I chose this topic to present in my “History of the K-16 Mathematics Curriculum” course, because I remember several teachers and professors mentioning that square root extraction had been taught as a standard topic up until a few years ago. Ever since, I’ve been curious how you can do so! I found that even though we do not need to find square roots manually anymore, a problem like this actually can connect ideas of length, area, trinomials, and number sense.

So, I looked back at two Arithmetic textbooks: Pike’s Arithmetic (1788) and Davies’ Arithmetic (1851). I chose these textbooks because Pike’s was rule-based and his stated goal was to present the rules, and use better illustrations and applications than had previously been available. Davies’ Arithmetic was published half a century later, after the “induction-synthetic” tug-of-war. Davies’ was still (or again?) rule-based, but the presentation method changed to: start with an overview, give definition, explain the propositions, give questions for students to answer mentally / orally, and to show connection of principles throughout each section. I felt that the biggest change between the two would be that Davies stressed he would explain the rules visually or concretely.

Over all, each textbook had some good and some bad: Davies did spend more time explaining to students what was happening in the rules, but Pike provided many more examples in a number of different fields. For example, above are three examples that require extraction of roots: arranging soldiers on a battle field, planting an orchard, and finding the diameter of a cistern pipe.

**Pike’s rules for extraction of the square root, illustrating with the example 54756.**

- Rule 1 – Distinguish the given number into periods of two figure each, by putting a point over the place of units, another over the place of hundreds, and so on, which points shew the number of figures the root will consist of.
- Place a dot over the 6 (units place), 7 (hundreds place), and 5 (the root will be three figures):

- Place a dot over the 6 (units place), 7 (hundreds place), and 5 (the root will be three figures):
- Find the greatest square number in the first, or left hand period, place the root of it at the right hand of the given number, (after the manner of a quotient in division) for the first figure of the root, and the square number, under the period, and subtract it therefrom, and to the remainder bring down the next period for a dividend.
- The greatest square number in 5 (left-most digit) is 4. The root of 4 is 2, so place the digit 2 on the right of 54756 (similar to division).

- The greatest square number in 5 (left-most digit) is 4. The root of 4 is 2, so place the digit 2 on the right of 54756 (similar to division).

- 5-4 = 1
- Bring down the next period for a dividend, means to bring down 47, so 147 is the new dividend.

- Place the double of the root, already found, on the left hand of the dividend for a divisor.
- The double of 2 is 4, so we divide 147 by 40. But – (in the next step) we find that 40 goes evenly into 147 three times, so the divisor will become 43.

- The double of 2 is 4, so we divide 147 by 40. But – (in the next step) we find that 40 goes evenly into 147 three times, so the divisor will become 43.
- Seek how often the divisor is contained in the dividend, (except the right hand figure) and place the answer in the root for the second figure of it, and likewise on the right hand of the divisor: Multiply the divisor, with the figure last annexed, by the figure last placed in the root, and subtract the product from the dividend: To the remainder join the next period for a new dividend.
- So, right now the divisor is 40 and the dividend is 147. 40 divides into 147, 3 times.
- So “3” is the second digit of the root, and we add “3” to the divisor 40 – that is, the new divisor is 43
- Now, we multiply 43 * 3 = 129, and subtract 147 – 129 = 18
- We bring down the next period (56) and append that to 18, so the new dividend is 1856

- Double the figure already found in the root, for a new divisor, (or bring down your last divisor for a new one, doubling the right hand figure of it) and from these find the next figure in the root as last directed; and continue the operation in the same manner, till you have brought down all the periods.
- The last divisor was 43, and we need to double the “right hand figure” which is “3”, so we have 460 as the new divisor.
- 460 divides into 1856, 4 times.
- So “4” is the third digit of the root, and we add “4” to the divisor 460 – that is, the new divisor is 464.
- Now, we multiply 464 * 4 = 1856.
- So the square root of 54756 is 234.

That wasn’t too bad, right? It’s difficult to see why Pike’s Rules work, but it does make sense if you follow the reasoning using some algebra.

