# The Addition and Subtraction Principles

Recall that if $A$ is a set and $A_1, A_2, ..., A_n \subset A$ such that $A = \bigcup_{i=1}^{n} A_i$ and for $j \neq k$ we have that $A_j \cap A_k = \emptyset$ for each $j, k = 1, 2, ..., n$ then the collection of sets $A_1, A_2, ..., A_n$ form a partition of $A$

We are now ready to look at two relatively simple principles known as the Addition principle and Subtraction principle which seem rather obvious, but we still go through the process of definition them below.

 Definition (The Addition Principle): If $A_1, A_2, ..., A_n \subset A$ form a partition of the finite set $A$ then the size of $A$ is equal to the sum of the sizes of $A_1, A_2, ..., A_n$, that is $\lvert A \rvert = \lvert A_1 \rvert + \lvert A_2 \rvert + ... + \lvert A_n \rvert = \sum_{i=1}^{n} \lvert A_i \rvert$.

For example, consider the set $A = \{ x, y, z \}$. If $A_1, A_2 \subset A$ are defined by $A_1 = \{ x \}$ and $A_2 = \{ y, z \}$. Clearly $A_1$ and $A_2$ form a partition of $A$ and we can clearly see that:

(1)
\begin{align} \quad \lvert A \rvert = 3 \end{align}

Furthermore:

(2)
\begin{align} \quad \lvert A_1 \rvert + \lvert A_2 \rvert = 1 + 2 = 3 \end{align}

Therefore we have that $\lvert A \rvert = \lvert A_1 \rvert + \lvert A_2 \rvert$.

## The Subtraction Principle

Recall that if $A$ is a set and $B \subset A$ then the set theoretic difference of $A$ minus $B$ is defined as:

(3)
\begin{align} \quad A \setminus B = \{ x : x \in A \: \mathrm{and} \: x \not \in B \} \end{align}

We are now ready to define the subtraction principle.

 Definition (The Subtraction Principle): If $A$ is a finite set and $B \subseteq A$ then the number of elements in $A \setminus B$ is equal to the number of elements in set $A$ minus the number of elements in set $B$, that is $\lvert A \setminus B \rvert = \lvert A \rvert - \lvert B \rvert$.

Note that if $A$ is a finite set and $B \subseteq A$ then $B$ is by extension a finite set too and so the formula given in the definition above is well-defined.

Consider the sets $A$ and $B \subset A$ defined as:

(4)
\begin{align} \quad A = \{ x : x \in \mathbb{Z} \: \mathrm{and} \: 1 \leq x \leq 100 \} \end{align}

and:

(5)
\begin{align} \quad B = \{ x : x \in \mathbb{Z} \: \mathrm{and} \: 51 \leq x \leq 100 \} \end{align}

We have that $\lvert A \rvert = 100$ and $\lvert B \rvert = 50$. Furthermore we note that:

(6)
\begin{align} \quad A \setminus B = \{ x : x \in \mathbb{Z} \: \mathrm{and} 1 \leq x \leq 50 \} \end{align}

Therefore $\lvert A \setminus B \rvert = 50$. As we can see, $\lvert A \setminus B \rvert = \lvert A \rvert - \lvert B \rvert$.