Download Extra-Credit Assignment #1

Extra-Credit Assignment # 1 / MATH 16100
Due: Tuesday, October 20, 2015, at 11:50 CDST
1. In this course, we may use freely the thirteen “Defining Properties of Z” axioms that are found in the
Properties of the Integers document. The question came up in class on October eighth, of whether or
not the product of two negative integers is necessarily a positive integer. Prove that this is in fact true.
You may use the following outline to help organize your proof (although you do not have to follow this
outline).
(a) Prove the additive cancellation law. That is, prove that
if a, b, c ∈ Z and a + c = b + c, then a = b.
(b) Prove that
if a ∈ Z, then 0a = 0.
(c) Prove that additive inverses of integers are unique. That is, prove that
if a, b, c ∈ Z and a + b = 0 = a + c, then b = c.
(We denote this common value of b and c simply by “−a”.)
(d) Prove that
if a ∈ Z, then (−1) a = −a.
(e) Prove that 0 is the only additive identity for the integers. That is, prove that
if z ∈ Z and a + z = a for all a ∈ Z, then z = 0.
(f) Prove that
if a ∈ Z, then a = − (−a) .
(g) Prove that multiplication is closed on the positive integers. That is, prove that
if a, b ∈ Z and a, b > 0, then ab > 0.
(h) Prove that
if a ∈ Z and a < 0, then − a > 0.
(i) Prove that
if a, b ∈ Z, then ab = (−a) (−b) .
(j) Prove that the product of two negative integers is a positive integer. That is, prove that
if a, b ∈ Z and a, b < 0, then ab > 0.
2. It is a common situation that one wants to compose more than two functions together. The question
of how to do this is more subtle than it appears. One has to define these compositions carefully and
prove an associative law in order to avoid having an overabundance of parentheses. In this problem,
we work through some of these details. In all parts of this problem, assume that n ∈ N, n ≥ 3, and,
for each j ∈ N, we have a set Xj ⊂ Z and a function fj : Xj → Xj+1 .
We begin with the question of defining composition of more than two functions. The definition is
1
recursive. We already have in Definition 1.27 of Script #1 a definition for the composition of two
functions. If we want to compose more, say, n functions, we define this n-fold composition by
fn ◦ fn−1 ◦ · · · ◦ f1 := fn ◦ (fn−1 ◦ fn−2 ◦ · · · ◦ f1 ) : X1 → Xn+1 ,
where we have to refer recursively to this definition in order to unwind the definition of fn−1 ◦ fn−2 ◦
· · · ◦ f1 . We can repeatedly “unwind” in this way to obtain, successively,
fn ◦ fn−1 ◦ · · · ◦ f1 := fn ◦ (fn−1 ◦ fn−2 ◦ · · · ◦ f1 )
:= fn ◦ (fn−1 ◦ (fn−2 ◦ fn−3 ◦ · · · ◦ f1 ))
:= fn ◦ (fn−1 ◦ (fn−2 ◦ (fn−3 ◦ fn−4 ◦ · · · ◦ f1 )))
..
.
:= fn ◦ (fn−1 ◦ (fn−2 ◦ (fn−3 ◦ (· · · (f2 ◦ f1 ))) · · · )) .
| {z }
n−2 of these
In particular, then,
f3 ◦ f2 ◦ f1 := f3 ◦ (f2 ◦ f1 ) .
(a) Prove directly that
f3 ◦ f2 ◦ f1 = (f3 ◦ f2 ) ◦ f1 .
This is the standard associative law for composition of three functions. (Hint: Prove that
(f3 ◦ f2 ◦ f1 ) (x) = ((f3 ◦ f2 ) ◦ f1 ) (x)
for each
x ∈ X1 .)
(b) Now prove by induction on n that, for all n ≥ 3,
fn ◦ fn−1 ◦ · · · ◦ f1 = (fn ◦ fn−1 ◦ · · · ◦ f2 ) ◦ f1 .
Note that part (a) above is your base case for your proof by induction.
(c) Try using your above results now to prove that, for example,
f4 ◦ (f3 ◦ f2 ) ◦ f1 = f4 ◦ f3 ◦ f2 ◦ f1 .
If you understand why this follows from what you have done so far, you will likely believe that
what you have done implies that, if we have an n-fold composition of the functions fn , fn−1 , . . . , f1 ,
with any parenthetical groupings whatsoever, the result is still equal to
fn ◦ fn−1 ◦ · · · ◦ f1 .
This allows us to write unambiguously n-fold compositions without parentheses, just as we write
unambiguously n-fold sums and products without them.
3. In your proof of Lemma 1.32 from Script #1, you should have proven separately that
• if f and g are injective, so is g ◦ f ; and
• if f and g are surjective, so is g ◦ f .
In this problem you extend these results. Suppose n, the sets Xj , and the functions fj are defined as
in Problem 2, and define
f := fn ◦ fn−1 ◦ · · · ◦ f1 : X1 → Xn+1
to be the composition of these n functions.
(a) Prove by induction on n that, if each fj is injective, then so is f .
(b) Prove by induction on n that, if each fj is surjective, then so is f .
(c) Conclude that if each fj is bijective, then so is f .
2