Sets Part1
From StpmWiki
Notes
- This part is going to focus solely on Set Algebra (specifically the laws of the algebra of sets). Other parts might be better covered in more appropriate sections (probability, number intervals, etc. or perhaps an introductory section to sets).
- I teach Set Algebra as the very first topic to my lower 6 students, the main reason being that it requires almost zero prior knowledge. It also has relatively few things to concentrate on and only one type of question (of proving). These factors make it the ideal gauge to the students ability to learn maths. In my opinion, any student who find this part difficult to deal with (even after trying out all the material here) should reconsider his options in taking STPM maths. Because it shows the inability to deal with laws and learning how to use them, and without that ability, there is really nothing left that can be done in paper 1.
Learning Objectives (Syllabus)
- understand the concept of set and set notation (not covered here)
- carry out operations on sets (not covered here)
- use the law of the algebra of sets
Prior Knowledge
- Almost none, except knowledge of Venn diagrams that helps in remembering the identity & inverse laws
Symbols and notations
:
:
:
:
:
:
:
Identity & Inverse Laws
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
- The extra explanation I give might or might not help understand them, so just ignore them if you already understand why each law are as such. You do have to find a way to make yourself understand why though, blindly memorizing them will bring you nowhere in the long run.
- The set
can be substituted with any other set or any combination of set
Of Apples and Oranges
In any formula, it is of utmost importance to identify the variable in it.
For example, in the law
, the part "
" whereas
, which means
is the (only) variable in this formula.
You can use "apple" or "orange" to represent those variables.
On the web, I will use
to represent "apple" and
to represent "orange".
Since this formula has only one variable, the formula becomes
.
Now, on your paper, draw the "apple" large enough for you to write inside it.
Now, if we write
inside the first "apple", means we must also write
inside the second apple.
Thus, we can conclude it is not just
, but
,
, and so on. So it really is ANY set union with null set gives us back THAT SAME set. So you really should remember it as "a set"
"Null set"="That same set".
The real value of using "apple" or "orange" is that you can draw them as big as you like, and you can write whatever inside it.
So you might get stumped with
at first.
But just draw an "apple" around the
, and what do you get?
which will be equals to
. Which just means
.
Try these examples :
Now lets look at
. There is still only ONE variable.
Now try :
Exercise 1
Simplify the following
Note : You MUST NOT refer back to the laws when doing this part. You HAVE TO train yourself. Give yourself sometime to look through the formulas first, then only start doing from beginning till end, and finally only check your answer.
1)
2)
3)
4)
5)
6)
7)
8)
9)
Commutative Laws
a)
b)
Associative Laws
a)
b)
Note:
Bracket Matters
- Order of brackets -
- Only 2 sets in a bracket, every pair of sets MUST have bracket
Example :
Combining Commutative and Associative
- Combining both, thus we CAN change position AND order ( meaning, we can rearrange the sets as we like, and put the brackets as we like), but ONLY IF SAME operation
- NEVER do that when different operation or when have complement outside the bracket
Example
Example 1:
- Thus, the working will look like this
Example 2:
Try it on your own first! Answer is
Working
Exercise 2
Simplify the following using set algebra
Note : Again you MUST NOT refer back to the laws or notes when doing this part. Train yourself. Give yourself sometime to look through the formulas first, then only start doing from beginning till end, and finally only check your answer (and also working if needed)
1)
2)
3)
4)
Distributive Laws
a)
b)
How to expand
Be VERY careful of the operations
and NOT
- Compare with
to easily remember it. You can even draw arrows (from
to
and from
to
) as guide.
If needed, rearrange using commutative law BEFORE expanding
=
