Understand · explore · practise
Prove a numerical root correct to decimal places using rounding boundaries and sign brackets. Explore bisection, midpoint error bounds, absolute and percentage error, and why rounded iterates or small residuals are not proofs.
Before you startLocating roots; rounding; inequalities
01 / Separate an estimate from a proof of accuracy
To prove α rounds to r at d decimal places, bracket α within r ± 0.5 × 10⁻ᵈ.
A calculator can suggest a root, but a decimal display does not prove that the true root rounds to those digits. Find a valid interval containing the root, then check that every possible value in that interval rounds to the claimed answer.
We use strict bounds around the halfway values, avoiding any ambiguity about how exact ties are rounded.
1 < α < 2. Width = 1; midpoint = 1.5.
f(1) = −2; f(2) = 1. The sign bracket is valid.
To prove 1.732 to 3 d.p., contain the root in (1.7315, 1.7325).
This bracket is not yet contained in the rounding interval.
Midpoint error is less than 0.5.
The view rescales to keep both intervals visible. Each selection performs the stated number of bisections from [1, 2], using unrounded values. Compare steps 10 and 11 at 3 d.p.
02 / Test the halfway boundaries
The halfway boundaries are 1.7315 and 1.7325
They are 0.0005 either side of 1.732.
f(1.7315) = −0.00190775
The lower boundary value is negative.
f(1.7325) = 0.00155625
The upper boundary value is positive.
f is continuous and strictly increasing on this positive interval
There is one root between the boundaries.
Therefore 1.7315 < α < 1.7325, so α = 1.732 to 3 d.p.
Every number strictly inside this rounding cell has the stated rounding.
What two halfway values surround 4.57 when rounding to 2 decimal places?
Half a hundredth is 0.005.
4.565 and 4.575. A strict root bracket contained between these values proves rounding to 4.57.
Give the halfway boundaries for −0.6821 correct to 4 decimal places.
Use ±0.00005 and put the lower value first.
−0.68215 and −0.68205. The negative sign does not change the width of the rounding cell.
03 / Use ordered bounds for a negative root
Use a = −2.2365 and b = −2.2355
Then a < b.
f(a) = 0.00193225; f(b) = −0.00253975
The signs run positive to negative, which is equally valid.
The polynomial is continuous and strictly decreasing on this negative interval
The bracket contains its unique negative root.
−2.2365 < α < −2.2355
Hence α = −2.236 to 3 decimal places.
For x² − 3, f(1.731) = −0.003639 and f(1.732) = −0.000176. Does this calculation prove the positive root is 1.732 to 3 d.p.?
These are not the halfway boundaries, and their signs agree.
No. Both values are negative, so they do not bracket the root. The claimed rounded value is correct, but it needs another argument, such as the halfway-boundary test in the worked example.
If −2.2364 < α < −2.2361, what is α to 3 d.p.?
Compare this whole interval with (−2.2365, −2.2355).
α = −2.236 to 3 d.p. Every value in the given interval lies inside that rounding cell.
04 / Refine a valid sign bracket by halving it
f(1) = −2 and f(2) = 1
Continuity gives a root between them.
Midpoint m = 1.5; f(m) = −0.75
The sign matches the left endpoint.
Keep [1.5, 2]
The endpoints of this half still have opposite signs.
Next m = 1.75; f(m) = 0.0625
Now keep [1.5, 1.75].
Continue until the bracket is narrow enough for the requested conclusion
If a midpoint value is exactly zero, an exact root has been found.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Apply two bisections to f(x) = x² − 6 on [2, 3]. Give the retained interval after each.
First test 2.5, then the midpoint of the retained half.
f(2.5) = 0.25 > 0, so retain [2, 2.5]. Then f(2.25) = −0.9375 < 0, so retain [2.25, 2.5].
Can the sign-change bisection argument be started with 1/x on [−1, 1]?
Check the proposed interval and its midpoint.
No. The function is discontinuous at zero, which is also the first midpoint. Opposite endpoint signs do not establish a root.
05 / Check containment, not just bracket width
After 10 halvings: 1.7314453125 < α < 1.732421875
The width is less than 0.001.
