Solutions to Assignment 4, Math 220
1 Determine which of the following sets are countably infinite and
which are uncountable:
a: Set of positive rationals
b: Set of all irrationals in (0, 1)
c: Set of all terminating decimals (a decimal whose digits are all
0 from√some point on)
d: {r + n : r ∈ Q and n ∈ N}
e: {x ∈ R−Q : x cannot be written as the square root of a nonnegative rational }
f: {x ∈ R : x is a solution to ax2 + bx + c = 0 for some a, b, c ∈
Q}.
Solution
a: countable
b: uncountable
c: countable
d: countable
e: uncountable
f: countable
2 Give an example of a countably infinite subset of the set of irrational
numbers which is dense in R.
Solution For example, one could take
√
S = Q + 2.
Every element of S is irrational, but S has the same cardinality as Q
and hence is countably infinite. Also, S is dense in R as it is just a
shifted version of Q.
3 For each sequence below determine whether it converges and, if it
converges, find its limit. No proofs are required.
a: an = n/(n + 1)
b: bn = (n2 + 3)/(n2 − 3)
c: cn = 2−n
d: tn = 1 + 2/n
e: xn = 73 + (−1)n
f: sn = 21/n
g: yn = n!
h: dn = (−1)n n
i: (−1)n /n
j: (7n3 + 8n)/(2n3 − 31)
k: (9n2 − 18)/(6n + 18)
l: sin(nπ/2)
1
2
m: sin(nπ)
n: sin(2nπ/3)
o: (1/n) sin n
p: (2n+1 + 5)/(2n − 7)
q: 3n /n!
r: (1 + 1/n)2
s: (4n2 + 3)/(3n2 − 2)
t: (6n + 4)/(9n2 + 7)
Solution
a: converges to 1
b: converges to 1
c: converges to 0
d: converges to 1
e: diverges
f: converges to 1
g: diverges
h: diverges
i: converges to 0
j: converges to 7/2
k: diverges
l: diverges
m: converges to 0
n: diverges
o: converges to 0
p: converges to 2
q: converges to 0
r: converges to 1
s: converges to 4/3
t: converges to 0
4 Give an example of
a: a sequence (xn ) of irrational numbers having a limit lim xn that
is a rational number.
b: a sequence (rn ) of rational numbers having a limit lim rn that
is an irrational number.
Solution
√
a: The sequence xn = 2/n consists of irrational numbers having
a limit lim xn = 0.
b: The sequence rn = (1 + 1/n)n of rational numbers has limit
lim rn = e.
5 Determine the following limits. No proofs are required, but show
any relevant algebra.
3
√
a: lim s√
n2 + 1 − n,
n where sn =
2
b: lim(√ n + n − n),
c: lim( 4n2 + n − 2n).
Solution
1
a: sn = √n2 +1+n
and so lim sn = 0
√
√
n
b: n2 + n − n = √n2 +n+n
and so lim( n2 + n − n) = 1/2
√
√
n
c: 4n2 + n − 2n = √4n2 +n+2n
and so lim( 4n2 + n − 2n) = 1/4.