Three and a half methods for finding square roots

The easiest way to find a square root in this technological age is to use a calculator. That’s a fine method if what you want to do is simply calculate a square root. However, if what you want to do is understand what a square root is, here are some methods for finding the value mechanically.

What IS a square root?

My students persistently confuse the terms “square” and “square root,” presumably because (a) the terms are so similar and (b) they don’t have a clear understanding of what the two terms mean. If your understanding of mathematics is purely mechanical, you have a weaker ground for distinguishing between similar terms.

In this case, the terms are similar because they’re related. We can go back to al-Khworizmi’s al-Jabr to see that he considered three types of related numbers: Unitless values, lengths, and areas. He focused in the first part of his book on the relationship between these three. Most relevant to this discussion, a square number is an area, while a square root is a length: The square root is the side length of a geometric square, while a square (number) is the area of the square.

This is confusing in modern English because we’ve conflated geometric and algebraic terms, and then largely set the geometric terms aside. The heavy emphasis in modern mathematics education on algebra makes it difficult to see that, for several millennia, mathematics was largely done in terms of geometry. Algebra is a relative latecomer, and our modern algebraic notation even more so.

So, to clarify, I’ll capitalize “square” when it’s a geometric object. A square is the area of a SQUARE whose side length is the square root. A SQUARE’s root is the length of its side. The term “root” here is not that far off from its use in plants: A tree’s root is the foundation on which the tree is supported; a SQUARE’s root is likewise its foundational support. Because of the properties of SQUAREs, if we know its root (side length), we can figure out its area (square), and vice versa.

We could have used this notion of root and generalized it across all regular polygons: An equilateral triangle’s root would be its side length, from which we could figure out its area. We could even have generalized it across all regular polyhedra (the root of a Platonic tetrahedron, for instance, would be the length of its edge, from which we can calculate its volume), and even up to the higher dimensions.

It happens that we did not do this. Instead, we generalized it only to the cube and higher dimensional analogs (such as the tesseract. There is a solid mathematical reasoning for this, given the importance of the SQUARE to area and the cube to volume, but it means that the word “root” appears in a limited mathematical context, and it’s easy to detach the term from its underlying concept.

Method 1: Brute Force

I was a part of the first generation of high school students that had ready access to calculators capable of providing square roots. Nonetheless, one day, one of my math teachers felt compelled to teach us how to find square roots mechanically. He used the slowest, but easiest, method: Brute force.

This is not a pointless exercise, because it reinforces a key concept. If \(a < b < c\), then \(\sqrt{a} < \sqrt{b} < \sqrt{c}\). So if we can find a number whose square is less than our square and another number whose square is more than it, we know that our target root has to be between them.

For instance, let’s say we want to find \(\sqrt{6}\). We know that \(2^2 = 4\) and \(3^2 = 9\), so \(2 < \sqrt{6} < 3\). We then try 2.5, and adjust as we go. This is a very long process, although it can be sped up with practice at making good guesses. Here are some rounds:

  • \(2^2 = 4\)
  • \(3^2 = 9\)
  • \(2.5^2 = 6.25\)
  • \(2.4^2 = 5.76\)
  • \(2.45^2 = 6.0025\)
  • \(2.44^2 = 5.9536\)

Method 2: Digit Pairs

By several reports, this is the method that was most often taught in schools before calculators pushed manual calculation of square roots out of education. Unfortunately, though, this is usually taught as an algorithm, which takes away its visual explanation. And while I like the visual nature of the explanation provided by Brett Berry, their use of a rectangle instead of a square detracts from the key point, and the strength of the method.

Before we talk about the algebra, consider the geometry. Let’s look at how this works for \(\sqrt{6}\). The largest square in the diagram to the right has an area of 6, so its side length (root) is \(\sqrt{6}\). The smallest square has a side length of 2 (a), and hence an area of 4.

There is a difference in area between these two squares. If we insert a second square, the difference in area will be fairly small. The area of the square whose side length is a + b is slightly less than the area of the square whose area is 6 (that is, a + b + some small value).

What is the area of the l-shaped region included in the middle square, excluding the smallest square? It consists of two rectangles with side lengths of a and b, and one square with a side length of b. That is:

\[(a + b)^2 = a^2 + 2ab + b^2\]

Now, that’s pretty close, but we can get closer still. We can add another rectangle with a known side length that is just slightly smaller than the target square. In the example, the new square is so close in size to the original square that we have to zoom in to see the difference.

The question is, though: How do we get the best estimates for a, b, c, and so on? Unlike with the brute force method, this method requires us to make each new square larger than the previous estimate but smaller than the goal.

So, let’s look at the algebraic algorithm.

First, write the square number with an even number of digits on each side of the decimal point. We’re going to be finding three digits after the decimal, so write it as 06.000000.

