Square roots and Heron’s method

What is it about?

The square root $\sqrt{a}$ (for $a \geq 0$ ) is the non-negative number whose square is $a$ : $\left(\sqrt{a}\right)^2 = a$ and $\sqrt{a} \geq 0$ . The number $a$ under the root sign is called the radicand. Geometrically, $\sqrt{a}$ is the side length of a square with area $a$ .

If you lay $a$ unit squares on squared paper as ever larger squares ( $1 + 3 + 5 + \ldots$ ), all squares fit exactly only for square numbers: $\sqrt{49} = 7$ , $\sqrt{144} = 12$ . Otherwise some are left over, and $\sqrt{a}$ lies between two whole numbers:

7^2 = 49 < 50 < 64 = 8^2 \quad\Rightarrow\quad 7 < \sqrt{50} < 8

If the radicand contains a square factor, you can partly take the root: $\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2} \approx 7.071$ .

Heron’s method: start with a rectangle of area $a$ , that is with the sides $x_0$ and $\frac{a}{x_0}$ . The new side is the mean of the two sides; the other side is again $a$ divided by the new side:

x_{n+1} = \frac{x_n + \frac{a}{x_n}}{2}

The area always stays $a$ , and the rectangle gets more and more square. Why? Cut a strip of width $\frac{x - a/x}{2}$ off the rectangle and put it on the short side. This gives almost a square with side $x_{n+1}$ – only a small corner is missing. So from the first step on, $x_{n+1}$ is always slightly above $\sqrt{a}$ and $\frac{a}{x_{n+1}}$ slightly below. This is very fast: the number of correct digits roughly doubles. For $\sqrt{2}$ with $x_0 = 2$ :

x_1 = 1.5 \qquad x_2 = \tfrac{17}{12} \approx 1.416667 \qquad x_3 = \tfrac{577}{408} \approx 1.4142157 \qquad \sqrt{2} \approx 1.4142136

Nested intervals: you trap $\sqrt{a}$ between numbers whose squares trap $a$ , and check by squaring:

1 < \sqrt{2} < 2, \qquad 1.4 < \sqrt{2} < 1.5 \;\;(1.96 < 2 < 2.25), \qquad 1.41 < \sqrt{2} < 1.42 \;\;(1.9881 < 2 < 2.0164)

Each digit makes the interval ten times shorter.

$\sqrt{2}$ is irrational: the diagonal of the unit square has length $\sqrt{2}$ , because the square on it is made of four half unit squares and has area $2$ . This length is not a fraction. If $\sqrt{2} = \frac{p}{q}$ were fully reduced, then $p^2 = 2q^2$ , so $p$ is even, $p = 2r$ ; then $4r^2 = 2q^2$ , so $q^2 = 2r^2$ and $q$ is even too – a contradiction to the reduced fraction. That is why the decimal expansion of $\sqrt{2}$ never ends and never repeats. In the same way, $\sqrt{a}$ is irrational for every whole number $a$ that is not a square number.

Try it

49 unit squares: press “Lay the square”. How do the squares $1, 4, 9, \ldots$ grow? How many unit squares are added each time? Compare with 50 unit squares.
Drag the green corner of the square. For which values of $a$ does everything fit exactly?
√2 with Heron: click “Next step” and watch the strip and the missing corner. The correct digits in the table are bold. How many are added per step?
A long rectangle: start with $10 \times 1$ . Drag the blue corner at step 0 and try other starting lengths. Does the start matter much?
Nesting √2: tap the subinterval that contains $\sqrt{2}$ . What does the check show if you tap the wrong one? Why does the nesting stop at once for √49?
√2 is not a fraction: go through the proof step by step and look at the near misses $\frac{p}{q}$ .

Tasks

Task 1: Between two square numbers

Load task

Between which whole numbers does $\sqrt{50}$ lie? Simplify $\sqrt{50}$ by partly taking the root.

Show solution

$7^2 = 49 < 50 < 64 = 8^2$ , so $7 < \sqrt{50} < 8$ . Also $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} \approx 5 \cdot 1.4142 \approx 7.071$ . Show in the simulation

Task 2: Heron by hand

Load task

Use Heron’s method to find two approximations of $\sqrt{10}$ , starting at $x_0 = 3$ .