=
How to factor
It's considerably harder to use the law from right to left, and we must firstly identify the forms which can be "factored" (not the correct term, but easier to imagine).
Lets first rewrite the laws:
a)
b)
Now, what do we see on the left hand side?
Example
Example : Simplify the following
Now try
Ans:
Using Distributive laws
Now let's put this new laws to use:
Example 1:
- Analyze
Example 2:
- Analyze
Exercise 3
Simplify the following using set algebra
Note : Learn to analyze the question and see why they can/need to be expanded/factored
1)
2)
3)
4)
5)
6)
7)
8)
Associative vs Distributive vs Identity
Let's take a breather from learning new things and spend some time differentiating the things we have learnt.
First, lets look at the fundamental differences between the associative(combined with commutative) and distributive laws.
We use associative
- when
- by
We use distributive
- when
- by
Thus
is done through
while
is done through
Identifying what laws the situation allow us to use will of course then determine what we do next.
So, for example, when we see
, we see different operations, thus we MUST use distributive law, and it becomes
.
It is surely WRONG to write
.
And if we see
, we see same operations, thus we simply rearrange and re bracket it to
.
It is fundamentally WRONG to make it
(as if expanding), even if it leads to the same answer.
If you have trouble (either understanding or agreeing) with the few above lines,
Comparison
Combining (associative + commutative) with distributive
Of Apples and Oranges (Part 2)
When we have a situation that don't suit the formula, we have two choices
- A) Derive a new formula
- B) Simplify the situation so that it fits the formula
Answer:
So let's look at the above problem.
There are 4 sets, but the order of operation does not allow us to simplify by "factoring" (distributive). Thus, we have to simplify it by looking at it as 3 sets, rather than 4.
We do this by grouping certain sets and assume it to be a single set for further manipulation. We can again use the apple "
" to help us imagine.
Thus, the question becomes
OR
which we can manipulate using associative and distributive laws respectively.
Before continuing, take note that at this stage, we HAVE NOT used any laws, we are just putting the symbol around the bracket which is already there. We CANNOT put the apple in the middle to get
, for example.
So now lets compare
with
. The first only allow us to rearrange and re bracket, whereas the second needs us to expand. Of course, we will choose
Let's see how we are going to simply
We see the two
's
Thus we know we can try the rearranging and rebracket method, but first, we must assume the last two sets as a single one
So in effect, what we have now is 3 different sets with the same operation
Where we have removed the front bracket (but remember the bracket inside the "apple" CANNOT be removed)
So now, what do we do? Consider the simpler case of
,
We have paired
with
since we know it will simplify.
So for
, which two sets do we pair to rebracket?
Since there is a
inside the "apple" , thus we pair
with
and we continue with distributive and simplify.
Thus, the full working will be
Now, try simplifying
Working
Exercise 4
Simplify the following using set algebra
Note : Make sure you understand fully the above two examples before attempting this.
1)
2)
3)
4)
5)
6)
Exercise 5
IMPORTANT NOTE
This is a revision exercise, and EVERY question here is just a REPEAT from the earlier exercise, however, they are jumbled up. As such, this is a practice of analyzing questions and choosing what method to use. It is VERY important that you try to do this exercise WITHOUT referring to any notes/ previous exercise. Doing so defeats the very purpose of this exercise. Instead, take some good time to look through ALL material before this, and then try these questions from beginning till the end, and see whether you can get all of them. Take note also that it is not only the answer you need to check, your working must be correct also. If you make a lot of mistake, its best to review the material again BEFORE continuing to part 2.
Simplify the following using set algebra
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
17)
18)


