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Section 3.3
69
x
=
!.
If
r
=1=
0, the quadratic formula tells us that this equation has solutions
r
+
1
±
J(r
+
1)2  2r
r
+
1
± v'r2
+
1
x
=
2r
=
2r':""' .
We claim that the solution
r+
1
Jr2
+
1
x
=
:
2r
is always in (0,1). First note that if r
>
0, then r
+
1
>
v'r2
+
1 since (r
+
1)2
>
r2
+
1. On the
other hand, if r
<
0, then r
+ 1
<
1
<
v'r2
+
1.
In
either case, we have shown that
r
+
1 v'r2
+
1
2r
>
O.
To show that this expression is also less than 1, we will prove the equivalent inequality
lrJr2 +1
2r
<
O.
If
r
>
0, then 1  r
<
1
<
v'r2
+ 1
giving the result.
If
r
<
0, then 1  r
>
v'r2
+ 1
because (1 
r)2
>
r2+
1, again giving the result. Since
f
is onetoone and onto, 'tis a onetoone correspondence.
18. Start at (0,0) and move as illustrated.
19. (a) 2, 2,4,
4, 8,
8, 16, 16,
...
(b)
1=2°,2,!,4,i,8,~,
...
(c) 1,4,7,10,13,16,19,
...
·
ree
·
r
re
• . r
r
r
•
r
r

•
r

•
•
• •
eee
ee
1
e
1 1
•
•
• •
•
(d) (1,1), (1, 2), (1,3), (2, 1), (2, 2), (3, 3), (3, 1), (3,2), (3, 3), (4,1),
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Quadratic Formula, Graph Theory

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