For the first step, find the largest square number that’s less than the first pair of digits. In this case, that’s 2.

Current estimate: 2. Remainder: \(06.00 00 00 – 2^2 = 2.00 00 00\). Include the next pair of digits as r.

For the next step, multiply your current estimate by 20 to get y, then find the highest integer x that satisfies the inequality \((x + y)x < r\). In this case, you want to solve for \((x + 40)x < 200\). Since \(44 \times 4 = 176\) and \(45 \times 5 = 225\), use 4.

Current estimate: 2.4. Remainder: \(2.00 00 00 – 4.4 \times 0.4 = 0.24 00 00\).

Notice that since y is two times ten times our previous estimate, we’re doing two things. First, we’re assuming that a is the higher decimal unit (in this case, the “ones place”). Since we’re looking for the tenths place, we multiply a by 10. Second, we’re taking into account that the area of the l-shaped region in the middle square is twice the area of the rectangles plus the difference squared. In other words, underneath the algorithm is the geometry.

Repeat until we have as many digits as we want. For the next round, \(y = 20 \times 24 = 480\) and r = 2400, so we want the largest integer such that \((x + 480)x < 2400\). Since \(484 \times 4 = 1936\) and \(485 \times 5 = 2425\), use 4 again.

Current estimate: 2.44. Remainder: \(0.04 64 00\). \(y = 20 \times 244 = 4880\) and r = 46400, so \((x + 4880)x < 46400\). Since \(4889 \times 9 = 44001\), the next digit is 9.

Current estimate: 2.449. Remainder: \(0.00 23 99\). We could of course keep going. Since \(\sqrt{6}\) is irrational, this will continue forever.

This is a faster technique than the brute force method, but it still relies on guessing next digits.

Method 3: The Babylonian Method

Consider all the rectangles with a given area (in this case, 6). What is the rectangle with the smallest perimeter? In every case, it is the SQUARE.

For instance, take an area of 16. We can make several rectangles with integer sides that have an area of 16: 1 x 16, 2 x 8, and 4 x 4. These have, respectively, perimeters of 34, 32, and 16. We could make rectangles with non-integer sides, such as .5 x 32, 1.5 x 10.6(6), and so on, but in every case, the perimeter will be more than 16 for non-SQUAREs.

Notice that, as the smaller of the two numbers gets larger (i.e., closer to the root), the perimeter gets smaller. This tells us that, because the arithmetic mean of two numbers is always between those two numbers, the arithmetic mean and its partner will be a better estimate for the square root than either of the two numbers in the original pair: 1 and 16 have a mean of 8.5; its partner is 1.88. Those numbers are tighter in on the root than 1 and 16 are.

So the concept behind this method is that we’re going to create a pair of bookends around the root, then use the arithmetic mean to move the bookmarks closer in. Each time, we will move the bookends closer in.

Let’s try this with \(\sqrt{6}\). This time, we’re going to use a number that’s close to \(6\); since we can, let’s use a factor, 3. 6/3 = 2, so our first bookend pair is (2, 3). The root is somewhere between those two.

To find the second bookend pair, find the mean of these numbers, then divide 6 by the result. \((2 + 3)\div 2 = 2.5\) and \(6 \div 2.5 = 2.4\), so our next pair of bookends is (2.4, 2.5).

Repeat: The mean of these is 2.45, and 6 divided by this is about 2.44898. We now have the value to two decimal places, since both of these round to 2.45. If we wanted more places, we could keep going.

For advanced students: Note that the square roots of the bookends is the geometric mean of those two numbers. What we’re doing, in more rigorous parlance, is using the fact that the geometric mean is always less than the arithmetic mean except when we’re dealing with a square (or a SQUARE).

Method Half:

The last idea isn’t for finding the square root of a prime, but rather for a composite in the cases where you know the roots of its factors.

One complaint I saw when researching this article is that high school teachers insist on having students keep things in radical form, ostensibly because the teachers don’t know how to find roots. I can’t speak for other teachers, but I encourage (and sometimes require) students to keep numbers in radical form because these numbers are irrational. Every time we estimate numbers, we introduce error. Sometimes radicals will multiply with radicals and, in so doing, become integers.

As an example of this, draw a right triangle with legs of 1 each. The hypotenuse is \(\sqrt{2}\). Now draw a segment perpendicular to this hypotenuse, with a length of 1. Draw a new right triangle; it’s hypotenuse is \(\sqrt{3}\). Do it again: The new hypotenuse will have a length of 2. The radical has gone away.

However, if you know an estimate for \(\sqrt{2}\) and for \(\sqrt{3}\), you can multiply them to get an estimate for \(\sqrt{6}\). In this case, using 1.414 and 1.732, we multiply them to get 2.449. This means you only really need to calculate the roots of primes manually, and keep a list handy.

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