# Rules for Algebraic Manipulation

There's a good change you spent some time in school learning how to solve equations through algebraic manipulations such as "add x to both sides". These rules and the way they are taught hint at a formal process of mathematical proof, but they are not as rigorous. Algebraic maipulations as taught in school can often lead to false conclusions and "paradoxes". Nevertheless, they can be made rigorous with fairly simple modifications.

## Square Root

This one is probably pretty familiar to most readers, so we'll start with square root. Normally when one says, for example "the square root of 49" it is understood that they mean 7, but it is also known that there are really two square roots of 49: 7 and -7. However there aren't always two square roots: 0 only has a single square root, and negative numbers have none (we are restricting our discussion to real numbers, not imaginary or complex).

When doing algebra with equations, one may wish to take the square root of both sides and it's important to remember that there might be two square roots.

(x-3)2 = 1

Take the square root of both sides:

x-3 = 1

x = 4

This only gives one solution, but there are actually two: the original equation is also true when x=2. TO get both solutions, we use a modified rule for square root:

Taking the Square Root : The statement A=B can be replaced with the two linked statements:

√A = √BOR   √A = - √B

Applying this rule to solve (x-3)2=1 yields both solutions:

(x-3)2 = 1
take the square root:
(x-3) = 1   OR   (x-3) = -1
x = 4   ORx = 2

## Division

Here's a simple example of algebra rules causing problems. Suppose we are given this equation to solve for x:

3x = 2x

The normal way to solve this is to subtract 2x from both sides, yielding the answer x=0. However suppose instead you decide to divide both sides by x:

3x/x = 2x/x
3 = 2

There clearly needs to be an additional requirement before we can "divide both sides by x", and it's usually taught that you can divide both sides by something as long as it's not zero.

3x/x = 2x/x     [as long as x is not 0]

This can also be handled by splitting the original equation into two, as with the square root rule above:

3x/x = 2x/xORx = 0

This method is better, because it doesn't result in a dead-end:

3 = 2   ORx = 0
Since 3 is not equal to 2, the other one must be true,
.'.   x = 0

So the rule for division becomes

Division by an Expression : For any expression E, the statement A=B can be replaced with the two statements:

A/E=B/EORE=0

## Infinity Minus Infinity

Here's an example of the opposite effect, in which sometimes you can perform an algebraic manipulation that usually is not allowed. Consider the sums:

1/1 + 1/5 + 1/9 + 1/13 + 1/17 + ...
1/3 + 1/7 + 1/11 + 1/15 + 1/19 + ...

both of these add up to infinity, but if we take the first and subtract the second:

1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - ...

we have a perfectly legal, converging infinite sum whose value is π/4.

C = 1 - 1 + 1 - 1 + 1 - ...
= 1 - (1 - 1 + 1 - 1 + ...)
= 1 - C

therefore, by albegra, we can add C to both sides to get

2C = 1

which gives C = 1/2. This is the simplest example of Cesaro summation, which has some useful applications.

Another Example

Derivation of 1 + k + k2 + k3 + ... for some constant K. Assume the sum has a value and call that value C.

C = 1 + k + k2 + k3 + ...

Multiply through by kL

Ck =     k + k2 + k3 + ...

Subtract the second from the first:

C - Ck = 1
C(1-k) = 1
C = 1/(1-k)   OR   1-k=0

If k is something like 1/2, it's easy to see that the sum 1 + 1/2 + 1/4 + 1/8 + ... is 2, and 1/(1-k) = 1/(1-1/2) is also 2, so the derivation is valid. But if k were, say, -2, then we'd get:

1 - 2 + 4 - 8 + 16 - ... = 1/3

There clearly needs to be an additional requirement before we can "subtract Ck from C". The problem has to do with the fact that in this example the infinite series sum (and therefore C and Ck) is infinite.

When this process is used in reverse, we can get even more "interesting" results. See the -1/12 article on my numbers pages for an example.

### Everybody is the Same Height

Let A and B be the heights of two people, and let M be the mean (average) of their heights. We will prove that A = B.

M = (A+B)/2
multiply by 2 :
2M = A+B
solve for A and for B :
A = 2M - B   and   B = 2M - A
reverse the order and change the signs of the second one :
A = 2M - B   and   A - 2M = -B
multiply them together :
A2 - 2AM = B2 - 2BM
add M2 to both sides :
A2 - 2AM + M2 = B2 - 2BM + M2
factor :
(A-M)2 = (B-M)2
take the square root :
A - M = B - M
A = B

Here the mistake is in the square root step, and is resolved easily. Everything is correct up to and including (A+M)2=(B+M)2. The corrected derivation proceeds:

(A-M)2 = (B-M)2
take the square root :
A-M = B-MORA-M = -(B-M)
add M to the first equation :
A = BORA-M = M-B

Now the conclusion is correct: either the two heights A and B are the same, or they are equally spaced from their mean M.

(This paradox is also called "An Elephant Weighs the Same as a Mosquito".)

### Proof That 1 Equals 2

This proof starts with any two equal values A and B:

A = B
multiply by A :
A2 = AB
subtract B2 :
A2 - B2 = AB - B2
factor :
(A+B) (A-B) = B (A-B)
divide by (A-B) :
A + B = B
substitute A=B (from step 1) :
B + B = B
simplify :
2B = B
divide by B :
2 = 1

The paradox appears after the divide step, and is resolved by using the division deduction rule. Starting with the step just before we divide:

(A+B) (A-B) = B (A-B)
divide by (A-B) :
A+B = BORA-B = 0
substitute A=B (from step 1) :
B+B = BORB-B = 0
simplify :
2B = BOR   0 = 0
divide by B :
2 = 1   OR   0 = 0

Once the division step is dome properly, the result is valid because although "1=2" is false, "0=0" is true so the overall result remains true.

mrob27
This page was written in the "embarrassingly readable" markup language RHTF, and was last updated on 2018 Feb 04. s.11