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Discrete Structures
Readings Check, section 5.1
Read Section 5.1, pages 248  252.
1) What is the notation A  B called? What set does it produce?
Cartesian product
2) Looking at example 5.3, what is |A  B  B  B| ?
24
3) What is a binary relation from A to B?
Any subset of A  B
4) Looking at example 5.5, and supposing that |C| = 4 and |D| = 5 how many possible relations are there from C
to D?
|CD| = 20
No. of relations = 220
5) In example 5.8 is (5, 28) in the relation R ?
No
6) Using the theorem on p. 252, how can (D  F)  (E  F) be rewritten?
(D  E)  F
Discrete Structures
Readings Check, section 5.2
Read Section 5.2, pages 252  258.
1) What is a function and how does it differ from a relation?
A function is a relation from A to B where each element of A appers as the first element of an ordered pair
exactly once.
2) In example 5.9 why are R1 and R2 not functions?
R1 has no image for 3, R2 has two images for 2
3) Reading definition 5.4, how does the codomain of f differ from the range of f ?
B is the codomain. Range of f is the subset of B containing all elements which have a pre-image under f.
4) What is the value of 4.25 ? What is the value of 4.25 ?
4.25 = 4
4.25 = 5
5) Suppose that |A| = 5 and |B| = 10. How many functions can possibly be made from A to B ? [see p. 255]
105
6) When is a function f : A  B called onetoone?
Each element of B apperars atmost once as the image of an element of A.
7) Again if |A| = 5 and |B| = 10, what is the number of onetoone functions from A to B ?
10!/5! = 30240
Discrete Structures
Readings Check, section 5.3
Read Section 5.3, pages 260  265.
1) Why is the case of example 5.20 not an onto function?
Every element of Z is not in the range of f.
ex. 3 has no pre-image under f.
2) Does an onto function also have to be onetoone? If so then explain why. If not then give an example of
an onto function that is not onetoone..
No. Greatest integer function.
3) Does a onetoone function also have to be onto? If so then explain why. If not then give an example of a
onetoone function that is not onto.
No.
f:Z
R , f(z) = z
4) Looking at example 5.23, if A = {x, y} then how many onto functions are there from A onto B? In other
words, answer the question at the end of the example when m = 2. Explain without reference to the formula
why the answer comes out like that.
0.
The maximum no. of elements in a function is the cardinality of A. So, if the function has to be onto, then A
should have atleast as many elements as B.
5) The blue box at the top of page 262 gives a formula for the number of onto functions from domain A onto
range B. With |A| = m and |B| = n, what does this formula equal in terms of distributions of objects into
containers? [be precise]
It gives the no. of ways to distribute m different objects into n different containers with no container empty.
6) What does the formula in the box on page 263 calculate? [That is, the formula for Stirling numbers of the
second kind.]
It gives the no. of ways to distribute m different objects into n identical containers with no container empty.
Weekly Summary, Week 4, Chapter 5, Discrete Structures
Name:
Due on Sunday, February 12, by midnight.
1) Give a recursive definition (similar to example 5.8) for the relation R containing the ordered pairs:
(0, 2), (2, 4), (4, 16), (6, 256), …
1. (0, 2) ϵ R
2. If (s, t) ϵ R
then (s+2, t2) ϵ R
2) Let A = {1, 2, 3} and B = {1, 2, 3, 4, 5}. How many onetoone functions f : A  B satisfy f(1) = 1 ?
12
3) Is it true or false that (A  B)  (A  C) = A  (B  C) for any sets A, B, C ? [Remember that for a
statement to be considered true, it must be true in all cases.]
True
4) Suppose that |A| = |B| = n.
a) If n = 1 how many onto functions f : A  B are possible?
1
b) If n = 2 how many onto functions f : A  B are possible?
2
c) If n = 3 how many onto functions f : A  B are possible?
6
d) In general, for |A| = |B| = n, how many onto functions f : A  B are possible?
[Hint: there is an easy pattern. Do the case of n = 4 if you still do not see it.]
n!
5) Consider the number 510,510 which factors into prime factors as 2(3)(5)(7)(11)(13)(17). In how many
ways can 510,510 be factored into 3 factors, all greater than 1?
[Note: this is an application of the formula for Stirling numbers of the second kind, as per example 5.28.
To avoid doing the actual calculation see the table at:
https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
No. of ways = S(7,3) =301