Math 320 / Homework 1 – ANSWER KEY
(1) (5 points) Prove that
√
Due: 9/18/15
3 is irrational.
√
Proof. Use√the contradiction method. Assume 3 is a rational number; that is, 3 = p/q, where p, q are two positive integers having no
common factors greater than 1. Then 3 = p2 /q 2 ; i.e., p2 = 3q 2 . We show
that 3 must be a factor of p. Note that p can be written as p = 3k + r,
where k is an integer and r is either 0, 1, or 2. Hence
p2 = 9k 2 + 6kr + r2 .
If r = 1 then we have p2 = 9k 2 + 6k + 1 = 3(3k 2 + 2k) + 1 = 3s + 1,
showing 3 is not a factor of p2 . If r = 2 then p2 = 9k 2 + 12k + 4 =
3(3k 2 + 4k + 1) + 1 = 3t + 1, again showing 3 is not a factor of p2 .
Therefore, we must have r = 0; that is, p = 3k. Hence p2 = 9k 2 = 3q 2 .
This implies q 2 = 3k 2 and thus 3 is a factor of q 2 . As shown above, this
would imply that 3 must be a factor of q as well. Therefore, we would
obtain that 3 is a factor of both p and q, a contradiction, as p and q
have no common factors.
√
This proves that 3 is not a rational number, so must be an irrational
number.
(2) (10 points) If a set A contains n elements, prove that the number of different subsets
of A is 2n . (Keep in mind that the empty set ∅ is considered to be a subset of every
set.)
Proof. We use the method of mathematical induction on n.
If n = 1, then A = {a}. The subsets of A are only ∅ and A; that is,
there are 2 = 21 different subsets of A.
Assume this proposition holds for n = k. Now assume A is a set of
k + 1 elements; say, A = {a1 , a2 , · · · , ak+1 } = A0 ∪ {ak+1 }, where A0 =
{a1 , a2 , · · · , ak }. Then any subset of A can be classified into two families:
(a) subsets containing ak+1 , (b) subsets not containing ak+1 .
Since any set in (a) is obtained by adding ak+1 to a set in (b); vice
versa, any set in (b) gives a set in (a) by adding ak+1 . Therefore the
numbers of subsets in the two families are the same. Let K be the
number of family (b). Then the total number of different subsets of
A is K + K = 2K. But the family (b) is exactly the family of different
subsets of set A0 = {a1 , a2 , · · · , ak }. Therefore, by induction, the number
K = 2k . Hence the number of subsets of A is 2K = 2 · 2k = 2k+1 ; this
proves the proposition when n = k + 1.
By induction, the proposition is true for all n ∈ N.
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(3) (10 points) Write a formal definition for the greatest lower bound inf A for a set A ⊆ R,
in the style of sup A. Then use the Archimedean Property to prove that inf{ n1 : n ∈
N} = 0.
(a) Definition: A number l is the greatest lower bound for a set A ⊆ R
if it meets the following two criteria:
(i) l is a lower bound for A; that is, a ≥ l for all a ∈ A;
(ii) if b is any lower bound for A, then l ≥ b.
In this case, we write l = inf A.
(b) Proof. First, since n1 > 0 for all n ∈ N, it follows that l = 0 is a
lower bound for the set A = { n1 : n ∈ N}.
We now verify condition (ii) in the definition (see above). So let b be
any lower bound for A, and we need to show that b ≤ 0. If not, suppose
b > 0. By the Archimedean Property (ii), there exists an n ∈ N such
that n1 < b. Since n1 ∈ A, this means that b is not a lower bound for A,
a contradiction to the assumption that b is a lower bound for A.
This completes the proof.
(4) (10 points) Let S be the set consisting of all sequences of digits 0 and 1. Show that S
is not countable.
Proof. Suppose, for the contrary, that S is countable. Then we can
list S as
S = {s1 , s2 , s3 , · · · , sk , · · · },
si 6= sj (∀ i 6= j),
where each sk is an element of S, that is a sequence of digits 0 and 1;
so we can write sk as
sk = (ak1 , ak2 , ak3 , · · · , akn , · · · ) with akn ∈ {0, 1}.
We use the Cantor diagonalization method to define a sequence
α = (b1 , b2 , b3 , · · · ) with
(
0 if ann = 1
bn =
, ∀ n = 1, 2, · · · .
1 if ann = 0
Then bn ∈ {0, 1}; so α is an element of S and hence α = sk for some
k ∈ N. This implies (b1 , b2 , · · · ) = (ak1 , ak2 , · · · ); that is, bn = akn for all
n = 1, 2, · · · . In particular, bk = akk ; but, by the definition of bk , bk 6= akk ,
and this gives the desired contradiction.
So S is not countable.
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(5) Let Q be the set of rational numbers and I be the set of irrational numbers.
(a) (5 points) Show that if a, b ∈ Q then ab ∈ Q and a + b ∈ Q.
Proof. Let a, b ∈ Q. We want to show ab and a+b are also elements
of Q. Write a = pq and b = st , where p, q, t, s are integers and q 6=
0, s 6= 0. Then
p t
ps + qt
pt
.
ab = , a + b = + =
qs
q s
qs
Since all numbers pt, qs and ps + qt are integers, it follows that
ab ∈ Q and a + b ∈ Q.
(b) (5 points) Show that if a ∈ Q and t ∈ I, then a + t ∈ I and at ∈ I as long as
a 6= 0.
Proof. Assume a ∈ Q and t ∈ I. Then a + t cannot be in Q since,
otherwise by part (a), it would imply that t = (a + t) + (−a) is in
Q.
If a 6= 0, then a1 ∈ Q. Note that t = (at) · a1 . Hence at cannot be in
Q since, otherwise, t = (at) · a1 ∈ Q, again by part (a).
(c) (5 points) Given any two real numbers a < b, show that there exists a number
t ∈ I such that a < t < b.
√
√
Proof. Consider two real numbers a− 2 < b− 2. By the√Density
of Q√in R, there exists
a− 2 < r <
√ a rational number r satisfying
√
b − 2. √
So a < r + 2 < b. Since r ∈ Q and 2 ∈ I, by part (a),
t = r + 2 ∈ I and satisfies a < t < b.
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