Show solution

$x_1 = \frac{3 + \frac{10}{3}}{2} = \frac{19}{6} \approx 3.16667$ and $\frac{10}{x_1} = \frac{60}{19} \approx 3.15789$ .

$x_2 = \frac{\frac{19}{6} + \frac{60}{19}}{2} = \frac{721}{228} \approx 3.162281$ . For comparison: $\sqrt{10} \approx 3.162278$ – four decimal places are already correct. Show in the simulation

Task 3: Finding intervals

Load task

Between which numbers with one and with two decimal places does $\sqrt{3}$ lie? Justify by squaring.

Show solution

$1.7^2 = 2.89 < 3 < 3.24 = 1.8^2$ , so $1.7 < \sqrt{3} < 1.8$ .

$1.73^2 = 2.9929 < 3 < 3.0276 = 1.74^2$ , so $1.73 < \sqrt{3} < 1.74$ . Show in the simulation

Task 4: Why is √a always in between?

Use the picture of the moved strip to explain: $x_{n+1}^2 = a + d^2$ with $d = \frac{x_n - a/x_n}{2}$ . What follows for $x_{n+1}$ and $\frac{a}{x_{n+1}}$ ? Watch the strip move

Show solution

The almost-square with side $x_{n+1}$ consists of the whole rectangle (area $a$ ) and the missing corner (area $d^2$ ). So $x_{n+1}^2 = a + d^2 \geq a$ , that is $x_{n+1} \geq \sqrt{a}$ . Since $x_{n+1} \cdot \frac{a}{x_{n+1}} = a = \sqrt{a} \cdot \sqrt{a}$ , it follows that $\frac{a}{x_{n+1}} \leq \sqrt{a}$ : $\sqrt{a}$ always lies between the two sides.

Task 5: How accurate is 577/408?

Load task

In Heron’s method for $\sqrt{2}$ (starting at $2$ ), $x_3 = \frac{577}{408}$ . How many decimal places are correct? Work out $\left(\frac{577}{408}\right)^2$ as exactly as you can without a calculator.

Show solution

$\frac{577}{408} = 1.4142156\ldots$ and $\sqrt{2} = 1.4142135\ldots$ : five decimal places are correct. We have $577^2 = 332\,929$ and $2 \cdot 408^2 = 2 \cdot 166\,464 = 332\,928$ . So $\left(\frac{577}{408}\right)^2 = 2 + \frac{1}{166\,464} \approx 2.000006$ – almost $2$ , but not exactly. Show

Task 6: No fraction hits exactly

Check without a calculator whether $\frac{17}{12}$ or $\frac{99}{70}$ can be equal to $\sqrt{2}$ . See the proof

Show solution

$17^2 = 289$ , but $2 \cdot 12^2 = 288$ ; $99^2 = 9801$ , but $2 \cdot 70^2 = 9800$ . Both times they differ by exactly $1$ – very close, but never equal: by the proof, $p^2 = 2q^2$ is impossible. Show all steps

Ideas for the classroom

Use: grade 9, topic “square roots” (definition of the root, approximation methods, irrational and real numbers, calculating with roots) – please check the exact assignment against your curriculum. The four views work well as an introduction (unit squares), main phase (Heron, intervals) and deeper look (proof).
Start without the formula: the moved rectangle makes Heron’s method visible before the formula is written down. It may be worth letting the students suggest first how a rectangle could be made “more square”.
Race: who gets six correct digits first – with Heron or with nested intervals?
For tasks: with “Show values” off (show=0), $\sqrt{a}$ and the approximations appear as “?” in the picture.
Model: bold digits in the table already agree with $\sqrt{a}$ ; the nested intervals are computed with whole numbers, so without rounding errors.
Common stumbling blocks:
- “ $\sqrt{2} = 1.41$ ” – that is only an approximation; $1.41^2 = 1.9881$ .
- The bounds of an interval are guessed but not checked by squaring.
- In Heron’s method the second side is averaged as well instead of being recomputed as $a : x$ – then the area is no longer right.
- $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ – false (e.g. $\sqrt{9 + 16} = 5 \neq 7$ ); only for products does $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ hold.
- Concluding “irrational” from “lots of digits look random” – only the proof shows that nothing repeats.