Traps and pitfalls in elementary algebraic manipulation

Question

Given the equation $x=x+2$: \begin{align} x &=x+2 \\\\ x-x &=2 \\\\ 0&=2 \end{align}

This impies there is no solution

Given the equation $x+1=0$: \begin{align} x + 1 &=0 \\\\ \frac{x+1}{x+1} &= \frac{0}{x+1} \\\\ 1&=0 \end{align} Clearly, the conclusion that there is no solution is invalid. Is there a set of algebraic manipulation dos and don'ts to avoid such errors?

Answer 1

First case:

You write:

$x=x+2 \rightarrow x-x=2$

Algebraically, you added $-x$ to both sides. The operation is correct. However, the starting point was not correct. I mean what does $x=x+2$ tell us? It says that there is a quantity equals itself plus 2. This statement is illogic. Hence, the problem is in applying the algebraic operation to an expression that is not logical. Such cases appear when proving some statements by supposing the opposite argument and proceed with algebraic operations to stop when such an illogic expression appears meaning that the assumption was false.

Second Case:

You write:

$x+1 = 0$

$\frac{x+1}{x+1} = \frac{0}{x+1} \rightarrow 1=0$

Division operation is undefined or illegal operation (well, maybe with one exception...see below) when you divide by explicit zero figure or an expression that is zero. This is the "defensive" approach to obtaining logical results that we were taught in school. One way to see why this is so, when you consider the relationship between division and subtraction. For example, $\frac{6}{2}=3$ can be seen as the unique number of subtracting operations of $(6-2)$ until zero is reached. In case of division by zero, the number of operations is not unique it could be one, two, etc.

This restriction in division is adopted by all computer programming languages use (to my knowledge).

Violating this restriction could possibly lead to invalid expression, and performing operations on invalid expression leads to illogic results.

In cases, one may be interested in the value of the function $f(x)$ when the value very near, but not equal, zero. In such cases, we use the limit techniques.

You can read more about this here: Wiki-Division By Zero.

I am aware of one case (very famous) where division by zero expression looks-like it works. This is the case of the Remainder Theorem. When you have a polynomial like $p(x)=x^2-1$ and you know that $x+1$ is a factor bit not the last factor of p(x), you can write $p(x)=\frac{x^2-1}{x+1}=x-1$. This is a valid case! It is not easy to see why this is valid without some knowledge in Analysis. For more on this you may want to look at Introduction to Division by Zero Calculus.