Ken Ward's Mathematics Pages

Arithmetic Algebra: Computing Square Roots

Main Arithmetic Algebra Page

Page Contents

1. Rationale
2. Application to Algebra
3. Square Roots of Numbers

Rationale

By examinining a known relationship in algebra, we might hope to figure out a process. Actually, a process we can generalise to other roots. We know: [1.1]

We wish to find:
[1.2]

First we note we can write down a on the answer line, the first bit of the square roots

Bringing down a², we subtract it from the original and we are left with 2ab+b²

We now figure out how the next number arises (we know it is b)

We need to double the answer so far (2a) and add a number, b, such that when this is multiplied by b, it is less than or equal to the remainder.

If necessary, we could use (a+b+c)² to ensure we understand the process.

We can note the following:

1. Find the first part of the square root (a)
2. Double the answer so far (2a) and bring it down.
3. Find a number, b, so when added to 2a, and the whole multiplied by b: (2a+b)·b is equal to the remainder (or just less)
4. If 2a+b equals the remainder, we have found the root
5. If there is a remainder, we repeat the process from step 2, bringing down 2(a+b) and seeking a new b

Application to Algebra

As an example, we seek to find using the square root radical algorithm above. The a's and b's below relate to the rationale above, and are indended as reminders and explanations.

		Root ...	Answer so far
	1+x	Find root of 1+x
	1	a2=1	1
	x	2a+b b=x/2	1+x/2
	x+x2/4	(2ab+b2)
2+x-x2/8	-x2/4	Remainder.   2a+b b=-x2/8	1+x/2-x2/8
	-x2/4-x3/8+x4/64	(2ab+b2)
2+x-x2/4+x3/16	x3/8+x4/64	Remainder. 2a+b, b=x3/16	1+x/2-x2/8+x3/16...

The square root of (1+x) to 10 terms is:
[2.1]

Square Roots of Numbers

Strangely, this is taught in Baby School, even though one can go through a lifetime's career in mathematical work without ever using it, or thinking of it, again!
In algebra, we seek a number b such that 2ab+b² is the largest number less than or equal to our remainder. In arithmetic, we change this formula to finding a b such that 20ab+b² fits our criterion. And the a's and b's are numbers less than 10. This is only important when considering the algebra, as it happens automatically in the arithmetic algorithm (when you, say, add 1 to 2 to get 21, you have effectively multiplied the number by 10, and doubled it when bringing it down).

Also, in arithmetic, we divide our number in two-digit groups from the units figure (on the principle that any number between 0 and 10, exclusive, squared is a number less than 100)

We seek the square root of 180625.

	425	Answer	Answer so far
	18 06 25
	16	Find a, so a2<=18 a=4	4
82	206	Remainder, bring down next pair (06). 2a+b Seek b, so 20ab+b2≤206 This is approximately 206/80 b=2	42
	164	20ab+b2=164
845	4225	Remainder Bring down next pair of numbers (25) 2a+b Seek b, so 20ab+b2≤4225 This is approximately 4225/840 b=5	425
	4225
	0		We are done. The answer is 425 exactly

Of course, in arithmetic, we use simply 2 and not 20, but when exploring arithmetic algebraically, taking a as a number between 0 and 10, we need to use 20.

When seeking b, we can assume it is small compared with 20a, and try dividing 20a into the remainder. If we are confused, we can figure out, for instance:
20ab+b²≤4225, or (20a+b)b≤4225 where a=42













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Faster Arithmetic - by Ken Ward