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Math 335: solutions to assignment 4
1. (3.2–12) A one-to-one correspondence between OWS and the natural numbers is given
by the function

1
x=1





2
x=3



3
x=5
f (x) =
.
(1)

4
x
=
7





5
x=8



x − 4 x ≥ 10
for x in OWS. The inverse correspondence is

1





3



5
f −1 (y) =

7





8



y+4
y
y
y
y
y
y
=1
=2
=3
=4
=5
≥6
(2)
for y a natural numbser.
2. (3.2–30) Label each package by a natural number n and each candy bar in a given
package by i = 1, 2. Then the collection of all candy bars is labeled by pairs (n, i). There is
a one-to-one correspondence betwen the pairs (n, i) and the natural numbers given by the
function
f (n, i) = 2n + i − 2.
(3)
For instance,
f (1, 1) = 1,
f (1, 2) = 2,
f (2, 1) = 3,
f (2, 2) = 4, . . . .
The inverse correspondence is given by
(
(m/2, 2)
m even
f −1 (m) =
.
((m + 1)/2, 1) m odd
(4)
(5)
Indeed,
(
f (m/2, 2) = m + 2 − 2 = m
m even
f (f −1 (m)) =
f ((m + 1)/2, 1) = m + 1 + 1 − 2 = m m odd
1
(6)
and, likewise, it can be checked that f −1 (f (n, i)) = (n, i).
Note. A common submitted solution was to state that there is a corresponence between
the even natural numbers and all natural numbers. This solution is incomplete without an
explanation of which set of numbers corresponds to which set of items (packages and candy
bars).
2. (3.2–31) It is clear that there are infinitely many peanuts. Let’s label each pile with a
natural number n and let’s label each nut in the n-th pile with a number k = 1, . . . , n. Then
each peanut is labeled by a pair (n, k). It can be verified that the function
n(n − 1)
+k
(7)
2
defines a one-to-one correspondence between such pairs and the natural numbers. For instance, notice that
f (n, k) =
f (1, 1) = 1,
f (2, 1) = 2,
f (2, 2) = 3,
f (3, 1) = 4, . . . .
(8)
2. (3.3–13) Suppose that the set of colourings has the same cardinality as the set of natural
numbers. Thus, we can enumerate all the colourings by natural numbers. Consider a given
such enumeration:
1 RRRBBRBR...
2 BRBBBRRR...
3 RBRBRBRB...
4 RBBBRBBB...
5 BRBBRRBB...
..
..
.
.
The diagonal elements of the above table, which have been highlighted, can be used to
construct the “diagonal colouring”: RRRBR. . . .
Swapping R’s and B’s in this colouring gives the colouring: BBBRB. . . . Since this
colouring differs from every colouring in the table above, this table does not contain all
colourings. Since this argument works for any given enumeration of colourings, we get a
contradiction and conclude that colourings cannot be enumerated by natural numbers.
2. (3.3–14) This is similar to the previous argument.
Suppose there is an enumeration
1 THHHHH. . .
2 TTHTHT. . .
3 HHHTTT. . .
4 THHTHT. . .
..
..
.
..
of all sequences of H’s and T’s by natural numbers. Consider the diagonal sequence TTHT. . . .
Swapping H’s and T’s in the diagonal sequence results in the sequence HHTH. . . , which cannot be found in the table above. Thus, these sequences cannot be enumerated by natural
numbers.
2
Alternative solution. There is a one-to-one correspondence between sequences of H’s and
T’s and sequences of R’s and B’s obtained by changing H’s into R’s and T’s in B’s. Since
the set of sequences of R’s and B’s has a greater cardinality than the natural numbers (as
shown in the previous question), so does the set of sequences of H’s and T’s.
2. (3.3–16) This is similar to the previous two arguments.
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