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206
Solutions to Exercises
Using the Binomial Theorem,
(1
+
v'2t
=
1 +
nv'2
+
(~)
(v'2)2
+
(~)
(v'2)3
+
(~)
( v''2)4
+(~)(v'2)5
+
(~)(v'2)6
+
...
+ (v'2)n
from which the result follows.
1 +
nv'2
+
2(~)
+
2v'2(~)
+
4(~)
+
4v'2(~)
+8(~)
+
8V2(~)
+ .
.. +
(V2)n
(c)
Xn+l
+
Yn+l
V2
=
(1
+ V2)n+l
=
(1
+
V2)(I
+
V2)n
=
(1
+
V2)(xn
+
YnV2)
=
Xn
+
YnV2
+
V2xn
+
2Yn
=
(xn
+
2Yn)
+
(xn
+
Yn)V2.
Equating coefficients of 1 and V2, we have the desired formulas.
15. (a) Using the Binomial Theorem,
(v'2

I)n
=
(v'2)n
+ +
(~)
(v'2)nl( 1) +
(;)
(v'2)n2( _1)2
+ .
.. +
(~)(v'2tk(I)k
+ .
.. +
(It.
When
n

k
is even,
(V2)nk
=
2
n;k,
which is an integer. When
n

k
is odd,
(V2)nk
=
nkl
V2(
V2)nkl
=
V2V2
2

which is of the form cV2 for some integer c.
It
follows that every
term in the expansion of (V2

I)n
is either an integer or of the form cV2. Collecting similar
terms gives the result.
(b) Since
1
<
2
<
4, we know
1
<
V2
<
2, hence 0
<
V2
 1
<
1.
So limn>
00 (
V2

I)n
=
o.
(c) Suppose V2 is rational. Then V2
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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
 Binomial Theorem, Graph Theory, Binomial

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