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1.
State both forms of the Archimedean property of
R
.
This is Theorem 1.4.2:
(i) Given any real number
x
, there is a natural number
n
such that
n > x
.
(ii) Given any real number
y >
0
, there is a natural number
n
such that
1
=n < y
.
2.
What do we mean when we say that the rational numbers are dense in the
real numbers
?
This is Theorem 1.4.3: Given two real numbers
a < b
, there is a rational
number
r
such that
a < r < b
.
3.
Prove that
T
1
n
=1
(0
;
1
=n
) =
;
.
A real number
y
is in
(0
;
1
=n
)
exactly when
0
< y <
1
=n
. If
y
2
(0
;
1)
,
then
y >
0
so, by the Archimedean property, there is a natural number
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Unformatted text preview: n such that 1 =n < y . Consequently, y is not in the interval (0 ; 1 =n ) for that n . So there can be no real numbers in T 1 n =1 (0 ; 1 =n ) . 4. Write p 2 as the sum of two irrational numbers . Perhaps the easiest way is p 2 = 1 2 p 2+ 1 2 p 2 . The number 1 2 p 2 is irrational because it is the product of the nonzero rational number 1 = 2 and the irrational number p 2 . Another way is p 2 = 2 p 2 + ( & 1) p 2 ....
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This note was uploaded on 05/26/2010 for the course MATH maa 4200 taught by Professor Dr. during the Spring '08 term at Miami University.
 Spring '08
 Dr.
 Algebra

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