CEC 14th - Past Year Paper Solution 2013-2014 Sem1
CE/CZ1001 – Discrete Mathematics
Solver: Li Haihui
Email Address: C130062@e.ntu.edu.sg
√
Assume the opposite (negation) is true, i.e.,
1. (a) First we prove that
can be denoted as
Then by definition, √
√
where both m and n are positive integers with no common factors.
√
Thus,
, which is contradictory with the fact that
√
Second we show that
contradiction,
.Therefore by
is true.
√ is rational.
√
.
{ }
Therefore,
(b) Assume that the length of identifiers is n bits, then it satisfies
.
Solving the inequality, we get
, i.e.
.
Since n should be an integer, the minimum value therefore is 4.
2. (a) We will prove that the set on each side is a subset of the other.
̅ ̅
For any x LHS,
̅ ̅
̅
̅
̅
̅
̅
̅ ]
[
] [
̅
̅
̅ ] [
[
̅
̅
̅ ]
(b) The 3 hypotheses are expressed in symbolic form as:
(1) A
P; (2) P
F; (3) F
E.
Likewise, the last statement can be expressed as E
A. It is not a conclusion of a valid
argument based on the above three hypotheses since there are cases when the conclusion is
false even though the premises are true. For example, let A, P, F and E be false, true, true
and true correspondingly. Though the three hypotheses are true in this case, the conclusion
is false.
3. (a)
Case 1:
| |
| |
| |
Therefore,
Case 2:
√
√
√
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CEC 14th - Past Year Paper Solution 2013-2014 Sem1
CE/CZ1001 – Discrete Mathematics
Therefore,
(b) The augmented matrix for the given linear system is
[
]
Apply Gauss-Jordan Elimination to transform the augmented matrix into a reduced form:
]→
[
→
[
]
→
[
]
→
[
[
]
]→
[
]
Therefore, the solution for the linear system is
(c) Note that |
.
Define a function
as
where
{
}.
| |, there is an element in Y that is the
By the Generalized Pigeonhole Principle, | |
remainder for at least 15 integers in A.
Any pair of integers with the same remainder should appear in Set B. Therefore, only the
following case can give the minimum size of Subset B: there is a set of A of 100 integers,
such that one element in Y is the remainder for 94 integers and the other 6 elements in Y are
the remainders for the rest 6 integers respectively. In this case, B has a minimum size of 94.
NOTE This question aroused controversy among candidates after exam. The solver thinks
that it is just a reasoning question with no much relation to Generalized Pigeonhole Principle.
Those in orange is just what the solver think the professor might initially want to test.
4. (a) (i) The composition
{
}.
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CEC 14th - Past Year Paper Solution 2013-2014 Sem1
CE/CZ1001 – Discrete Mathematics
(ii)
transitive.
For all
For all
is an equivalence relation because it is reflexive, symmetric and
if
is reflexive;
and
, then
For all
if
, then
is symmetric.
(b) (i) Assume the number of vertices in G that have a degree of 2 is n.
The total degree of Graph G equals twice the number of edges in G, that is,
.
An equation can then be derived:
.
Solving, we get
(ii) Graph G contains an Euler circuit because G is connected (as given) and every vertex
of G has even degree of either 2 or 4.
(iii) NOTE PYP correction: V(G) = {a, b, c, d, e, f, g, h, i, j}
Since
The vertex set cannot be edge-partitioned into two
subsets with no edge between any two vertices in the same subset.
(c) Given
and
, the composition of f and g is
.
Consider an arbitrary chosen
,
is onto
(
)
is onto
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Thank you and all the best for your exams! 
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