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MATH 3360 Applicable Algebra. HW 14 Solutions
Problem 24.4
Let
x
=
√
2 +
√
3, then
x
2
= 5 + 2
√
6. Subtracting 5 from both sides
and squaring, gives
x
4

10
x
+ 25 = (
x
2

5)
2
= 4
·
6 = 24. So
√
2 +
√
3
is a solution to the following minimal polynomial
p
(
x
) =
x
4

10
x
+ 1.
Problem 24.9
(i) One possible answer for a primitive root is
x
+ 1.
(ii) If you picked
x
+ 1 for a primitive root, then the minimal poly
nomial is
q
(
x
) =
x
2
+
x
+ 2. You can check this, by noting that
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Unformatted text preview: q ( x + 1) = ( x 2 + 2 x + 1) + ( x + 1) + 2 = x 2 + 1 ≡ 0 mod x 2 + 1. Problem 4.1 We know that a 3error correcting code has minimal distance of d = 3 · 2 + 1 = 7. Also, we know that 2 ∈ Z 11 has order 10. So one possible generator polynomial is g ( x ) = ( x2)( x2 2 )( x2 3 )( x2 4 )( x2 5 )( x2 6 ) = = ( x2)( x4)( x8)( x5)( x10)( x9) . 1...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 BILLERA
 Algebra

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