Notice above that Davies did not simply give his rules, but included a visual (square), some thinking questions (*What is the greatest square of a single figure? In what places of figures will the square of the tens be found?*), and an overall explanation: *The square of two figures is equal to the square of the tens, plus twice the product of the tens by the units, plus the square of the units. * So, if a two-digit number is ab, where a is the first digit and b is the second (not multiplied but digits of a number), then (ab)^2 = (a0)^2 + 2*a0*b + (b)^2.

- Point off the given number into periods of two figures each, counted from the right, by setting a dot over the place of units, another over the place of hundreds, and so on. [Similar to Pike’s so far.]
- Find the greatest square in the first period on the left, and place its root on the right after the manner of a quotient in division. Subtract the square of the root from the first period, and to the remainder bring down the second period for a dividend.
- Double the root already found and place it on the left for a divisor. Seek how many times the divisor is contained in the dividend, exclusive of the right-hand figure, and place the figure in the root and also at the right of the divisor.
- Multiply the divisor, thus augmented by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. But if the product should exceed the dividend, diminish the last figure of the root.
- Double the whole root already found, for a new divisor, and continue the operation as before, until all the periods are brought down.

Davies illustrates the meaning of his rules by providing a visual of a perfect square, in this case, illustrating that the square root of 1296 is 36.

- That is, first the student would find the largest square less than 12, which is 9, so start with 900 in the large square.
- But the square root of 900 is just 30. What’s left? 1296 – 900 = 396.
- From the visual, we can see that we’ll have two rectangles with one length 30 and the other is unknown, but the same value for each. So multiple 30*2 = 60. So how many times can 60 go evenly into 496? 6 times.
- Multiply 60*6 = 360, so the two side rectangles have area 180, and all that is left is the 6*6 = 36.
- So the square root of 1296 = 900 + 2*180 + 6^2 = 36^2.

Recall from above that, (36)^2 = (30)^2 + 2*30*6 + (6)^2 = 900 + 2*180 + 36.

Let’s see if Davies’ visual can help us make sense of Pike’s process with 54756. First of all, why put dots over the digits?

From Pike, we know the dots divide the number into “periods” of 56, 47, and 5 (moving from right to left). Another way to think of the periods is as the numbers 50000, 4700, and 56. The first number, 50000 gives us a way to find the area and side lengths of the largest square:

- So, even though we do not need to find square roots manually anymore, a problem like this actually can connect ideas of length, area, trinomials, and number sense. Remembering that we are anticipating that the root of a five digit number must be three digits, say
*a*represents the hundreds digit,*b*represents the tens digit, and*c*represents the units digit. Then the root of 54756 is*(a+b+c)*. - So we find
*a*by asking: What is the largest perfect square less than 5? 4 is, so we have the area of the dark blue square is*a*^2 = 40000 = (200)^2, so*a*= 200. Subtracting off the area of this square from the area of the whole square, we have 54,756-40000=14,756. - We’ll find
*b*by asking: How many times does the double of*a*(400) go into 14,756? - So, we divide 14,756 by 400.

But, remembering that*b*represents the tens place of our final answer, it will be some digit followed by a zero. So, even though 400 goes into 14,756 evenly 36 times, we’ll round down to*b*= 30. Now we find the size of the square consisting of*a*^2 + 2**a***b*+*b*^2 = 40000 + 12000 + 900 = 52,900. So 54,756 – 40,000 = 14,756. And 14,756 – 12000 – 900 = 1856. - Finally, find
*c*by asking: How many times does the double of*(a + b)*, that is, 2*(200 + 30) = 460, go into 1856? 460 goes into 1856 evenly 4 times, so*c*= 4. And our final answer is 234.

Note that each time we have simply found the missing length of a rectangle, and found thus that:

(*a + b + c*)^2 =*a*^2 + 2**a*b*+*b*^2*+ 2**(*a + b*)**c*+*c*^2. So, even though we do not need to find square roots manually anymore, a problem like this actually can connect ideas of length, area, trinomials, and number sense.