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 B \cap \left( B' \cup A \right)  =A \cap B

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

 \begin{align}
B \cap \left( B' \cup A \right)
& =\left(B \cap B' \right) \cup \left(B \cap A \right) \\
& =\phi \cup \left(B \cap A \right) \\
& =B \cap A \\
& =A \cap B \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

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

 \begin{align}
B \cap \left( B' \cup A \right)
& =\left(B \cap B' \right) \cup \left(B \cap A \right) \\
& =\phi \cup \left(B \cap A \right) \\
& =B \cap A \\
& =A \cap B
\end{align}

\therefore B \cap \left( B' \cup A \right)  =A \cap B  \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}

OR

 \begin{align}
\mathrm{L.H.S} 
& = B \cap \left( B' \cup A \right)\\
& =\left(B \cap B' \right) \cup \left(B \cap A \right) \\
& =\phi \cup \left(B \cap A \right) \\
& =B \cap A \\
& =A \cap B = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

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.


\begin{array}{rcl}
B \cap \left( B' \cup A \right) & = & A \cap B\\
\left(B \cap B' \right) \cup \left(B \cap A \right) & = & A \cap B\\
\phi \cup \left(B \cap A \right) & = & A \cap B\\
B \cap A & = & A \cap B\\
A \cap B & = & A \cap B \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{array}

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


\begin{array}{rcl}
B \cap \left( B' \cup A \right) & = & A \cap B\\
\left(B \cap B' \right) \cup \left(B \cap A \right) & = & \phi \cup \left(B \cap A \right) \\
& = & B \cap A \\
& = & A \cap B \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{array}

See the mistake?

  • B \cap \left( B' \cup A \right)  =A \cap B ... student copied the question, decided to start from left ...
  • \left(B \cap B' \right) \cup \left(B \cap A \right) =\phi \cup \left(B \cap A \right) ... 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 \left(B \cap B' \right) \cup \left(B \cap A \right) =A \cap B, which is not the question. We MUST copy the side that we want to start before working it out.


B \cap \left( B' \cup A \right)  =A \cap B ... if we do decide to copy the question as reference, do leave a line before the answer


 \begin{align}
B \cap \left( B' \cup A \right)
& =\left(B \cap B' \right) \cup \left(B \cap A \right) \\
& =\phi \cup \left(B \cap A \right) \\
& =B \cap A \\
& =A \cap B \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

Yet another mistake is this :

 \begin{align}
B \cap \left( B' \cup A \right)
& =\left(B \cap B' \right) \cup \left(B \cap A \right) \\
& =\phi \cup \left(B \cap A \right) \\
& =B \cap A \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

It may seem trivial, but it isn't. Yes, of course B \cap A = A \cap B, 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
    • Even if you can think few steps ahead, you still have to write down each step. For example \left(A \cap A' \right) \cup \left(A \cap B \right)= A \cap B 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.
  • Use ONLY the basic laws to prove the identity, NEVER use own laws or other identities
    • Say, for example, we might think, are the following correct?
      • A-\left(B \cap C \right) = \left(A-B \right)\cap\left(A-C \right)
      • A \cap \left(B-C \right) = \left(A \cap B \right) - \left(A \cap C \right)
      • Answer : It does not matter, since we AREN'T allowed to use it in our working anyway. (if you are interested though, you can use basic laws to show the first is wrong but the second is true). Even if you use a correct identity (without proving it using basic laws), it simply means you are skipping steps.
  • When we have a choice of what to do, keep an eye on the other side to know which to take
    • For example, if we are to prove \left(A' \cap B' \right) \cap \left(A \cup B \right)=\phi, starting from LHS, it seems we are going to expand the B' \, into \left(A \cup B \right). However, we can also apply De Morgan's to the first bracket, which will bring us to the RHS much faster.
  • Start from the side which is more complicated, however, if stuck, start from the other side (refer to section below).
  • When something seems to go wrong, stop and check above workings, else you might be wasting a lot of time running in circles.

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
  •  \begin{align}
\mathrm{L.H.S} 
& = ... \qquad \\ 
& = ... \\
& = \heartsuit
\end{align}
\begin{align}
\mathrm{R.H.S} 
& = ... \\
& = ... \\
& = \heartsuit
\end{align}

\therefore \mathrm{L.H.S}  = \mathrm{R.H.S}  \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}

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
    •  \begin{align}
\mathrm{L.H.S} 
& = ... \qquad \\ 
& = ... \\
& = ... \\
& = ... \\
& = \heartsuit
\end{align}
\begin{align}
\mathrm{R.H.S} 
& = ... \\
& = ... \\
& = \heartsuit
\end{align}
    • Just take whichever side was longer, and just copy the other steps from the other side (from bottom to top, of course)
    •  \begin{align}
\mathrm{L.H.S} 
& = ... \qquad \\ 
& = ... \\
& = ... \\
& = ... \\
& = \heartsuit \\
& = ... \\
& = ... \\
& = \mathrm{R.H.S}  \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \\ 
\end{align}
    • Important Note : DON'T waste your time to copy the whole thing again. Just continue from whichever side, and cut away the other side which is no longer needed. (Which is why you should do it side by side)
  • 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
    •  \begin{array}{rcl}
\mathrm{L.H.S} & = &  \mathrm{R.H.S} \\
... & = & ... \\ 
... & = & ... \\ 
... & = & ... \\ 
... & = & ... \\ 
\heartsuit & = & \heartsuit  \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} 
\end{array}
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)\left(A \cap B \right) \cup \left(A' \cup B'\right)=\xi

