Honors Introduction to Analysis I
Homework IV
Solution
February 23, 2009
Problem 1
A rational function is a ratio of two polynomials with real coefficients,
R
(
x
) =
P
(
x
)
/Q
(
x
)
,
Q
= 0
.
Equality between rational functions and the operations of addition and multiplication are defined similarly to the
case of the usual fractions (i.e. the rational numbers). Prove that the set of rational functions with these operations
is a field. You may assume standard facts about fractions and polynomials.
Solution.
Just check all the axioms, and take into account that the 0 polynomial is the polynomial with all the
coefficients 0.
Problem 2
Using only the axioms of an ordered field, prove that the field of complex numbers, with the usual
operations, cannot be ordered (you need to show that it is impossible to select a subset P of “positive” complex
numbers consistently with the properties listed in Theorem 2.2.2).
Solution.
The idea of the proof is that in an ordered field,
x
2
is positive for any nonzero element
x
. Indeed,
according to the axioms of an ordered field, either
x
or

x
is positive. Since
x
2
= (

x
)
2
, it is always a product of
two positive elements, hence itself positive. Now apply this to the complex numbers
x
=
i
(imaginary unit) and
x
= 1. We obtain that
i
2
=

1 and 1
2
= 1 are both positive. This is a contradiction: for example, their sum is 0,
but the sum of two positive elements must be positive. Hence the field of complex numbers cannot be ordered.
Problem 3
Compute the sup, inf, limsup, liminf, and all the limit points of the following sequences:
•
x
n
= 1
/n
+ (

1)
n
•
x
n
= 1 + (

1)
n
/n
•
x
n
= (

1)
n
+ 1
/n
+ 2 sin(
nπ/
2)
Solution.
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 '08
 PROTSAK
 Polynomials, Addition, Rational Functions, Order theory, Limit of a sequence, lim sup

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