Mathematics notes, chapter 2
Standard Form
Textbook Mathematics Form 3 (DLP), pages 30–49
What this chapter covers
- Explain the meaning of significant figures and determine the number of significant figures of a number (2.1)
- Round off numbers to a given number of significant figures and state calculator answers to a required number of significant figures (2.1)
- Recognise numbers in standard form A × 10ⁿ, where 1 ≤ A < 10 and n is an integer, and change between single numbers and standard form (2.2)
- Add, subtract, multiply and divide numbers in standard form (2.2)
- Solve problems involving numbers in standard form, including metric prefixes (2.2)
2.1Significant figures
A measurement or estimate is only as exact as the digits we trust. Significant figures (s.f.) tell us how accurately a number is stated. They are counted from the first non-zero digit.
Significant figures are the digits of a number that are stated accurately to the level of accuracy required. Counting starts at the first non-zero digit.
| Type of digit | Significant? | Example |
|---|---|---|
| Every non-zero digit | Yes | 8 435 has 4 s.f. |
| Zero between non-zero digits | Yes | 5 008 has 4 s.f.; 12.051 has 5 s.f. |
| Zero before the first non-zero digit | No (it only fixes the place value) | 0.0809 has 3 s.f. |
| Zero at the end of a decimal | Yes (it shows the accuracy) | 1.2700 has 5 s.f. |
| Zero at the end of a whole number | No, unless the accuracy says so | 15 000 to the nearest thousand has 2 s.f.; to the nearest ten has 4 s.f. |
Rules and examples after textbook p. 34.
Example 1 (after textbook p. 34): number of significant figures
(a) 40 080 (nearest one): 5 s.f.(b) 0.003 050: digits 3, 0, 5, 0 → 4 s.f.
(c) 76 000 (nearest hundred): 3 s.f.
(d) 0.2000: 4 s.f. (the three end zeros show accuracy)
Do not count the zeros in 0.0045 as significant. They only hold the place value. But 0.004 50 has 3 s.f. because the last zero is at the end of a decimal.
Rounding off to a number of significant figures
Use the same idea as rounding to a place value: look at the digit after the last significant figure.
- If it is 5 or more, add 1 to the last kept digit. If it is less than 5, leave the last kept digit unchanged.
- In a whole number, replace the dropped digits with zeros so the place values stay correct.
- In a decimal (digits after the decimal point), simply drop them.
Example 2 (after textbook pp. 35–36): round off to 3 s.f. and 1 s.f.
(a) 68.79 to 3 s.f.: 4th digit 9 ≥ 5, so 68.8(b) 68.79 to 1 s.f.: 2nd digit 8 ≥ 5, so 70 (not 7)
(c) 0.008 024 to 3 s.f.: 3rd s.f. is 2, next digit 4 < 5, so 0.00802
(d) 24 761 to 2 s.f.: next digit 7 ≥ 5, so 25 000
After rounding a whole number, the answer must still be the same size: 24 761 → 25 000, not 25.
Rounding 0.008 025 to 3 s.f. gives 0.00803: the 5 after the last kept digit pushes the 2 up to 3. Check which digit follows the last one you keep.
Calculator answers
Carry all the digits on the calculator and round only at the end. Then state the answer to the number of significant figures asked.
Example 3: 8.59 ÷ 2.1 − 1.26 correct to 3 s.f.
8.59 ÷ 2.1 = 4.0904…; 4.0904… − 1.26 = 2.8304… → 2.83(a) How many significant figures has 0.070 30? (b) Round 5 261 to 2 s.f. (c) Round 0.005 709 to 2 s.f.
Show answerHide answer
(a) 7, 0, 3, 0 → 4 s.f.
(b) 5 300
(c) 0.0057
(b) 5 300
(c) 0.0057
2.2Standard form
Scientists use very big numbers (distance between stars) and very small ones (width of an atom). Standard form writes them neatly.
Standard form is a single number written as A × 10n, where 1 ≤ A < 10 and n is an integer.
- A number 10 or more has a positive index.
- A number less than 1 has a negative index.
- A number from 1 up to (but not including) 10 has index 0, e.g. 4.5 = 4.5 × 100.
Definition after textbook p. 37.
Single number to standard form
Place the decimal point after the first non-zero digit. The index is the number of places the point moved: positive if it moved left, negative if it moved right.
| Single number | Point moves | Standard form |
|---|---|---|
| 45 000 | 4 places left | 4.5 × 104 |
| 6 700 000 | 6 places left | 6.7 × 106 |
| 0.07 | 2 places right | 7 × 10−2 |
| 0.000 062 5 | 5 places right | 6.25 × 10−5 |
12.5 × 103 and 0.8 × 10−4 are not in standard form because A is not between 1 and 10. Write 1.25 × 104 and 8 × 10−5.
