The Algebra of Number Patterns: A Simple Problem.

Take any number, and multiply it by itself e.g. for the number six, 6 × 6 = 36. Now reduce the original number by one i.e. here to 5; then increase the original number by one i.e. to 7; and find the product of these two new numbers: 5 × 7 = 35. If we repeat this procedure for different numbers, as shown below, we notice that each set of answers (in each column) are alike in some way...

We notice that the answers on the bottom line are one less than the corresponding answers on the top line. This is the case for any number manipulated in this way. If we now adjust the top line answers by taking one away from them, they are now equal to the answers on the bottom line. This relationship can be shown as follows for the numbers one to four:

(1)(1)-1 = 2(0); (2)(2)-1 = 3(1); (3)(3)-1 = 4(2); (4)(4)-1 = 5(3).

If we represent the number we originally choose to manipulate by 'x', the relationship is denoted by (x)(x)-1 = (x+1)(x-1), or x²-1 = (x+1)(x-1). The expression in red is often used in algebra and is called "the difference of two squares". It is one method of solving a quadratic equation.

The purpose of the above demonstration is to show that examining number relationships can lead to useful algebraic expressions. In many cases a complicated expression originates from a person noticing different number patterns.

Changing Operations

Now consider the following four number patterns:

(11)×3 = 11+3 = 41; (12)×4 = 12+4 = 52; (13)×5 = 13+5 = 63; (14)×6 = 14+6 = 74

It would be possible to conclude from the above that multiplication and addition are identical operators, for you have the same answer whether you multiply or add. Of course this isn't true in general. But how is it true for the above, which can be represented by the following:

Simple algebraic steps show that the three expressions above, which are separated by the equal signs, are identical.

This numerical pattern expresses a special relationship. It is saying that you can start with a mixed number of the form (1 + (1/n)), multiply it by the quantity n+1, or add it to the quantity n+1, and you will get the same result.

The Three-Digit Reverse and Subtraction

"Take any three digit number, with no two digits the same, such as 532. Reverse the digits: 235. Then subtract the smaller from the larger: 532 - 235 = 297. If you repeat the steps described with any three-digit number, you will find that in every answer the middle digit is 9"

Here is the algebraic proof for the above. Our starting number can be represented by 'abc', where a is the hundreds digit, b the tens digit, and c the units digit. Our starting number can be written as

100a + 10b + c.

If we reverse the digits, the new number can be written as

100c + 10b + a.

Assuming a is larger than c, we now subtract the smaller from the larger:

(100a+10b+c) - (100c+10b+a) = 100a - 100c + c - a

= 100(a - c) + c - a

= 100(a - c) - (a - c)

It looks as if (a - c) represents our new hundreds digit and - (a - c) represents our new units digit. But - (a - c) represents a negative number (we assumed a is greater than c), and it's not satisfactory to write a positive number by using a negative units digit. We alter the situation by adding 100 and subtracting 100 in the expression obtained above:

100(a - c) - (a - c) = 100(a - c) - 100 + 100 - (a - c)

Rearranging the above to obtain a hundreds digit, a tens digit, and a units digit,

100(a - c - 1) + 9(10) + [10 - (a - c)].

We now have our result in a desirable form. We can see that the tens digit is always 9, and the sum of the hundreds and units digits is always 9:

(a - c - 1) + [10 - (a - c)] = 10 - 1 = 9.