April 2018 – Mathemafrica

0.4. Cartesian product

We know we can use binary operations to add two numbers, x and y: $x+y, x-y, x \times y, x \div y.$ Furthermore there are other operations such as $\sqrt{x}$ or any other root and exponents. Operations can involve other mathematical objects other than numbers, such as sets.

$def^n$ Given two sets, A and B, we can define multiplication of these two sets as the Cartesian product. The new set is defined as

$A \times B = \{(a,b): a \in A, b \in B\}$

Before looking at abstract examples, consider this case:

e.g.1. Assume there is a student in a self-catering residence and they want to make food preps for the first four days in the week. They want to know how many possible combinations they can make using fruits (between grapes and apples) and meals (pasta and meatballs, chicken wrap).

To solve this, let $A = \{ \text{ grapes, apples } \} \text{ and } B = \{ \text{pasta and meatballs, chicken wrap} \}$

Then the possible meal options are: (grapes, pasta and meatballs), (grapes, chicken wrap), (apples, pasta and meat balls) and (apples, chicken wrap).

The Cartesian Product of sets A and B would be:

$\text{A x B} = \{( \text{ grapes, pasta and meatballs}), (\text{ grapes, chicken wrap }), (\text{ apples, pasta and meat balls }), (\text{ apples, chicken wrap}) \}$

We can think of the above example in more abstract terms.…

By Yolisa| April 14th, 2018|Uncategorized|3 Comments

0.3. power sets

Recall powers (or exponents) of numbers: $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$

Similarly, sets have the power operation to create new sets.

$def^n$ If A is a set, then the power set of A is another set denoted as

$\mathbb{ P }(A) = \text{ set of all subsets of A } = \{ x: x \subseteq A \}$

Recall: A is a subset of B if every element in A is also in B. Furthermore, if A is a finite set with n-elements, then we can find the number of subsets in A by using this formula:

$2^n$

To find the power set of A, we write a list of all the subsets of A first – remembering that:

the empty set is a subset of every set,
and every set is a subset of itself

Let’s look at some examples:

e.g.1. $A = \{1, 2, 3 \}$

Using the formula $2^n$ , we know that there are $2^3 = 8$ possible subsets of A, namely:

$\varnothing, \{1, 2, 3 \}, \{1 \}, \{2 \}, \{3 \}, \{1, 2 \}, \{2, 3 \} \text{ and } \{1, 3 \}$

Hence the power set is the set that contains all the above subsets:

$\mathbb{ P }(A) = \{ \varnothing, \{1, 2, 3 \}, \{1 \}, \{2 \}, \{3 \}, \{1, 2 \}, \{2, 3 \}, \{1, 3 \} \}$

Note: The cardinality (size) of $\mathbb{ P }(A) = 8 = 2^3$ where size of A= 3 elements

e.g.2.

…

By Yolisa| April 10th, 2018|Uncategorized|1 Comment

https://giphy.com/gifs/infinite-boxes-vG1Dgq3JRXLMc

Consider a set $A = \{2, 3, 7\} \text{ and } B = \{2, 3, 4, 5, 6, 7\}.$ Note that every element in set A is also found in set B, however, the reverse is not true (B contains elements 4, 5 and 6 which are not in A)

Consider another case, $A = \{2n: n \in \mathbb{ N }\} = \{ 0, 2, 4, 6, ... \} \text { and } B = \mathbb{Z} = \{ ..., -2, -1, 0, 1, 2, ... \}.$ Again, we can see that every element in set A is also found in set B and similarly, everything in B cannot be found in set A. B contains negative and odd integers, which are not in A.

To describe this phenomena, mathematicians defined subsets:

$def^n$ Suppose A and B are sets. If every element in A is an element of B, then A is a subset of B and we denote this as $A \subseteq B$

If B is not a subset of A, as in the above cases, then there exists at least one element, say $x \in B \text{ such that } x \notin A. \text{ We denote this as } B \subsetneq A$

e.g.1. $\{2, 3, 5, 7, ... \} \subseteq \mathbb{ N }$ but $\{\frac{1}{3}, 2, 5, 7, ... \} \subsetneq \mathbb{ N }$ since $\frac{1}{3} \in \mathbb{ Q }$

e.g.2. $\mathbb { N } \subseteq \mathbb { Z } \subseteq \mathbb{ Q } \subseteq \mathbb{ R }$

e.g.3. $(\mathbb{ R } \times \mathbb{ N }) \subseteq (\mathbb{ R } \times \mathbb{ R })$ since $(\mathbb{ R } \times \mathbb{ N }) = \{(x, y): x \in \mathbb{ R }, y \in \mathbb{ N }\}$ and $( \mathbb{ R } \times \mathbb{ R }) = \{ (x, y): x \in \mathbb{ R }, y \in \mathbb{ R } \}$ Hint: look at what sets y is in