# Intermediate Algebra/Algebraic Axioms

This section is unfinished.

In mathematics, you built up a series of rules and concepts to solve for equations. At most, they were series of equations, but nevertheless, an application of some rules to get something out. In higher mathematics, you start by unraveling these built up rules and notions - focusing on reconstructing them with what is know and what can be assessed. This is the start of that journey.

## Establishing the Axioms

When you begin the task of proving everything through other proofs, you will come across a problem - you can't prove everything without needing to simply end it somewhere. What is this fundamental point of beginning called? An axiom.

Axioms can be thought of as rules that require no proof. Focusing on algebra, we can start from these initial axioms. Note that the variables ${\displaystyle a,b,c}$ are being treated as nothing more than numbers.

 Associative Law ${\displaystyle (a+b)+c=a+(b+c)}$ Commutative Law ${\displaystyle a+b=b+a}$ Identity Law ${\displaystyle a+0=a}$ Inverse Law ${\displaystyle a+(-a)=0}$

These axioms suppose some symbols to mean specific things. ${\displaystyle 0}$, although it has always been treated as a number in algebra, is here treated as the addition identity. ${\displaystyle -a}$ likewise, although treated as a negation of a number in algebra, is here being treated as the inverse of ${\displaystyle a}$. In a sense, they have been robbed on the meaning we commonly associated with them.

Here's another facet of these axioms: they all work with each other. This is supposed to mean that each law does not contradict with one another. Observe that, for example, you can create a new association through the use of the commutation law and association law.

{\displaystyle {\begin{aligned}(a+b)+c&=a+(b+c)\\&=a+(c+b)\\&=(a+c)+b\end{aligned}}}

From here, you can start building some properties of algebra that might have been considered fundamental, for example ${\displaystyle a+x=a\Rightarrow x=0}$ - or the idea that adding some number to any number that results in no change must mean that the number you added is 0.

{\displaystyle {\begin{aligned}a+x&=a\\a+(-a)+x&=a+(-a)\\0+x&=0\\x&=0\end{aligned}}}

This proof required all four axioms.

However, algebra with only addition would be easy, so let's add another set of axioms for another algebraic operator. The next one on the list that you might have thought was subtraction, since they're paired together in algebra after all. But, we can do better by using operations without new axioms - after all, subtraction obeys the association, identity, and inverse law for addition anyways. How? Try out some numbers; this is left as an exercise for you to experiment with.

About the commutative law; well it's intuitively not allowed, right? Try out some numbers. Unless ${\displaystyle a=b}$, it won't work. Well, this can actually be inferred, much like how adding some number to any number that results in no change must mean that the number you added is 0. It requires multiplication though, specifically redefining the laws of addition for multiplication.

 Associative Law ${\displaystyle (a\cdot b)\cdot c=a\cdot (b\cdot c)}$ Commutative Law ${\displaystyle a\cdot b=b\cdot a}$ Identity Law ${\displaystyle a\cdot 1=a}$ Inverse Law ${\displaystyle a\cdot a^{-1}=1}$

We will also add an axiom that connects both axioms together.

 Distributive Law ${\displaystyle a\cdot (b+c)=ab+ac}$ "Equality Law" ${\displaystyle 0\neq 1}$

Now, how can using multiplication and addition justify the subtraction axiom? Easy! Apply the distributive property over it, after adding stuff all over it.

{\displaystyle {\begin{aligned}a-b&=b-a\\a-b+a+b&=b-a+a+b\\a+a+b-b&=b+(-a+a)+b\\a+a&=b+b\\a\cdot (1+1)&=b\cdot (1+1)\\a&=b\end{aligned}}}

Now, here's a nifty rule that comes from including multiplication - you can prove that negative numbers multiplied by negative numbers equals the positive counterpart. This deserves further explanation.

First, we will prove the easier statement,

{\displaystyle {\begin{aligned}(a+b)+c&=a+(b+c)\\&=a+(c+b)\\&=(a+c)+b\end{aligned}}}