BODMAS Rule:
BODMAS is about simplifying an expression by firstly removing the brackets in the order i.e. (), {}, []. Removal of brackets is followed by addition, subtraction, multiplication, division, square roots, cube roots, powers, cancellation of numerator/ denominator and so on.
Example 1:
Simplify using BODMAS rule: 25 – 48 ÷ 6 + 12 × 2
Solution:
25 – 48 ÷ 6 + 12 × 2
= 25 – 8 + 12 × 2, (Simplifying ‘division’ 48 ÷ 6 = 8)
= 25 – 8 + 24, (Simplifying ‘multiplication’ 12 × 2 = 24)
= 17 + 24, (Simplifying ‘subtraction’ 25 – 8 = 17)
= 41, (Simplifying ‘addition’ 17 + 24 = 41)
Example 2:
Simplify using BODMAS rule: 78 – [5 + 3 of (25 – 2 × 10)]
Solution:
78 – [5 + 3 of (25 – 2 × 10)]
= 78 – [5 + 3 of (25 – 20)], (Simplifying ‘multiplication’ 2 × 10 = 20)
= 78 – [5 + 3 of 5], (Simplifying ‘subtraction’ 25 – 20 = 5)
= 78 – [5 + 3 × 5], (Simplifying ‘of’)
= 78 – [5 + 15], (Simplifying ‘multiplication’ 3 × 5 = 15)
= 78 – 20, (Simplifying ‘addition’ 5 + 15 = 20)
= 58, (Simplifying ‘subtraction’ 78 – 20 = 58)
Example 3:
Simplify using BODMAS rule: 52 – 4 of (17 – 12) + 4 × 7
Solution:
52 – 4 of (17 – 12) + 4 × 7
= 52 – 4 of 5 + 4 × 7, (Simplifying ‘parenthesis’ 17 – 12 = 5)
= 52 – 4 × 5 + 4 × 7, (Simplifying ‘of’)
= 52 – 20 + 4 × 7, (Simplifying ‘multiplication’ 4 × 5 = 20)
= 52 – 20 + 28, (Simplifying ‘multiplication’ 4 × 7 = 28)
= 32 + 28, (Simplifying ‘subtraction’ 52 – 20 = 32)
= 60, (Simplifying ‘addition’ 32 + 28 = 60)
Modulus of a real number:
In many engineering calculations you will come across the symbol “| |”. This is known as the modulus.
The modulus of a number is its absolute size. That is, we disregard any sign it might have.
Examples:
- The modulus of −8 is simply 8.
- The modulus of −\(\frac{1}{2}\) is \(\frac{1}{2}\).
- The modulus of 17 is simply 17.
- The modulus of 0 is 0.
So, the modulus of a positive number is simply the number.
The modulus of a negative number is found by ignoring the minus sign.
The modulus of a number is denoted by writing vertical lines around the number.
Note:
The modulus of a negative number can be found by multiplying it by −1 since, for example, −(−8) = 8.
This observation allows us to define the modulus of a number quite concisely in the following way.
\[ | a | =
\begin{cases}
a & \quad \text{if } a > 0\\
-a & \quad \text{if } a < 0
\end{cases}
\]
Examples:
1. | 9 | = 9
2. | − 11 | = 11
3. | 0.25 | = 0.25
4. | − 3.7 | = 3.7
Virnaculum (or Bar):
When an expression contains Virnaculum, before applying the ‘BODMAS’rule, we simplify the expression under the Virnaculum.
Example 1:
Simplify: \(78 \,- [24 \,- {16 \,- (5 \,– \overline{4 \,- 1})}]\)
Solution:
\(78 \,- [24 \,- {16 \,- (5 \,– \overline{4 \,- 1})}]\)
= 78 – [24 – {16 – (5 – 3)}] (Removing vinculum)
= 78 -[24 – {16 – 2}] (Removing parentheses)
= 78 – [24 – 14] (Removing braces)
= 78 – 10
= 68.
Example 2:
Simplify: \(197 \,- [1/9 \{42 \,+ (56 \,- \overline{8 \,+ 9})\} \,+ 108]\)
Solution:
\(197 \,- [1/9 \{42 \,+ (56 \,- \overline{8 \,+ 9})\} \,+ 108]\)
= 197 – [1/9 {42 + (56 – 17)} + 108] (Removing vinculum)
= 197 – [1/9 {42 + 39} + 108] (Removing parentheses)
= 197 – [(81/9) + 108] (Removing braces)
= 197 – [9 + 108]
= 197 – 117
= 80
Example 3:
Simplify: \(95 \,- [144 \,÷ (12 \,\times 12) \,- (-4) \,- \{3 \,– \overline{17 \,- 10}\}]\)
Solution:
\(95 \,- [144 \,÷ (12 \,\times 12) \,- (-4) \,- \{3 \,– \overline{17 \,- 10}\}]\)
= 95 – [144 ÷ (12 x 12) – (-4) – {3 -7}]
= 95 – [144 ÷ 144 – (-4) – {3-7)]
= 95 – [1 – (-4) – (-4)] [Performing division]
= 95 – [1 + 4 + 4]
= 95 – 9
= 86