(Originally written for my Intro to Homological Algebra Summer Seminar Group at SJC)

One of the first things you learn in Algebra, is how to manipulate equations.  Given the equation:

2x + 3 = 4,

we quickly learned how to subtract 3 from both sides of the equation.  In this post, I would like to go through the very formal definitions which make this possible.

The Definition of a Group and Why We Can Cancel

Recall that a group is a set, B, together with a binary operation B \times B \xrightarrow{\mu} B, which is associative, has a unique identity, and has unique inverses.  Before going forward, it is important to understand what it means to have a function, \mu, and what it means to have pairs in B \times B as its input.  For a quick review of some foundations for this post, check out the section Review: Sets, Cartesian Products, Equality, and Functionsbelow.

OK, once you feel like you understand Cartesian Products and functions, let’s say what addition is!  So addition on a set, B, by the definition of a group, is a binary operation, i.e. a function,

B \times B \xrightarrow{\mu} B

Let’s unpack this.  So for two elements, p and q in the set B, we have a way to add them together to produce a third element, call it y \in B:

\mu(p,q)=y

but we normally just write:

p+q=y

OK, but let’s look at what it means to be a function, and what this has to do with the whole point of this post!  Recall we started with the equation

2x + 3 = 4

Well suppose that the set, B is the set of all of our numbers.  Then 2x+3 and 4 are two elements in B, but that are equal.  Two other numbers that are equal are -3 and -3.  Therefore, we have two ways to write the same pair:

(2x+3, -3) and (4, -3)

Since their first components are equal and their second components are equal then we have

a = (2x+3, -3) = (4, -3)

Now, remember our function, \mu: B \times B \to B?  For the moment let’s call B \times B the set A, so that A = B \times B.  Well then being a function, \mu : A \to B,  means that  if (a, b_1) , (a,b_2) \in \mu then we must have that b_1 = b_2.  Note that for our setup, we have

b_1 = \mu(a) = (2x+3)+(-3) and b_2 = \mu(a) = (4) + (-3)

But since b_1 = b_2, then we must have that

(2x+3) + (-3) = (4) + (-3)

So now you finally know that the reason you can add to both sides is that addition is a function!!!  Anyway, now you could use the associative property of our group,

(2x)+(3 + (-3)) = 4 + -3

and then the additive inverse property,

(2x)+0 = 4 + -3

and now the identity property,

(2x) = 4 + -3

to finally see why we can cancel something from both sides of an equation.

Concept Check: You should feel free to move on from the ideas in this post when you can easily understand and show that, similarly, we can cancel the 4 being multiplied on the right hand side of the equation

2x + 3 = x \cdot 4

Review: Sets, Cartesian Products, Equality, and Functions

Assuming we have basic understanding of set theory, we need to first consider the set of all pairs of elements from a set A and a set B, which is called the Cartesian Product, and written A \times B.  More formally, we define the Cartesian Product by:

A \times B := \{ (a, b) \ \vert \ a \in A \text{ and } b \in B \}

Note that the only way two pairs (a,b) and (x,y) are allowed to be equal is if both a=x and b=y.  As it turns out, in order to say what a function is, we will need Cartesian products.

First we define a relation, which is simply a subset of a Cartesian Product.  That is to say, a relation, R, between A and B, is simply some subset

R \subset A \times B

For example, if I consider the set of all teams participating in the world cup this year, call it T, and the set of all locations of the stadiums being used in the world cup this year, call it L, then I can consider the relation of locations in which certain teams have played during the tournament this year as the subset A \subset T \times L.  So for some examples:

  • (Egypt, Ekaterinburg), (Egypt, St. Petersburg), (Senegal, Moscow), (Portugal, Sochi), and (Russia, Moscow) are all elements of A while
  • (Iran, Moscow) is not an element in A, since Iran did not play in Moscow during this World Cup (I looked it up).

Notice how for my relation, A, it allows for both the T-values and L-values to repeat (you would often call these the x-values and y-values, respectively, in your experience).  Since you were told presumably that for a function, the first component is not allowed to repeat, then you have just observed that not all relations are functions.  A function is a special type of relation, that is:

function f: A \to B is a relation, f \subset A \times B, in which if both (a,b) and (a, b') are in the set, f, then it must be that b = b'.

Note that usually you simply write an element (a,b) \in f as: f(a) =b.  But it’s important to realize that technically, the definition using the pairs is the more careful and formal one.

Concept Check: You should feel free to move on from this definition of a function using Cartesian Products if you can explain why this definition implies the vertical line test which we were all taught when we first learned to distinguish functions from non-functions.

OK, now you are ready to move back to the original discussion.

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