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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:squareRoots.gif [1.1]

We wish to find:
squareRoots2.gif [1.2]

First we note we can write down a on the answer line, the first bit of the square roots
squareRoots4.gif
squareRoots3.gif
squareRoots5.gif
          squareRoots6.gif

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)
squareRoots4.gif
squareRoots3.gif
squareRoots5.gif
squareRoots7.gif           squareRoots6.gif

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 squareRoots10.gif 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.

squareRoots8.gifRoot ...Answer so far
1+xFind root of 1+x
1a2=11
squareRoots9.gif    x2a+b
b=x/2
1+x/2
x+x2/4(2ab+b2)
2+x-x2/8 -x2/4Remainder.  
2a+b
b=-x2/8
1+x/2-x2/8
-x2/4-x3/8+x4/64(2ab+b2)
2+x-x2/4+x3/16x3/8+x4/64Remainder.
2a+b,
b=x3/16
1+x/2-x2/8+x3/16...

The square root of (1+x) to 10 terms is:
squareRoot(1+x).gif [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.

425AnswerAnswer so far
18 06 25
16Find a, so a2<=18
a=4
4
82206Remainder, bring down next pair (06).
2a+b
Seek b, so 20ab+b2≤206
This is approximately 206/80
b=2
42
16420ab+b2=164
8454225Remainder
Bring down next pair of numbers (25)
2a+b
Seek b, so 20ab+b2≤4225
This is approximately 4225/840
b=5
425
4225
0We 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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