Squeezing √2
Tonight's idea
\(\sqrt{2}\) is the number whose square is 2. The question is how to trap it. The test is always the same: square a guess and compare with 2.
\(1^2 = 1\) and \(2^2 = 4\), so \(\sqrt2\) is between 1 and 2. \(1.4^2 = 1.96\) and \(1.5^2 = 2.25\), so it is between 1.4 and 1.5. \(1.41^2 = 1.9881\) and \(1.42^2 = 2.0164\): between 1.41 and 1.42. Each new decimal place takes up to nine squarings to find.
Bisection is tidier. Keep a trap \([\text{low}, \text{high}]\) with \(\sqrt2\) inside. Square the midpoint. If the square is above 2, the midpoint becomes the new high; if below, the new low. The trap halves every step.
The Babylonian method is faster still. Take any guess \(x\). Then \(x\) and \(2/x\) multiply to 2, so they sit on opposite sides of \(\sqrt2\): if one is too big the other is too small. Their average is a better guess.
New guess \(= \dfrac{1}{2}\left(x + \dfrac{2}{x}\right)\). Repeat.
Picture a rectangle of area 2 with sides \(x\) and \(2/x\). Averaging the sides makes it squarer; the limit is a square of area 2.
Watch one: Babylonian from \(x = 1.5\)
\(2/1.5 = 1.3333\), so the next guess is \(\tfrac12(1.5 + 1.3333) = 1.41667\). Again: \(2/1.41667 = 1.41176\), average \(1.41422\). Again: \(1.41421356\). Compare with \(\sqrt2 = 1.41421356237\ldots\)
| Step | Guess | Correct digits |
|---|---|---|
| 0 | 1.5 | 1 |
| 1 | 1.41667 | 3 |
| 2 | 1.4142157 | 6 |
| 3 | 1.414213562375 | 12 |
The number of correct digits roughly doubles each step. Bisection earned one digit every three or four steps. Computers find square roots this way, or by close relatives of it.
Your turn
- Trap \(\sqrt{10}\) between consecutive tenths, then between consecutive hundredths. Show the two squarings that settle each trap.
- Bisection for \(\sqrt5\), starting from the trap \([2, 3]\). Do four steps. Write the midpoint, its square, and the new trap each time. How wide is the trap at the end?
- Babylonian for \(\sqrt{10}\) from \(x = 3\). Do two steps, keeping six decimal places. \(\sqrt{10} = 3.16227766\ldots\). How many digits are right after each step? Compare with question 1.
- (a) Explain why, if \(x\) is bigger than \(\sqrt2\), then \(2/x\) must be smaller than \(\sqrt2\). (b) So after any Babylonian step you get a trap for free: \(\sqrt2\) lies between the new guess and 2 divided by it. Write that trap after the first step from \(x = 1.5\). How wide is it?
Stretch: with fractions, no decimals
Start the Babylonian method at \(x = 1\) and keep every guess as a fraction: \(1, \tfrac32, \tfrac{17}{12}, \ldots\) Find the next two. For each one, compute \(\text{top}^2 - 2 \times \text{bottom}^2\). What do you notice, and what does it say about how close the fraction is to \(\sqrt2\)?
Answers
- \(3.1^2 = 9.61\) and \(3.2^2 = 10.24\), so \(3.1 \lt \sqrt{10} \lt 3.2\). \(3.16^2 = 9.9856\) and \(3.17^2 = 10.0489\), so \(3.16 \lt \sqrt{10} \lt 3.17\).
- Midpoint 2.5, square 6.25, too big: \([2, 2.5]\). Midpoint 2.25, square 5.0625, too big: \([2, 2.25]\). Midpoint 2.125, square 4.515625, too small: \([2.125, 2.25]\). Midpoint 2.1875, square 4.78515625, too small: \([2.1875, 2.25]\). Width \(0.0625\). (\(\sqrt5 = 2.236\ldots\))
- \(10/3 = 3.333333\), average \(3.166667\): two digits right (3.1). Then \(10/3.166667 = 3.157895\), average \(3.162281\): right to five decimal places (3.16228). Two steps beat the nine squarings of question 1.
- (a) Dividing 2 by a larger number gives a smaller result. If \(x > \sqrt2\), then \(2/x \lt 2/\sqrt2 = \sqrt2\). (b) First step gives \(1.41667\), and \(2/1.41667 = 1.41176\). So \(1.41176 \lt \sqrt2 \lt 1.41667\), a trap of width about \(0.005\) after one step. (The guess is the upper end: the average of two numbers whose product is 2 is never below \(\sqrt2\).)
Stretch. \(\tfrac12\left(\tfrac{17}{12} + \tfrac{24}{17}\right) = \tfrac{577}{408}\), then \(\tfrac{665857}{470832}\). The starting guess \(\tfrac11\) gives \(1 - 2 = -1\). For every fraction after it, \(\text{top}^2 - 2 \times \text{bottom}^2 = 1\): \(3^2 - 2 \cdot 2^2 = 1\), \(17^2 - 2 \cdot 12^2 = 1\), \(577^2 - 2 \cdot 408^2 = 332929 - 332928 = 1\). So \(\left(\tfrac{577}{408}\right)^2 = 2 + \tfrac{1}{408^2}\), above 2 by only \(1/166464\). The bottoms grow fast, so the overshoot shrinks fast.