Download UCC Maths Enrichment - Logic 1. For every proposition p, the

UCC Maths Enrichment - Logic 1.
For every proposition p, the negation p0 is a proposition which is FALSE whenever the
proposition p is TRUE, and TRUE whenever p is false:
p p0
T F
F T
The first 2 exercises explore the correct way to negate ”p AND q”, ”p OR q”, ”p(Everybody)”,
”p(Somebody)”.
When people try to negate a statement like ”p AND q”, they often exagerate and the
result is not a negation. This can be modelled as a ”role playing game” where Alice and Bob
are making hypotheses about two characters: e.g. the Prime-Minister and the Queen. Ben is
supposed to always negate Alice. After each dialogue, the Prime-Minister and the Queen reveal
themselves we see who was right, who not. Whenever Alice is right, Bob should be wrong, and
vice-versa, but Bob doesn’t always manage that.
For example:
• Alice: The prime minister and the queen are in a good mood this morning.
• Bob: The prime minister and the queen are not in a good mood this morning.
• The queen appears, smiling. The prime-minister is frowning.
Conclusion: Both Alice and Bob are wrong, so Bob’s statement is not the negation of Alice’s
statement (Bob exagerated).
1. For every proposition p, let ¬p denote its negation.
Complete the truth tables below:
p
T
T
F
F
q p and q ¬(p and q) ¬p or ¬q
T
F
T
F
What can you say about the statements ¬(p and q) and (¬p or ¬q)?
p
T
T
F
F
q p or q ¬(p or q) ¬p and ¬q
T
F
T
F
What can you say about the statements ¬(p or q) and (¬p and ¬q)?
1
2
p
T
T
F
F
q ¬p ¬q p ⇒ q ¬p or q ¬q ⇒ ¬p
T
F
T
F
What can you deduce from the table above?
2. For each of the following statements:
a) Which of these is the negation the proposition Everybody can fool Harry.
i)
ii)
iii)
iv)
Everybody is unable to fool Harry.
Somebody can fool Harry.
There is someone who can fool Harry.
Nobody can fool Harry.
Find the negations of the following statements:
b) Fred and George can fool Harry.
c) There is no one who can fool everybody.
d) Everyone can be fooled by somebody.
e) No one can fool himself/herself.
f) There is someone who can fool exactly 2 people.
g) ”You may fool all the people some of the time; you can even fool some of the people all
of the time; but you can’t fool all the people all the time.”(Abraham Lincoln).
h) ”If you try to fool me, you’re a fool”.
√
√
√
3. Show that none of 3 and 2 + 3 can be written as a fraction of two whole numbers.
4. a) The sum of two positive integers is 11. Show that one of them is greater than 5.
a) The sum of two positive distinct integers is 12. Show that one of them is greater than 6.
5. Use sets to solve the following puzzle: Marys ideal man is tall, dark and funny. She
knows four men: Alec, Bill, Carl and Dave.
•
•
•
•
•
Only 3 of the men are tall, only 2 are dark, and only 1 is funny.
Each of the 4 men has at least 1 of the required traits.
Alec and Bill have the same complexion.
Bill and Carl are the same height.
Carl and Dave are not both tall.
Only one of the 4 men has all of the characteristics Mary requires. Which one?
6. a) There are 11 bridges between 6 houses. No two bridges connect the same two houses.
Show that one can use these bridges to travel between any two houses.
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b) There are 12 bridges between 6 houses. No two bridges connect the same two houses.
Show that one can use these bridges to travel between any two houses and back, without passing
over any of the bridges twice.
7. a) Let l be a line, O a point on it, and A and B two points in plane on different sides
of the line. If the angles made by OA and OB with l are equal, show that either l is the angle
[ or A, O and B are on the same line.
bisector of the angle AOB
b) Let l be a line, O a point on it, and A and B two points in plane on the same side of
the line. If |AO| + |OB| is the minimum length of a trip from A to B meeting l at some point,
show that the angles made by OA and OB with l are equal.
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Anca Mustata, UCC
UCC Maths Enrichment - Logic 2 with some hints
1. For each of the following propositions, write its negation:
Each person can be fooled by at least two people.
There exists someone who can fool at most two people.
