This section of our page will serve as our forum to transmit any mathematical concept that doesn't fit in with the other sections (Logic and Math History). However, at the time of this page's creation we were not sure how much of these concepts people would pick up. So, here is our teaching of a basic algebraic concept that will be useful in many situations. Please use the link at the bottom of the page to e-mail us, and tell us what you think. We will use any recommendations that are provided in the construction of further sections of this page.

Basic Algebra

This section deals with a basic form of algebra. Here, you are shown how to solve equations with one variable (x), using functions, or mappings of x. In further editions of this section, we may explain in detail what a function entails, but for now it is sufficient to say that a function of the variable x takes x and does something with it. It may be seen as a sort of machine--when you put something in you always get something out, but what that is is decided by what kind of machine you are using.

In this section, we will be using functions that everyone must be acquainted with--addition, subtraction, multiplication, and division. To denote the function x+4, you would write [+4]x (read plus 4 x). The function x /9 would be written [/9]x ( read divided by 9 x). Using this method, you can also perform multiple operations on x, which is where the fun begins. In this case, the innermost function would be resolved first, then the next, and the next, etc. For example, the expression:

"([*8][+5][-9][/3]x)"

would be read, and performed "x divided by 3, minus 9, plus five, times 8". Any number of functions can be stacked upon the end of an x, indicating any number of operations upon the variable.

Before we get to solving these equations, another concept must be explained. The way in which the equations are solved involves working backwards, and applying the inverse of each function that you applied to get to x. Firstly, what is an inverse? Well, basically, an inverse is an opposite. There are two kinds of inverses that we must deal with here--the additive inverse, and the multiplicative inverse. The additive inverse takes a positive number or function and makes it negative, while it takes a negative number or function and makes it positive. Therefore, the additive inverse of 4 would be -4, and the additive inverse of [+8] would be [-8]. The multiplicative inverse takes a number and changes it into its reciprocal. So, the multiplicative inverse of 4 would be 1/4 (since 4 could be written as 4/1), and the multiplicative inverse of the function [*5] would be [*1/5].

Now that you have that down, there are two other functions that we must deal with: [Inv.] (read inverse times) and [Inv+] (read inverse plus). But first, a few things must be explained. As you probably know, the operations addition, and multiplication, are commutative. That means that x+4 would be the same as 4+x. But subtraction and division are not commutative, so what would happen if the function was, say, 4-x. That is where [Inv+] comes in. [Inv+] takes a function, say 5-x, and negates x, and then inverts the operation. Now, this is probably pretty confusing, so it is best illustrated in a diagram:

[Inv.] does the same thing, except that it takes the reciprocal of x, and then multiplies it by the original number in the function. It looks like this:

OK. Now that you know all about these functions, it's time to start solving problems. Algebraic equations with one variable are solved most easily by using arrow diagrams. These depict the functions that are applied to x to get an answer. Then they show what the inverse of each of the functions are. These are then applied to an answer to get back to x. Technique: First, you list the functions in the order they occur (closer to x = first). Then you work backwards, applying the inverses of each function, starting with the last one applied to x. For these purposes, the four regular operations are performed as normal, and [Inv.] and [Inv+] make the reciprocal and negation of number, respectively. The last number you get is your answer. To check your work, you can plug that value into the original equation, and see if it works out to the answer.