  • 
\begin{align}
\mathrm{L.H.S} 
& = \left(A \cap B \right) \cup \left(A' \cup B'\right) \\
& = \left(A \cap B \right) \cup \left(A \cap B\right)' \\
& = \xi \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

2)A-\left(B' \cup A \right) = \phi

  • 
\begin{align}
\mathrm{L.H.S} 
& = A-\left(B' \cup A \right) \\
& = A \cap \left(B' \cup A \right)' \\
& = A \cap \left(B \cap A' \right) \\
& = \left(A \cap A' \right) \cap B \\
& = \phi \cap B\\
& = \phi \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

3)A-B=B'-A' \,

  • 
\begin{align}
\mathrm{R.H.S} 
& = B'-A' \\
& = B' \cap A \\
& = A \cap B' \\
& = A-B \\
& = \mathrm{L.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

4)\left( A \cap B\right)- \left( A' \cap B\right)= A \cap B

  • 
\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}

5)\left(A \cup B \right)' \cup \left(B-A \right)=A'

  • 
\begin{align}
\mathrm{L.H.S} 
& = \left(A \cup B \right)' \cup \left(B-A \right) \\
& = \left(A' \cap B' \right) \cup \left(B \cap A' \right) \\
& = \left(A' \cap B' \right) \cup \left(A' \cap B \right) \\
& = A' \cap \left(B' \cup B \right) \\
& = A' \cap \xi \\
& = A' \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

6)\left(A-B \right) \cup A' =\left(A \cap B \right)'

  • 
\begin{align}
\mathrm{L.H.S} 
& = \left(A-B \right) \cup A' \\
& = \left(A \cap B' \right) \cup A' \\
& = A' \cup \left(A \cap B' \right) \\
& = \left(A' \cup A \right) \cap \left(A' \cup B' \right) \\
& = \xi \cap \left(A' \cup B' \right) \\
& = \left(A' \cup B' \right) \\
& = \left(A \cap B \right)' \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

7)\left(A' \cup B \right)-\left(A \cup B' \right)=A' \cap B

  • 
\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}

8)\left[A \cup \left(A' \cap B \right) \right]' \cap B =\phi

  • 
\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}

9)B \cup \left[\left(A' \cup B \right) \cap A \right]'=\xi

  • 
\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}

10)A-\left(B\cap C \right) = \left(A-B \right)\cup\left(A-C \right)

  •  \begin{align}
\mathrm{L.H.S} 
& = A-\left(B\cap C \right) \\
& = A \cap \left(B\cap C \right)' \\
& = A \cap \left(B' \cup C' \right) \\
& = \left( A \cap B' \right) \cup \left( A \cap C' \right) \\
& = \left( A -B \right) \cup \left( A -C \right) \\
& = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}
  • RHS to LHS works well too

11)A-\left(B\cup C \right) = \left(A-B \right)\cap\left(A-C \right)

  •  \begin{align}
\mathrm{L.H.S} 
& = A-\left(B\cup C \right) \\
& = A \cap \left(B \cup C \right)' \\
& = A \cap \left(B' \cap C' \right)\\
& = A \cap B' \cap C'
\end{align}

\begin{align}
\mathrm{R.H.S} 
& =\left(A-B \right)\cap\left(A-C \right) \\
& = \left(A \cap B' \right) \cap \left(A \cap C' \right) \\
& = \left(A \cap A \right) \cap \left(B' \cap C' \right) \\
& = A \cap \left(B' \cap C' \right) \\
& = A \cap B' \cap C'
\end{align}
    • \therefore \mathrm{L.H.S}  = \mathrm{R.H.S}  \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}

  • OR  \begin{align}
\mathrm{R.H.S} 
& =\left(A-B \right)\cap\left(A-C \right) \\
& = \left(A \cap B' \right) \cap \left(A \cap C' \right) \\
& = \left(A \cap A \right) \cap \left(B' \cap C' \right) \\
& = A \cap \left(B' \cap C' \right) \\
& = A \cap \left(B \cup C \right)' \\
& = A-\left(B\cup C \right) \\
& = \mathrm{L.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
\end{align}

12)A \cap \left(B-C \right)=\left(A \cap B \right)-\left(A \cap C \right)

  •  \begin{align}
\mathrm{L.H.S} 
& = A \cap \left(B-C \right) \\
& = A \cap \left(B \cap C' \right) \\
& = A \cap B \cap C'
\end{align}

\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 B \cap C' \\
\end{align}
    • \therefore \mathrm{L.H.S}  = \mathrm{R.H.S}  \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}

  • OR
\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}

13)\left(A-B \right)-C=\left(A-C \right)-\left(B-C \right)

  •  \begin{align}
\mathrm{L.H.S} 
& = \left(A-B \right)-C \\
& = \left(A \cap B' \right) \cap C' \\
& = A \cap B' \cap C'
\end{align}

\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) \\
& = A \cap B' \cap C'
\end{align}
    • \therefore \mathrm{L.H.S}  = \mathrm{R.H.S}  \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}

  • OR
\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}
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