Homework 2 Key
1. (3 points)
(a) Take the hint: number the elements of A as a1 , a2 , . . . and those
of B as b1 , b2 , . . .. The elements of A × B are then pairs of the
form
(ai , bj )
where i = 1, 2, . . . and j = 1, 2, . . .. To see that A × B is countably infinite, use the figure on page 29 as inspiration and list the
elements of A × B as
(a1 , b1 ); (a2 , b1 ), (a1 , b2 ); (a3 , b1 ), (a2 , b2 ), (a1 , b3 ); · · ·
Each element of A × B appears in a unique position in the list.
In fact, if you’re a glutton for punishment you can work out that
(ai , bj ) appears in position
k=
(i + j)2
i
3j
− −
+ 1.
2
2
2
The set A × B is thus countable, since we can list its elements.
(b) We know from part (a) that if each Ak is countably infinite then
A1 × A2 is countably infinite. Consider this the base case for an
induction. Now suppose that it is true that for any such sets
A1 × A2 × · · · Am is countable infinite. Then A1 × Am × Am+1 is
countably infinite, for
A1 × · · · Am × Am+1 = B × Am+1
where B = A1 × · · · × Am . Since B is countably infinite, so is
B × Am+1 , which establishes the claim, for any FINITE n.
2. (4 points)
√
√
2
(a) Note that α = 2√satisfies
x
−
2
=
0,
α
=
5 satisfies x2 − 5 = 0.
√
The choice α = √ 2 + 5 is a bit trickier,
√ but not much. Note
2
2
that α = 7 + 2 10, so that α − 7 = 2 10. Square both sides
again to see that (α2 − 7)2 = 40, so α satisfies x4 − 14x2 + 9 = 0.
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(b) Let Bn denote the Cartesian product
Bn = |Z × Z ×
{z· · · × Z} = {(a0 , a1 , . . . , an ); ak ∈ Z}.
n copies
From the previous problem part (b), Bn is countably infinite. Consider the subset B̃n of those elements of Bn in which the ak are relatively prime. Then B̃n is countable, and also the set B = ∪n≥1 Bn
is countable. We consider an element p of B as a polynomial in
the obvious way. Each element of p corresponds to a finite number
of algebraic numbers, and each algebraic number corresponds to
a unique such polynomial. Let r(p) denote the roots of p ∈ B
(note r(p) is finite). Then the set of all algebraic numbers can be
written as
∪p∈B r(p),
a countable union of countable sets, hence countable.
(c) If the reals were all algebraic then R would be countable; this is
false, hence transcendental numbers must exist!
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