MAT 300 Mathematical Structures March 31, 2004
Test 2, sample solutions
1. a. Arelation from a set Ato a set Bis a subset of A×B.b. A relation ∼on a set Ais
reflexive if ∀a∈A, a ∼a.c-e. Suppose (X, ≤) is a partially ordered set. An element m∈X
is called a maximal element if ∀x∈X,x≥mimplies x=m. An element `∈Xis called a
largest element if ∀x∈X,`≥x. A subset S⊆Xis bounded if there exist a, b ∈Xsuch that
∀x∈S, a ≤x≤b.f. A relation ffrom a set Ato a set Bis a function if for every a∈A
there exists exactly one b∈Bsuch that (a, b)∈for afb, which is usually written as f(a) = b,
g. A function ffrom a set Ato a set Bis one-to-one if for every b∈Bthere exists at most one
a∈Asuch that f(a) = b.
2. Proof by induction on n. If there is only one disk, then it is clearly allowed to move it from
open peg to another. Now suppose n∈Z+and it is possible to move a tower of ndisks to a
different peg, and consider the case of (n+ 1) disks, ordered from largest to smallest on one peg.
By induction hypothesis one may move the top ndisks to a second peg since the presence of the
largest disk does not affect the legality of any of the moves of the nsmaller disks. Then move the
largest disk to the third peg. Finally, again using the induction hypothesis move the nsmaller
disks from the second peg to the third peg.
3. Define the relation ∼on Zby x∼yif and only if 5|(y−x). This relation is reflexive because
for every x∈X, 5|0 = x−x. It is symmetric, because whenever x, y ∈Zsuch that x∼y, i.e.
5|(y−x) then also 5|(x−y), i.e. y∼x. Finally suppose x, y, z ∈Zare such that x∼yand
y∼z. This means that there exist integers p, q ∈Zsuch that (y−x)=5pand (z−y)=5q.
Hence (z−x) = 5(p+q). Since (p+q) is again an integer, this shows that 5|(z−x) or x∼z,
establishing transitivity of ∼.
4. Suppose f:A7→ Band {Bα:α∈Λ}is an indexed collection of subsets of B. To show that
T
α∈Λ
f−1(Bα)⊆f−1 T
α∈Λ
Bα!, suppose x∈T
α∈Λ
f−1(Bα). This means that x∈f−1(Bα) and
hence f(x)∈Bαfor every α∈Λ. This implies f(x)∈T
α∈Λ
Bαand therefore x∈f−1 T
α∈Λ
Bα!
5. Suppose A,Band Care sets and g:A7→ Band f:B7→ Care functions.
a. Suppose both fand gare onto and consider any c∈C. Since fis onto, there exists b∈Bsuch
that f(b) = c. Since gis onto, there exists a∈Asuch that g(a) = b. Hence (f◦g)(a) = f(g(a)) = c
showing that f◦gis onto.
b. Conversely suppose that f◦gis onto. We show that fis onto. Let c∈C. Since f◦g
is onto, there exists a∈Asuch that (f◦g)(a) = f(g(a)) = c. Define b=g(a)∈B. Then
f(b) = f(g(a)) = c.
Bonus. Suppose f:A7→ Ais onto. Prove by induction on nthat fn:A×Ais onto for every n∈Z+.
When n= 1, then fn=fwhich is onto. Now suppose that n∈Z+and fnis onto. Then by 5.a.
fn+1 =f◦fnis onto because it is the composition of two functions that are both onto.