MGF 1106: Exam 1 Supplementary Assignment
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On General Statements Being True or False
On the exam, you were given some true/false questions involving general statements in
set theory. Before stating the new problems, we will begin by taking a look at how such
true/false questions work in general, including a look at a few of the problems that appeared
on the exam (or similar).
Question: Is the following statement true or false?
The sky is blue.
Solution: You wake up tomorrow to find the following email in your inbox:
Dear Earthling,
I am a bizarre alien from a distant planet called Glorbus 9. I used my telepathic
tail to divine your email address. All my friends here on Glorbus are jerks and
I’m headed to Earth for some time away. I’m in a weird mood and I want to
see something pretty. If I don’t I will be very upset. Tell me, is the Earth sky
blue? That is my favorite color. If you advise me well on this, I will grant you
any wish you desire upon my arrival. (However, if you mislead me, I will swap
your fingers and toes in anger!)
Sincerely,
Xeebfavv Zizwazz
What do you reply?
Of course, you might say, ”Yo, Xeebfavv, take it from me, this sky is blue. You’re gonna
love it! P.S. I want a Ferrari!”
... But, what if Xeebfavv shows up on a cloudy day? Or at sunset? Or at night?! If
he sees a sky that isn’t blue, you’ll have to start endorsing your checks with your feet,
which will be very embarrassing at the bank.
Your best bet may be to reply, ”I’m sorry Mr. Zizwazz, no, the sky isn’t always blue.
Hope that helps.”
Indeed, ”The sky is blue.” is FALSE as a general statement, because it’s not always true.
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Question: Is the following statement true or false?
A⊆A∪B
Solution: This, of course, was the first problem on the exam. The statement may be
written in words as follows:
For any two sets A and B, the union A ∪ B always contains A.
In the form of an abstract Venn diagram, the statement claims the following:
Hopefully the picture helps make this obvious; indeed, this general statement about unions
is TRUE. By definition, the union of any set A and any set B is the set consisting of any
and all elements which can be found in either A or B. Thus, all the elements in A also
appear in the union A ∪ B, so A ⊆ A ∪ B.
We can ‘test’ this by defining two sets A and B, however we want. For example, let
A = {1, 3, 5, 6, 9}
B = {2, 3, 6, 8}
The union of these two sets is
A ∪ B = {1, 2, 3, 5, 6, 8, 9}
and, indeed,
A = {1, 3, 5, 6, 9} ⊆ {1, 2, 3, 5, 6, 8, 9} = A ∪ B
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Question: Is the following statement true or false?
A⊆A∩B
Solution: This is similar to the second problem on the exam. The statement may be
written in words as follows:
For any two sets A and B, the intersection A ∩ B always contains A.
In the form of a Venn diagram, the statement claims that:
Again, I hope the picture helps you conclude that this statement is FALSE. It is actually true for some choices of A and B (see below), but it is not true for all choices of A
and B, so it is not true as a general statement.
To give a concrete example of how this may fail, let:
A = {1, 2, 3}
B = {3, 4, 5}
The intersection is A ∩ B = {3}, so A 6⊆ A ∩ B. Again, because the statement fails in at
least one case, it is false as a general statement.
Note, however, that this statement is actually true for some choices of A and B. For
example, letting
A = {1, 2, 3}
B = {1, 2, 3, 4, 5}
In this case, A ∩ B = A, so, yes, A ⊆ A ∩ B.
(Another for this to work is to let A = ∅. Then A is a subset of any set!)
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Shaded Area Problems
Example: (from lecture)
Express the shaded area using ∪, ∩, and/or 0 .
Solution: The shaded area is A, plus that extra little ‘chunk’ outside of A. To add to
a set, we can use unions.
Note, however, that writing down what that little chunk is will be annoying. A better
way to see this is as A, plus B ∩ C. A already includes some of B ∩ C, but there’s no harm
in throwing it in ‘again’, and writing this is much easier;
So the shaded area is given by (parentheses are critical! )
A ∪ (B ∩ C)
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Example: Express the shaded area using ∪, ∩, and/or 0 .
Solution: Here, the area is B, minus the ‘pacman mouth’. A convenient way to delete
portions of a set is to intersect with a compliment.
But, again, the pacman mouth is not convenient to work with. The best way to see this
area is that is everything in B:
which is not in A ∩ C:
So ...
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... again, in words, the shaded area is:
everything which is in B and not in A ∩ C;
So, the shaded area is given by
B ∩ (A ∩ C)0
(You can also see this by shading (A ∩ C)0 and intersecting (overlapping) with B. Doing this would be good practice!)
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Name:
MGF 1106: Exam 1 Supplementary Assignment
1. (10 points) Say whether each statement is true or false. In addition, if it is true, give an
example to illustrate. If it is false, give an example where the statement fails.
a) A ⊂ A ∪ B
b) A ∩ B ⊆ B
c) A ∩ B ⊂ A ∪ B
d) A ∩ C ⊆ B ∪ C
e) If A ⊆ B, then A ⊆ A ∩ B.
2. (10 points) Express each shaded area using ∪, ∩, and/or 0 .
(see next page for hints)
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Hints for Problem 1:
Remember, a statement like ”A ⊂ A ∪ B” is TRUE only if it is true for every two sets
A and B. If you can find any particular pair of sets A and B for which it fails, the statement is FALSE.
a) What if B ⊆ A?
b) Draw the statement as a Venn diagram (as in examples above).
c) Careful, note that it is ”⊂”, not just ”⊆”. What if A ∩ B = A ∪ B?
d) Draw a Venn diagram.
e) If A ⊆ B, then what is A ∩ B =?
Hints for Problem 2:
a) View as the union of two pieces; one piece is easy, the other was on the exam.
b) View as two pieces (which may overlap).
c) View as three pieces.
d) View as two pieces.
e) One piece is easy, ignoring that one, how do you get the other?
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