Chapter 4: Indices, Surds, Standard Form and Bounds
Indices, Surds, Standard Form and Bounds
Understand compact number notation and use bounds to reason about accuracy.
AI teacher
Start the short explanation, then use the visual model and practice before returning to the workbook.
Voice: Browser voice
Workbook pages
Junior Maths Book 1 pages 51-68
Use numerical indices, standard form, estimation, surds, rationalising, rounding, bounds and error intervals.
AI explanation
The AI teacher links indices to repeated multiplication, standard form to place value and powers of 10, surds to exact square-root form, and bounds to measurement uncertainty.
- 1An index tells how many times a base is multiplied by itself.
- 2Standard form writes very large or small numbers as a number between 1 and 10 times a power of 10.
- 3A surd is an exact irrational root that has not been rounded.
- 4A bound describes the range of values a rounded measurement could have.
Animated teaching
See the method
A decimal point moves across place-value columns, a surd splits into square factors, and a measurement expands into an error interval.
72 000 = 7.2 x 10^4
Why this method works
- Index laws follow from counting repeated factors.
- Standard form works because moving decimal places is the same as multiplying or dividing by powers of 10.
- Bounds work because rounding hides a small interval of possible true values.
Common mistakes
- Writing 42 as 4 x 2 instead of 4 squared.
- Using a standard-form first number that is not between 1 and 10.
- Assuming square root of a sum equals sum of square roots.
- Using the rounded value as if it were exact in a bounds question.
Multiple methods
- Standard form can be checked by moving the decimal or by estimating place value.
- Surds can be simplified by looking for square factors or by prime factors.
- Bounds can be found with half the rounding unit or by drawing an interval.
Worked examples
Standard form
Write 0.00046 in standard form.
Method: Move the decimal until the first number is between 1 and 10.
- 1.0.00046 becomes 4.6.
- 2.The decimal moved 4 places right.
- 3.Use a negative power of 10.
Answer: 4.6 x 10^-4
Bounds
A length is 4.8 cm to the nearest 0.1 cm. Find the lower and upper bounds.
Method: Use half the rounding unit.
- 1.Half of 0.1 cm is 0.05 cm.
- 2.Lower bound = 4.8 - 0.05.
- 3.Upper bound = 4.8 + 0.05.
Answer: 4.75 cm <= length < 4.85 cm
Interactive practice
Try before you look
Question 1
Write 72 000 in standard form.
Question 2
Simplify sqrt(75).
Question 3
A mass is 12 kg to the nearest kg. What is the lower bound?
Review prompt
Explain why sqrt(20) is not equal to sqrt(16) + sqrt(4).
Write a two-sentence answer in your notebook before marking the workbook section complete.
Challenge question
A cube has side length 4.8 cm to the nearest 0.1 cm. Explain why the upper bound for its volume uses 4.85^3, not 4.8^3.
Hint: The measured side could be slightly larger than 4.8 while still rounding to 4.8.