In my previous article I discussed a variation on multiplication by hand allowing us to, in a single step, write down the result of multiplying two numbers. This was an exercise in showing off some stuff about our place value system and could be a segue into some Algebra basics.

In this article, I’ll take a look at square roots.

Typically, students are taught about square roots via familiar numbers. For example, the square root of \(25\) is \(5\) since \(5 \times 5 = 25\). And eventually students are explained that a general notion of a square root of a number is “what number times itself gives the number we’re square rooting?” or something similar to that effect.

Interestingly, very few students of arithmetic ever really learn to compute a square root by hand. From a utility standpoint, which seems to dominate math education narrative from time to time, there isn’t really a need to compute square roots by hand. And that is often enough reason to jettison it as a topic of study. College math majors may learn of a number of algorithms for computing / approximating square roots. Probably the first set of methods students see are in Calculus — either the bisection method or Newton’s method. Then in an Analysis class they may have to prove something about the Babylonian method. And that’s often about it. There are other hand computation methods, but they are seldom shown to students.

Here’s a different way for finding the square root of a number. More precisely, I will show how to find the largest integer \(x\) such that for a given integer \(y\) we will have \(x \le \sqrt{y}\). For the remainder of this article if I write something to the effect of “Find the square root of some integer”, I mean it as described in the previous sentence.

Let’s get started by way of example.

Find \(\sqrt{589}\).

Since \(589\) is a three-digit number, its square root must be a two-digit number (why?). Thus, if \(a,b\) are digits, and \(ab\) is the two-digit number formed by the concatenation of \(a\) and \(b\), then we want \(ab \times ab \le 589\).

Unlike multiplication, with finding square roots, we will proceed from the most significant position to the least significant position. And this is where the fun begins.

As you may have seen from the previous article I linked to earlier, the hundreds place can be found by \(10s \times 10s + \mbox{ carry from \(2 \times 10s \times 1s\)}\). In other words, we want the largest digit \(a\) such that \(a^{2} + \mbox{ carry}(2\cdot a\cdot b) \le 5\). Thus, our only choices are \(a \in \{0,1,2\}\) and since \(b\) is currently unknown, the carry is also unknown and at minimum is zero. Thus, \(a = 2\) is the best choice. Therefore, our solution to \(\sqrt{589}\) currently looks like \(2b\).

We are left with finding the best \(b\). Since our number looks like \(2b\), at worst we have a difference of \(189\) to make up (\(500 – 400\) since \(20 \times 20 = 400\)). Thus, we want the largest digit \(b\) such that \(2\cdot 2\cdot b \le 18\). The \(b\) that satisfies this best is \(b = 4\). And indeed \(24^{2} = 576 < 589\) while \(25^{2} = 625 > 589\), therefore the best integer approximation from below to \(\sqrt{589}\) is \(24\).

So, hopefully, that made sense.

But it was a tiny problem. Let’s try something that will require three digits. I mash on my number pad and get \(38439\). So, let’s find the best integer approximation from below to \(\sqrt{38439}\).

Since \(38439\) is a five-digit number, its square root is a three-digit number (why?) which we will write as \(abc\), where \(a,b,\mbox{ and } c\) are digits. Since \(a\) is in the hundreds place, we have that \(100s \times 100s = 10000s\). Thus, and reasoning in the same way as in the two-digit multiplication, we know that the carry from the thousands place can be at minimum zero, which means that we want the largest \(a\) such that \(a^{2} \le 3\). Therefore, \(a = 1\). So, our number now looks like $$1bc$$

Multiplying what we have, we know that we have a deficit of \(28439\) to cover (\(38439 – 10000\)). Therefore, \(2\cdot 100\cdot (10\cdot b) \leq 28439\), which implies that \(b = 9\) (since \(b\) is a digit and maxes out at \(9\) in base 10). And our number now looks like $$19c$$

Incorporating the newly found digit we have added another \(26100\) (\(2 \times 100 \times 90 + 90^{2} = 26100\)). Thus, we have a deficit of \(2339\) to cover as best as possible with \(c\). Therefore, we want the largest digit \(c\) such that \(2\cdot 100 \cdot c + 2\cdot 90 \cdot c +c^{2}\le 2339\) (note we can ignore the \(90^{2}\) part since we already subtracted out its effect). With a little bit of Algebra (or if your students haven’t had Algebra yet, then trial error starting at \(c = 0\) incrementing by one), we find that \(c = 6\) and our number is $$196$$ Indeed \(196^{2} = 38416 < 38439\) while \(197^{2} = 38809 > 38439\).

At the core of this, all that is happening is a step-by-step unwinding of $$(100a + 10b + c)^{2}$$
where we have
$$\begin{align}
(100a + 10b + c)^{2} & = 10000a^{2} + 2000ab + 100b^{2} + 200ac + 20bc + c^{2}\\
& \le 38439
\end{align}$$

We can see immediately that \(a = 1\) has to be the only choice giving

$$\begin{align}
(100 + 10b + c)^{2} & = 10000 + 2000b + 100b^{2} + 200c + 20bc + c^{2}\\
& \le 38439\\
\end{align}$$
yielding
$$\begin{align}
2000b + 100b^{2} + 200c + 20bc + c^{2} & \le 28439\\
\end{align}$$

which then implies that \(b = 9\) is the best choice, and so forth.

I showed a buddy of mine, who is a math teacher, this way of computing square roots and he put me to the test. He mashed on the calculator and gave me a 14-digit number to square root. It took about 20 minutes of furious arithmetic, but I was able to get the best lower bound approximation in the integers!

I have briefly shown this to students because I can’t burn that much class time as it takes very careful and repeated explanation. But those students who are really curious come by after class for a more detailed explanation.

If you want to show this to students, I would recommend having them futz around with two-digit squares for a while. Three-digit squares can be a bit tedious to work out, but are still a fun exercise for the enterprising student. For students of Algebra or beyond, this is a fun exercise to keep the arithmetic and expansion skills sharp as well as a reminder of how place value works. For students new to multiplication, I would say, hold off on this until they’re super comfortable with multiplication and place value; and even then, stick to finding two-digit squares. It’s a good point of discussion for place value and can be a fun math puzzle regardless.

As for utility, it is limited to none. But who cares. This is for curiosity.