If Fred cannot fool Ron, then Ron can’t be fooled by anybody.
If Ron can fool Harry, then he can fool Fred and George.
Solution:
There exist someone who can be fooled by at most one person.
Everybody can fool at least three people.
Fred cannot fool Ron and there is someone who can fool Ron. Ron can Fool Harry, but he cannot
fool Fred or George.
2. Each entry in the table indicates the number of steps you can take outside that square,
moving either up, or down or left or right. Starting from the center of the table, find a route to
the End (E) point.
5
1
1
4
2
1
1
2
5
2
3
2
1
3
2
2
4
2
1
1
E
3
5
3
3
Idea: this exercise is meant to suggest moving backwards from solution to the start of
the problem. It should motivate the method of proof by contradiction: p ⇒ q is equivalent to
not(q) ⇒ not(p).
3. The ages of ten people born in ten different years add up to 156. Prove that at least
one of them is over 20.
Idea: assume none of them is over 20. Then the oldest is at most 20, the one before at
most 19, ..., last is at most 11. Add up (possibly using the trick of symmetric pairing in the
sum) to at most 155.
4. Let x and y be two positive numbers such that
x + 2y < 6 or 2x + y < 9.
a) Prove that x < 4 or y < 1.
b) Prove that x + 3y < 7 or 3x + y < 13.
Formulate and prove a statement similar to b).
Idea: The negation of b) Is x + 3y ≥ 7 and 3x + y ≥ 13. Proof by contradiction:
Case 1. Assume x + 3y ≥ 7 and 3x + y ≥ 13 and x + 2y < 6. You can then use −x − 2y > −6
in conjunction with each of the first two inqualities to eliminate x and get 1 < y < 1.
Case 2. Assume x + 3y ≥ 7 and 3x + y ≥ 13 and 2x + y < 9, etc.
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5. Consider the following axioms of Euclid:
i) Through any two distinct points in plane there passes a unique line.
ii) Any line contains infinitely many points.
iii) Given a line l and a point P in plane, there exists a line through the point which is perpendicular on l, namely, making equal angles with l on both sides.
iv) Consider two lines a and b in plane intersected by a third line d. If these lines form two
angles, both found on the same side of a, b and d respectively, and the sum of which is less than
the sum of two right angles, then these two lines intersect at a point.
Using these, prove that through any point in plane not found on a given line, there exists
a unique line parallel to the initial line. Have you used all the axioms above? If not, which ones
haven’t you used?
6. Let A, B, C and D be sets of points (x, y), with x, y ∈ {1, 2, ..., 9}. We know:
All points (x, y) of A satisfy: If x is even, then y is odd.
All points (x, y) of B satisfy: If y is odd, then x is even.
C is the set of all points such that x is even and y is odd, or x is odd and y is even.
D is the set of all points such that x is even or y is odd.
i) What is the maximum possible number
of elements for each
T
T of the sets A, B, C, D?
ii) Which of the following are true: a) A B ⊆ C? b) C ⊆ A B? c) D ⊆ C?. Prove your
answers.
Idea: A could include all cases when x is odd, as well as those when x is even and y is odd,
etc...
7. Each kid in a classroom asks how many other kids in class have the same first name as
her/him. Then he/she asks how many kids share the second name with her/him. Among the
answers, all numbers from 0 to 6 are heard. If no two kids share both the first and second name,
can you prove that there are at least 19 kids in this class? Is it possible for a class of 19 kids to
satisfy all the requirements above?
8. Let A, B, C be three distinct sets, eachTwith
T 60 elements. The union of all three sets
has 80 elements. Show that the intersection A B C has at least 5 elements, or provide a
counterexample.
9. Using triangle congruences, it is not difficult to prove that a triangle is isosceles if and
only if it has two equal heights, or medians. It is more difficult to prove that a triangle with two
equal bisectors is isosceles. Try it, by contradiction.
10. Let ABC be a triangle, and A0 , B 0 , C 0 be points on the sides BC, AC and AB
respectively. We say that the triangle A0 B 0 C 0 is inscribed in ABC. Prove that the triangle
A0 B 0 C 0 has minimum perimeter among all such triangles if and only if AA0 , BB 0 and CC 0 are
heights.
Anca Mustata, UCC