Standard form to single number
Positive index: move the point right (the number is 10 or more). Negative index: move the point left (the number is less than 1). Fill empty places with zeros.
Example 4 (after textbook p. 38): change to single numbers
(a) 4.17 × 105 = 417 000(b) 3.6 × 10−5 = 0.000 036
Metric prefixes
| Prefix | Symbol | Value | Standard form |
|---|---|---|---|
| tera | T | 1 000 000 000 000 | 1012 |
| giga | G | 1 000 000 000 | 109 |
| mega | M | 1 000 000 | 106 |
| kilo | k | 1 000 | 103 |
| milli | m | 0.001 | 10−3 |
| micro | µ | 0.000 001 | 10−6 |
| nano | n | 0.000 000 001 | 10−9 |
| pico | p | 0.000 000 000 001 | 10−12 |
| femto | f | 0.000 000 000 000 001 | 10−15 |
Prefix table after textbook p. 39.
Example 5 (textbook p. 38): metric units
(a) 0.0057 nanometre in metre: 5.7 × 10−3 × 10−9 = 5.7 × 10−12 m(b) 216 gigabytes in bytes: 2.16 × 102 × 109 = 2.16 × 1011 bytes
(a) Write 0.000 408 in standard form. (b) Write 9.1 × 104 as a single number. (c) Express 75 micrometres in metres, in standard form.
Show answerHide answer
(a) 4.08 × 10−4
(b) 91 000
(c) 75 × 10−6 = 7.5 × 101 × 10−6 = 7.5 × 10−5 m
(b) 91 000
(c) 75 × 10−6 = 7.5 × 101 × 10−6 = 7.5 × 10−5 m
Adding and subtracting
The indices must be the same first. Change the smaller index to the larger one, then add or subtract the A values and keep the index.
Example 6 (textbook p. 40): 7.02 × 104 + 2.17 × 105
7.02 × 104 = 0.702 × 1050.702 × 105 + 2.17 × 105 = 2.872 × 105
Example 7 (textbook p. 41): 8.21 × 10−4 + 1.49 × 10−5
1.49 × 10−5 = 0.149 × 10−48.21 × 10−4 + 0.149 × 10−4 = 8.359 × 10−4
Never add the indices when adding numbers: 2 × 103 + 3 × 103 = 5 × 103, not 5 × 106.
Multiplying and dividing
- Multiply: multiply the A values, add the indices. (A × 10m) × (B × 10n) = AB × 10m+n
- Divide: divide the A values, subtract the indices. (A × 10m) ÷ (B × 10n) = (A ÷ B) × 10m−n
- If the new A is 10 or more, or less than 1, adjust it so the answer is in standard form.
Example 8 (after textbook p. 42): operations
(a) (2.5 × 103) × (4 × 105) = 10 × 108 = 1 × 109(b) (5.9 × 105) ÷ (2 × 102) = 2.95 × 103
(c) (6.8 × 10−3) ÷ (4 × 10−6) = 1.7 × 103
Use the calculator in scientific mode (Mode, Sci) to check your answer. Always still show the working in the exam.
Calculate in standard form. (a) 4.5 × 106 + 3 × 105 (b) (3 × 10−2) × (5 × 104) (c) (8.4 × 103) ÷ (2.1 × 10−2)
Show answerHide answer
(a) 4.5 × 106 + 0.3 × 106 = 4.8 × 106
(b) 15 × 102 = 1.5 × 103
(c) 4 × 103−(−2) = 4 × 105
(b) 15 × 102 = 1.5 × 103
(c) 4 × 103−(−2) = 4 × 105
Problems in standard form
Follow the textbook steps: understand the problem, plan a strategy, carry it out, then write a conclusion with units.
Example 9 (textbook p. 43): reams of paper
One ream has 800 sheets, each 9.4 × 10−3 cm thick. Total thickness of n reams is 225.6 cm.Thickness of one ream = 800 × 9.4 × 10−3 = 7.52 cm.
n = 225.6 ÷ 7.52 = 30 reams.
Example 10: very small length, with units
A virus has diameter 1.2 × 10−7 m. How many viruses side by side fit across 3 × 10−3 m?(3 × 10−3) ÷ (1.2 × 10−7) = 2.5 × 104 viruses.
Check units first. Convert kilometres, grams and so on to the same unit before you add, subtract or divide.