Sets Part3
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Contents |
Proving Identities
As stated before we started this topic, STPM questions is in the form of proving given identities using ONLY set algebra. Before going on, MAKE SURE you have already mastered all the materials in Part 1 & Part 2. It doesn't only mean remembering the laws, you must know how and when to use them (exercise 5 is a good test). Otherwise you will struggle with the steps here, and thus, MISS THE WHOLE POINT of this part.
The structure of the proof
Let's use a simple example to illustrate how we are going to lay down the prove.
Question : Using set algebra, prove that
After deciding which side we want to start from
- COPY that side
- Work until we get the other side EXACTLY
- Write "... proved" at the end. (if question is show, write shown, of course)
So in this case, it will look like
To add clarity (especially if we are not starting from the left-hand-side), we can rewrite the identity, or we can add R.H.S to represent right-hand-side and L.H.S to represent left-hand-side. Thus
OR
Common mistakes involving structure
When answering questions involving "prove", it is important to get the structure correct, and not just the calculation. A wrong structure could result in zero marks even with the correct calculation
A common mistake is using the prove when we are asked to prove it.
is a WRONG way to write the proof. Always remember this general rule
"WHEN ASKED TO PROVE SOMETHING, THAT SOMETHING IS ALWAYS THE LAST LINE IN OUR ANSWER, NEVER THE FIRST."
This problem is compounded by the fact that some of us would normally copy the question before we start our answer. If you still want to copy the question as reference, do leave a line before starting our answer.
Another related common mistake (much more common) is this
See the mistake?
... student copied the question, decided to start from left ...
... but he did not copy the original left hand side, instead directly write its expansion.
This is still serious, since technically it still means the student have not answered the question. Note that the first line is just the question, NOT part of the answer, and thus, in this case the student have only proved
, which is not the question. We MUST copy the side that we want to start before working it out.
... if we do decide to copy the question as reference, do leave a line before the answer
Yet another mistake is this :
It may seem trivial, but it isn't. Yes, of course
, but the point is you have to write it down to show you know it. Otherwise, the above answer is incomplete.
Do's and Don'ts
- MUST use correct structure of a proof (refer above)
- DO NOT skip steps
- Use ONLY the basic laws to prove the identity, NEVER use own laws or other identities
Left/Right sides
The standard form will either be
- Start from LHS to prove RHS
- Start from RHS to prove LHS
Starting from both sides
We can also use this alternative form
Start from LHS, start from RHS, do until both is the same
- The working will thus look like this
Remember to write the last line.
When to use
While we would normally only use the third form in difficult questions, there is much benefit in learning it.
- Sometimes we get stuck when trying from one side. And we check for careless mistakes and can't find any. There is NOTHING wrong to just stop there for a while, and START from THE OTHER SIDE and see whether it will come to the same answer as the first side had. Try to do the two workings side by side if you have enough space on the test pad. I always use this analogy : Boyfriend is at A. Girlfriend is at B. Boyfriend can't go to B cause its too far and he don't know the way. Girlfriend can't go to A cause its too far and she don't know the way. So BOTH just made their way to C which is somewhere between A and B.
- Most of the time, however, if you are able to do it, it now means we can already see the full way from one side to the other. In other words, from
- Just take whichever side was longer, and just copy the other steps from the other side (from bottom to top, of course)
- Important Note :
- However, it is still perfectly acceptable to just go ahead and say LHS=RHS, and it's actually the expected method in certain difficult questions. BUT in STPM, its best to take the extra time to cut one side and continue from the other just to be safe.
- Some people give me the reaction that this is not a perfect answer, not the perfect way to think. Perhaps they think a smart person will surely be able to do from one side all the way to the other. That is NOT the way a smart people think. Trust me.
- However, do not confuse using this method with the common mistake I stated before, which was
which is certainly WRONG.
Exercise 9
As usual, make sure you have understood the material before attempting. Take some time to fully understand how you are supposed to lay out your answer.
Using the laws of algebra of sets, show the following.
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
is a correct statement, but you do need to write the steps in between. Remember, we are not trying to find the answer, we are showing its' proof.
, starting from LHS, it seems we are going to expand the
into
. However, we can also apply De Morgan's to the first bracket, which will bring us to the RHS much faster.





![\begin{align}
\mathrm{L.H.S}
& = \left( A \cap B\right)- \left( A' \cap B\right) \\
& = \left( A \cap B\right) \cap \left( A' \cap B\right)' \\
& = \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 \\
& = A \cap B \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}](/images/math/3/1/4/314f1d7f211242124e46468e5996b490.png)


![\begin{align}
\mathrm{L.H.S}
& = \left(A' \cup B \right)-\left(A \cup B' \right) \\
& = \left(A' \cup B \right)\cap \left(A \cup B' \right)' \\
& = \left(A' \cup B \right)\cap \left(A' \cap B \right) \\
& = B \cap \left[ A' \cap \left(A' \cup B \right) \right]\\
& = B \cap A'\\
& = A' \cap B \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}](/images/math/4/8/5/4857052e20ba3146e464396b5cf3a0d4.png)
![\begin{align}
\mathrm{L.H.S}
& = \left[A \cup \left(A' \cap B \right) \right]' \cap B \\
& = \left[A' \cap \left(A' \cap B \right)' \right] \cap B \\
& = \left(A' \cap B \right)' \cap \left(A' \cap B \right) \\
& = \phi \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}](/images/math/d/6/1/d61b683aff94dc6fa3b7933f711c48cc.png)
![\begin{align}
\mathrm{L.H.S}
& = B \cup \left[\left(A' \cup B \right) \cap A \right]' \\
& = B \cup \left[\left(A' \cup B \right)' \cup A' \right] \\
& = \left(A' \cup B \right) \cup \left(A' \cup B \right)' \\
& =\xi \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}](/images/math/2/a/2/2a28ab82f52236a84276cb667141247b.png)


![\begin{align}
\mathrm{R.H.S}
=\left(A \cap B \right)-\left(A \cap C \right)
& = \left(A \cap B \right) \cap \left(A \cap C \right)' \\
& = \left(A \cap B \right) \cap \left(A' \cup C' \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 \left( B \cap C'\right) \\
& = A \cap \left( B - C\right) \\
& = \mathrm{L.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}](/images/math/9/1/3/9132f4de429e6640e8a6c942d26929ac.png)
![\begin{align}
\mathrm{R.H.S}
& =\left(A-C \right)-\left(B-C \right) \\
& =\left(A \cap C' \right) - \left(B \cap C' \right) \\
& =\left(A \cap C' \right) \cap \left(B \cap C' \right)' \\
& =\left(A \cap C' \right) \cap \left(B' \cup C \right) \\
& = A \cap \left[ C' \cap \left(B' \cup C \right) \right] \\
& = A \cap \left[ \left( C' \cap B'\right) \cup \left( C' \cap C\right) \right] \\
& = A \cap \left[ \left( C' \cap B'\right) \cup \phi \right] \\
& = A \cap \left( C' \cap B'\right) \\
& = \left(A \cap B' \right) \cap C' \\
& = \left(A \cap B' \right) - C \\
& = \left(A-B \right)-C \\
& = \mathrm{L.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}](/images/math/f/3/0/f30ee30eb01077edd63d5aaa86b38822.png)

