Number

Rounding and bounds

Explore accuracy and included or excluded endpoints.

Predict, test and explain

Step 1 of 3

Predict

What does 3.75 round to, to 1 decimal place, using halfway rounds up?

7.36⟶7.47.36\longrightarrow 7.4
7.357.35
7.457.45
7.36⟶7.47.36\longrightarrow 7.4

Scroll the diagram sideways to see every label.

Error interval7.35≤x<7.457.35\leq x<7.45

The filled lower endpoint is included. The open upper endpoint is excluded: that value rounds to the next step. Halfway rounds up for these positive values.

Check the working
QuantityValue
Rounded7.4
Lower bound, included7.35
Upper bound, excluded7.45

InvestigateWhy is one endpoint filled and the other open?

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Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

A positive length is recorded as 3.7 cm to the nearest 0.1 cm. Give its error interval, using halfway rounds up.

Hint

Find half the rounding step on either side of 3.7.

Reveal answer

The lower bound is included; the upper bound is excluded because 3.75 rounds to 3.8.

3.65≤l<3.75(cm)3.65\leq l<3.75\quad\text{(cm)}

Build an explanation

Hint 1

Identify the rounding unit. To the nearest 0.1, the unit is 0.1 and half the unit is 0.05.

Hint 2

Subtract half the unit for the lower bound and add half the unit for the upper bound.

Hint 3

Include the lower bound and exclude the upper bound: lower bound ≤ actual value < upper bound.

Worked example

This example uses fixed values, separate from the diagram controls.

A length L is recorded as 7.4 cm to the nearest 0.1 cm. Write its error interval.

  1. Half the rounding unit is 0.1 ÷ 2 = 0.05 cm.
  2. The lower bound is 7.4 − 0.05 = 7.35 cm.
  3. The upper bound is 7.4 + 0.05 = 7.45 cm. This endpoint rounds to 7.5 cm, so it is excluded.

Answer: 7.35 cm ≤ L < 7.45 cm

Connect this idea

Watch out: The upper bound is excluded. L = 7.45 cm does not round to 7.4 cm under the usual school rounding rule.