... s, n ∈ Z.
Take an (r + s)-set and split it into an r-set and an s-set. Choosing an n-subset
amounts to choosing a k-subset from the r-set and an (n −k)-subset from the s-set for
various k.
16
CHAPTER ... −I
A
n
)
=
J {1, ,n}
(−1)
|J|
I
A
J
,
Summing over s ∈ S we obtain the result
A
1
∪ ··· ∪ A
n
=
J {1, ,n}
(−1)
|J|
|A
J
|,
which is equivalent to the required result.
Ju...
... For example, try out the following goals.
| ?- X is 5 + 7.
| ?- X is 5 - 4 + 2.
| ?- X is 5 * 45.
| ?- X is log(2.7).
| ?- X is exp(1).
| ?- X is 12 mod 5.
The expression on the right ...
| ?- p(a).
Perform each test separately with only the given clauses as input.
a. p(b). b. p(c).
p(a) :- p(b). p(a) :- p(b).
p(a) :- p(c). p(a) :-...
... corresponding subset.
A correspondence with these properties is called a one-to-one correspondence (or bijec-
tion). If we can make a one-to-one correspondence b etween the elements of two sets, then
they ... important method of counting: we established a corre-
spondence between the objects we wanted to count (the subsets) and some other kinds of
objects that we can count easily (the numbers...
... t
1
, t
2
, . . . , t
ℓ
then k = ℓ and for each i, 1 ≤ i ≤ k, there exists
j, 1 ≤ j ≤ k such that p
i
= q
j
and s
i
= t
j
.
16 CHAPTER 1. PRELIMINARIES
Proof. We prove the result using the strong ... the remainder when a · b is divided by m.
(a) Prove that
i (mod n) + j (mod n) = i + j (mod n) and i (mod n) · j (mod n) = ij (mod n).
(b) In Z
5
, solve the equation 2x ≡ 1 (mod 5)....
... one-to-one?
d) Is the function onto?
11. Let f : A → B and g : B → C be functions. Prove each of the following:
a) If f and g are one-to-one, then g ◦ f is one-to-one.
b) If g ◦ f is one-to-one, ... one-to-one.
c) If f is onto and g ◦ f is one-to-one, then g is one-to-one.
d) If f and g are onto, then g ◦ f is onto.
e) If g ◦ f is onto, then g is onto.
f) If g ◦ f is onto and g is one-to-one,...