Method 1 - Numerical

This is evaluated as following. Brackets are used to make the order of operations more explicit.

\( 17 \div 8 + 5  = (17 \div 8) + 5 = 2.125 + 5 = 7.125 \)


Solution - Method 2 - Using Fractions

\( 17 \div 8 + 5  = \left( \frac{17}{8} \right) + 5 = 2\frac{1}{8} + 5 = 7\frac{1}{8} \leftrightarrow 7.125 \)


Question 3

\( 32 + 15 \div 4 \)

Solution - Method 1 - Numerical

\( 32 + 15 \div 4 = 32 + \left( 15 \div 4 \right) = 32 + 3.75 = 35.75 \)


Solution - Method 2 - Using Fractions

\( 32 + 15 \div 4 = 32 + \left( \frac{15}{4} \right) = 32 + 3\frac{3}{4} = 35\frac{3}{4} \leftrightarrow 35.75 \)


Question 4

\( 3 + (42 - 8) \div 2 \)

NOTE: This equation already has a bracketed expression, so that the subtraction is performed before the division.


Solution

\( 3 + (42 - 8) \div 2 = 3 + ((42 - 8) \div 2) = 3 + ((34) \div 2) = 3 + (17) = 20 \)


Question 5

\( 52 - 18 \div (27 \div 22) \)

This expression is more complex. Note the brackets in the division, otherwise the equation is potentially unclear. The initial bracketing is a follows.

\( 52 - (18 \div (27 \div 22)) \)


Solution - Method 1 - Numerical

\( 52 - (18 \div (27 \div 22)) =  52 - (18 \times (22 \div 27)) = 52 - (396 \div 27) = 52 - 14.6667 \) = 37.33 (4 sf).


Solution - Method 2 - Using Fractions

\( 52 - (18 \div (27 \div 22)) =  52 - \frac{18}{\left( \frac{27}{22}\right)} = 52 - 18 \times \frac{22}{27} = 52 - 2 \times \frac{22}{3} = 52 -  \frac{44}{3}   = 37\frac{1}{3} \leftrightarrow \) 37.33 (4 sf).


Question 6

\( (42 + 5) \times (22 - 32) \)


Solution

\( (42 + 5) \times (22 - 32) = 47 \times -10 \) = -470.


Question 7

\( 3 - 10 \times 0 \div 4 - 4 \)

NOTE: This expression is potentially unclear, due to the mix of multiplication and division without brackets. The expression will be read from left to right.


Solution

\( 3 - 10 \times 0 \div 4 - 4 = 3 - ((10 \times 0) \div 4) - 4 = 3 - (0 \div 4) - 4 = 3 - 0 - 4 \) = -1.


Question 8

\( \frac{\sqrt{16} - 2}{4} \)


Solution

\( \frac{\sqrt{16} - 2}{4} = \frac{(\sqrt{16}) - 2}{4} = \frac{4 - 2}{4} = \frac{1}{2} \) or 0.5.


Question 9

\( (62 - 22 \div 4 + (2 + 24))2 \)

NOTE: The 2 on the end means "multiplication by concatenation". Some calculators interpret \( (1+2)3 \) as \( (1 + 2) \times 3 \). If your calculator does not, it may show an Error. In this case, you will have to modify the expression to \( (62 - 22 \div 4 + (2 + 24)) \times 2 \).

Solution - Method 1 - Numerical

\( (62 - 22 \div 4 + (2 + 24))2 = (62 - (22 \div 4) + (2 + 24)) \times 2 = (62 - 5.5 + 26) \times 2 = 82.5 \times 2 = 165 \)


Solution - Method 2 - Fractions

\( (62 - 22 \div 4 + (2 + 24))2 = (62 - (22 \div 4) + (2 + 24)) \times 2 = (62 - \frac{11}{2} + 26) \times 2 = 82\frac{1}{2} \times 2 = 165 \)


Question 10

\( (42 \div 7 + 18 - \sqrt{25} - 4) \times 0 \div 25674399876 \)

NOTE: This expression is potentially unclear, due to the mix of multiplication and division without brackets. The expression will be read from left to right.


Solution

This is the first expression where me might wish to perform some pre-optimisation. We will think of the expression as the following. We will eliminate the division first.

\( (42 \div 7 + 18 - \sqrt{25} - 4) \times 0 \times \frac{1}{25674399876} \)

All the highest level operations are now multiplication, so we can now re-arrange it as we please. A good re-arrangement is shown below.

\( 0 \times (42 \div 7 + 18 - \sqrt{25} - 4) \times \frac{1}{25674399876} \)

The reason this is done is that multiplying any finite expression by zero gives a result of zero. So we can solve this without taking any further steps.

\( (42 \div 7 + 18 - \sqrt{25} - 4) \times 0 \div 25674399876 = 0 \times (42 \div 7 + 18 - \sqrt{25} - 4) \times \frac{1}{25674399876} = 0 \)

Last modified: Friday, 8 May 2020, 5:08 PM