The lower endpoint is below 1.7315
The bracket still contains numbers that round to 1.731 and numbers that round to 1.732.
After 11 halvings: 1.73193359375 < α < 1.732421875
Both bounds lie inside (1.7315, 1.7325).
This bracket proves α = 1.732 to 3 d.p.
The position of the interval matters as well as its width.
Does 2.34545 < α < 2.34555 prove a unique rounding to 3 d.p.?
The halfway boundary 2.3455 lies inside the interval.
No. Values on its two sides round to 2.345 and 2.346 respectively, even though the bracket width is only 0.0001.
If 0.87652 < α < 0.87658, what can you state to 3 d.p.?
Compare with the boundaries 0.8765 and 0.8775.
α = 0.877 to 3 d.p. The entire bracket is above 0.8765 and below 0.8775.
06 / Bound the midpoint error after n halvings
Initial width W → width W/2ⁿ after n halvings; midpoint error < W/2ⁿ⁺¹.
The true root lies strictly inside a retained sign bracket. Its distance from the midpoint is therefore less than half the bracket width. Count the initial interval as n = 0.
This is an absolute-error bound. It is not automatically a proof that every value rounds to the same number of decimal places.
Starting from an interval of width 1, give the width and midpoint-error bound after 7 halvings.
Use 2⁷ and 2⁸.
The width is 1/128 = 0.0078125. The midpoint error is less than 1/256 = 0.00390625.
Starting with width 1, how many halvings guarantee midpoint error below 0.0001?
Require 1/2^(n+1) ≤ 0.0001.
13 halvings suffice: 1/2¹⁴ ≈ 0.00006104. At 12 halvings, the bound 1/2¹³ ≈ 0.00012207 is too large.
07 / Calculate absolute error when a reference is known
Absolute error = |1.732 − √3|
Take a nonnegative distance.
The error is approximately 0.00005080757
The exact root is available here for comparison.
The 3 d.p. rounding bound is 0.0005
The actual error is smaller than this general bound.
For roots without a known exact expression, a validated bracket still gives an error bound
You do not need to know the true root to bound its location.
A valid root bracket is 3.14 < α < 3.16. Give its midpoint estimate and an absolute-error bound.
Take the midpoint and half the width.
The estimate is 3.15 and its error is less than 0.01.
08 / Put an absolute error in proportion
Percentage error = |approximation − reference| / |reference| × 100%, for reference ≠ 0.
For 1.732 as an approximation to √3, the percentage error is about 0.002933%. Always identify the reference used. If the exact value is unavailable, a comparison with a numerical reference is only as reliable as that reference.
Percentage error is undefined for a zero reference and can be large near zero even when the absolute error looks small.
An approximation is 11.8 and the exact value is 12. Find the absolute and percentage errors.
Use 12, not 11.8, in the denominator.
The absolute error is 0.2. The percentage error is (0.2/12)×100% = 1⅔%.
The exact value is 0.002 and the approximation is 0.003. Find the percentage error.
The absolute error is 0.001.
(0.001/0.002)×100% = 50%. A small absolute error need not be a small relative error.
09 / Keep working precision and verify the final claim
Carry unrounded calculator values from one iteration to the next, then round only the answer you are asked to report. Premature rounding changes the iteration itself and may obscure endpoint signs.
Two successive values that display the same digits are useful evidence of numerical stability, but that alone does not prove convergence or the accuracy of the true root. A validated sign bracket inside the rounding cell gives the needed mathematical conclusion.
A small residual |f(x)| measures how close the output is to zero. It does not by itself measure how close x is to a root; the scale and slope of the function matter.
For f(x) = 10⁻¹²(x − 100), find the root and the residual at x = 0. Does this small residual prove that zero is an accurate root approximation?
A nonzero scale factor does not change the root.
The root is 100. At x = 0, |f(0)| = 10⁻¹⁰, yet the input error is 100. A tiny residual alone is not an accuracy guarantee.
10 / State exactly what has been verified
Section 1 of 10 · Separate an estimate from a proof of accuracy