Babylonian Square Root Method – Iterations
Heron’s iteration x←(x+S/x)/2 nails √2 in three steps: 1 → 1.5 → 1.4167 → 1.41422.
- Every iterate and its error shown.
- Converges fast.
Figures are estimates from this calculator’s standard formula — adjust the inputs in the tool below for your exact number.
How the Babylonian Method Calculator Works
The Babylonian (Heron’s) method refines a square-root guess by averaging it with S divided by it. If the guess x is too high, S/x is too low, and their average is closer to √S than either. For S = 2 starting at x₀ = 1: x₁ = (1 + 2)/2 = 1.5; x₂ = (1.5 + 2/1.5)/2 = 1.41667; x₃ = 1.4142169 — already within 0.0000031 of √2 ≈ 1.4142136. The calculator lists every iterate, its square (to show the residual error), and the number of correct digits, which roughly doubles per step — quadratic convergence, the same engine inside Newton’s method applied to f(x) = x² − S. Any positive seed works; poor guesses just add a step or two (x₀ = 7 for √2 catches up by the fourth iterate). The method doubles as the classic divide-and-average algorithm taught before calculators.
- 1
Enter S and a starting guess
Any positive guess; 1 or S/2 are common defaults
- 2
Average x with S/x
Each iterate xₖ₊₁ = (xₖ + S/xₖ)/2 is printed
- 3
Watch errors halve their digits
Correct digits double per iteration
- 4
Stop at tolerance
The run halts when |xₖ² − S| falls below your threshold
Use Cases
Numerical Methods Class
Demonstrate quadratic convergence on √2
Mental Math Training
Two hand iterations get √10 to 3.1623
Embedded Systems
Integer square roots via divide-and-average loops
Tips
- 1
Averaging x and S/x always moves toward √S
- 2
Digits double each iteration — three steps beat six decimal places
- 3
The seed hardly matters; bad guesses cost one extra step
Common Mistakes
Averaging x with x/2 instead of S/x
Stopping after one iterate and expecting full precision
Using a negative seed, which sends the iteration complex
Confusing the method with digit-by-digit longhand square rooting
FAQs
How fast does the method find √2?
Why does averaging work?
Does the starting guess matter?
Can it compute cube roots?
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