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MATH 290 – WRITING ASSIGNMENT 2 – SOLUTIONS
1. Let
{
a
1
, a
2
, a
3
, ...
}
be a sequence of real numbers. Use induction to prove that for all
natural numbers
n
,

a
1
+
a
2
+
a
3
+
···
+
a
n
 ≤ 
a
1

+

a
2

+

a
3

+
···
+

a
n

.
Solution:
The proof will be by induction. If
n
= 1 then the result reduces to

a
1
 ≤ 
a
1

which is obvious. If
n
= 2 then the result is

a
1
+
a
2
 ≤ 
a
1

+

a
2

which was
proved in Writing Assignment 1. Hence the result holds for
n
= 1 and
n
= 2.
Let
n
be a natural number with
n >
2 and assume that the result holds for
1
,
2
, ..., n
. We must prove that it holds for
n
+ 1. Note that

a
1
+
a
2
+
a
3
+
···
+
a
n
+
a
n
+1

=

(
a
1
+
a
2
+
a
3
+
···
+
a
n
) +
a
n
+1

and that since the result holds for
n
= 2,

(
a
1
+
a
2
+
a
3
+
···
+
a
n
) +
a
n
+1
 ≤ 
a
1
+
a
2
+
a
3
+
···
+
a
n

+

a
n
+1

.
(Note that this is why we had to establish the base case
n
= 2 above and not
just
n
= 1.) Since the result holds also for
n
,

a
1
+
a
2
+
a
3
+
···
+
a
n
 ≤ 
a
1

+

a
2

+

a
3

+
···
+

a
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 Fall '08
 Alligood,K
 Advanced Math, Real Numbers, Natural Numbers

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