means we take ALL areas that are in
means we take ONLY areas that are in 
![[(A\cap B')\cap C]](/images/math/3/4/b/34bae099e6c021d0d77b95163d3e9e7d.png)
if cannot see. 









and b) as
but DONT do this in working (only do it if the final answer given in that form)
![\left[ \left( A \cup B' \right) \cap C \right] \cup \left[ A \cap \left( B \cup C \right)' \right]](/images/math/4/b/1/4b1d46a1819fb1359e9e142f374d28cb.png)


or 

, 3 sets with different operation


and we rearrange it to
, lastly we put the brackets surrounding the 

... here we rearrange and "re-bracket" the sets. Remember we do this only because ALL operations are the SAME here.







or
. Others DO NOT work.
or
or
etc. DO NOT work.
or
?




, we know it is wrong, cause it will give us 



...don't forget the earlier laws!















![=A \cap \left[ B \cap \left( A \cup B' \right) \right]](/images/math/0/8/6/0867abcf13acb6a8cfcaa0b8dd87864a.png)
![=A \cap \left[ \left( B \cap A \right) \cup \left( B \cap B' \right)\right]](/images/math/7/0/6/70671391be36a8407a98e20bb5b12772.png)
![=A \cap \left[ \left( B \cap A \right) \cup \phi \right]](/images/math/0/2/3/02353a2e663d5a26e3766075fea9d130.png)


Did you get it? If not, try again on your own first!
... this would be the most important step. Make sure you understand the reason we do it![=B \cup \left[ \left( A' \cup A\right) \cap \left( A' \cup B\right) \right]](/images/math/7/d/5/7d5c1b36d321aba07a4e6fdeb04aeb47.png)
![=B \cup \left[ \xi \cap \left( A' \cup B\right) \right]](/images/math/7/9/9/799a7c8bcc7e48505ae0d989b26f9646.png)


![\begin{align}
\left( A \cap B \right) \cap \left( A \cup B' \right)
& = A \cap \left[ B \cap \left( A \cup B' \right) \right] \\
& = A \cap \left[ \left( B \cap A \right) \cup \left( B \cap B' \right)\right] \\
& = A \cap \left[ \left( B \cap A \right) \cup \phi \right] \\
& = A \cap \left( B \cap A \right) \\
& = \left(A \cap A \right) \cap B \\
\end{align}](/images/math/4/1/7/417f93cfa7638b74ab2c8c56afff830f.png)
![\begin{align}
\left( A \cup B \right) \cup \left( A \cap B' \right)
& = A \cup \left[ B \cup \left( A \cap B' \right) \right] \\
& = A \cup \left[ \left( B \cup A \right) \cap \left( B \cup B' \right)\right] \\
& = A \cup \left[ \left( B \cup A \right) \cup \xi \right] \\
& =A \cup \left( B \cup A \right) \\
& = \left(A \cup A \right) \cup B \\
& = A \cup B \\
\end{align}](/images/math/6/2/f/62f345a2ce2ab0e00ba1737676a1e8c5.png)

![\begin{align}
\left( A' \cap B \right) \cup \left( A \cup B \right)
& = B \cup \left[ A \cup \left( A' \cap B \right) \right] \\
& = B \cup \left[ \left( A \cup A' \right) \cap \left( A \cup B \right)\right] \\
& = B \cup \left[ \xi \cap \left( A \cup B \right) \right] \\
& = B \cup \left( A \cup B \right) \\
& = \left(B \cup B \right) \cup A \\
& = B \cup A
\end{align}](/images/math/5/5/8/558318dc40ebe00a973697e2041cf59c.png)

![\begin{align}
\left( B \cup C \right) \cup \left( B' \cap A \right)
& = C \cup \left[ B \cup \left( B' \cap A \right)\right] \\
& = C \cup \left[ \left( B \cup B' \right) \cap \left( B \cup A \right) \right] \\
& = C \cup \left[ \xi \cap \left( B \cup A \right) \right] \\
& = C \cup \left( B \cup A \right) \\
& = A \cup B \cup C
\end{align}](/images/math/1/7/f/17f42a10a382a217bb23f04de9273b5e.png)

![\begin{align}
\left( C \cup B \right) \cap \left( B' \cap A \right)
& = A \cap \left[ B' \cap \left( C \cup B \right)\right] \\
& = A \cap \left[ \left( B' \cap C \right) \cup \left( B' \cap B \right) \right] \\
& = A \cap \left[ \left( B' \cap C \right) \cup \phi \right] \\
& = A \cap \left( B' \cap C \right) \\
& = A \cap B' \cap C
\end{align}](/images/math/9/3/d/93d88a8a4e95d37afe815f64609671c3.png)

![\begin{align}
\left( A \cup C \right) \cap \left( A' \cap B \right)
& = B \cap \left[ A' \cap \left( A \cup C \right)\right] \\
& = B \cap \left[ \left( A' \cap A \right) \cup \left( A' \cap C \right) \right] \\
& = B \cap \left[ \phi \cup \left( A' \cap C \right) \right] \\
& = B \cap \left( A' \cap C \right) \\
& = A' \cap B \cap C
\end{align}](/images/math/e/5/9/e59f2110d4fdd3b4ecbcc28ad2e5409b.png)