Suppose that your firend is thinking of an animal, and you guess that he is thinking of a platypus. 

You say:

"The animal is both duck-billed and a mammal."

In other words, you are saying that \( A \cap B\) is true. 

You are saying that the animal will be in the green section of the figure below.




Suppose your friend tells you that he is thinking of an animal and that this animal is either duck-billed, or a mammal.

In other words he is telling you that \( A \cup B \) is true. 

 Assume that your friend is telling the truth.  

He is telling you that whatever animal he is thinking of, it will be somewhere in the green area in the figure below.



Given two fraction, say, \(\frac{1}{3},\) and \(\frac{4}{5},\) we can multiply them together by multiplying the two numerators (the numbers upstairs) and the two denominators (the numbers downstairs), separately.

So for example:

\[ \frac{1}{3} \times \frac{4}{5} = \frac{ ( 1 \times 4 ) }{( 3 \times 5)} = \frac{ 4}{15}.\]

Or in general:

\[ \frac{P}{M} \times \frac{ Q}{N} = \frac{ P \times Q }{ M \times N}.\]

When we multiply fractions we are really taking fractions of fractions. Imagine that we have divided a pie into five equal parts. One slice falls on the floor and so we have four slices left; i.e., we have \(\frac{4}{5}\) of the original pie. Now even though we had cut out pie into five pieces, because we had expected five people to show up for our party, only three people actually show up.

We want to divide what is left of the pie up evenly among our three guests, and so each guest gets

\[ \frac{1}{3} \times \frac{4}{5} = \frac{ ( 1 \times 4 ) }{( 3 \times 5)} = \frac{ 4}{15}\]

of a pie.

In practice, this means that we would take each of the pieces we originally cut, and cut them all into three parts. Then we would give each guest four of these smaller pieces.

Because we had originally cut the pie into five pieces, and then cut each one of these pieces (except the one that fell on the floor) into three pieces, each little piece is equal to one fifteenth of a pie. Then we give each guest for each of these smaller pieces.

\[ \frac{ 4}{15}.\]

Suppose that you are playing a geussing game with a friend. You have both agreed that you will guess an animal. 

Your friend thinks of an animal and you try to guess what it is.

Let's assume, for the sake of this lesson, that your friend always tells the truth. 

He tells you that the animal that he is thinking of is a duck-billed animal. 

If we call the set of all animals C, and the set of all duck-billed animals A, then you friend has told you that the animal he is thinking of is in A

If x is the unkown animal, then the statement:

"x is a duck-billed animal"

is true.

That is: A is true.

We represent this fact  graphically thus:


Suppose we have two pies.

One pie is cut into three equal pieces and the second one is cut into five equal pieces.

Now suppose we want to add a piece from the first pie to a piece from the second pie. How much pie do we have?

As an equations this reads:

\[ \frac{1}{3} + \frac{1}{5} = ? \]

As a mathematical expression, there is nothing wrong with:

\[ \frac{1}{3} + \frac{1}{5}\]

but intuitively we don't really know how much pie this is.

We need to be able to compare \(\frac{1}{3}\) to \(\frac{1}{5}\) and to be able to add these two quantities together.

What we need is a common denominator.

We can view the expression:

\[ \frac{X}{Y} \]

as asking, and answering, the question:

How many \(Y\)s are there in each \(X\)?

So for example if \(X = 1 \) and \(Y = 2,\) 

then the fraction:

\[ \frac{X}{Y} = \frac{1}{2} ,\]

is asking:

How many twos are there in one?

And the answer is one-half.

The negation of the set (A and B):

\[ \lnot \left( A \cap B \right)\]

is called A nand B.


If all we know is that A is not equivalent to B, we can write this as:

\[ \lnot \left( A \leftrightarrow B \right)\]

\[ = \lnot \left[  \left( \left( \lnot A \right) \cup B \right) \cap \left( \left( \lnot B \right) \cup A \right) \right] \]


Now suppose that we know that  A implies B, and that  B implies A

In symbols this is:

\[ \left( A \rightarrow B \right) \cap \left( B \rightarrow A \right)\]

\[ = \left( \left( \lnot A \right) \cup B \right) \cap \left( \left( \lnot B \right) \cup A \right)\]

\[ A \leftrightarrow B.\]

In this case we say that A is equivalent to B.

This relation is symmetric, so B is also equivalent to A.


Suppose we are looking for an animal, but all we know about it is that it either does not have a duck bill, or it is a mammal.

We write this as

"(not A) or B"

\[ = \left(  \lnot A \right) \cup B\]

\[ = A \rightarrow B.\]