Another Field 13 We'll let Q(√2) represent the set of numbers that can be written in the form a+b√2, where 'a' and 'b' and rational numbers (i.e., 'a' and 'b' are from the set Q). For example, +¹3³√2, 7.2 – 6√2, and 15 (= 15 + 0√2) are three numbers that would be in the set Q(√2), but √√3+ 4√2 or 3 + √2 would not be in the set Q(√2) (neither √3 nor are rational numbers). A number such as 3+√ would be in the set Q(√2) since - can be re-written in the form a+b√2 using the following steps: 3+5√2 7 7 3-5√2 21-35√2 = 3+5√2 3+5√2 3-5√2 9 - 25(2) * 21-35√2 -41 a. (Q(√2), +, *) is a commutative ring, and b. (Q(√2) - {0}, *) is a group. 21 35 41 41 √2 Your task: Show that the system (Q(√2), +, *) is a field, where + is the ordinary addition operation and is the ordinary multiplication operation. So, you need to show that:

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Chapter2: Second-order Linear Odes
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Another Field
We'll let Q(√2) represent the set of numbers that can be written in the
form a+b√2, where 'a' and 'b' and rational numbers (i.e., 'a' and 'b' are from
13
2
the set Q). For example, +¹3√2, 7.2-6√2, and 15 (= 15 + 0√2) are three
numbers that would be in the set Q(√2), but √√3+ 4√2 or 3 + √2 would not
be in the set Q(√2) (neither √√3 nor are rational numbers). A number such
3+5√2 would be in the set Q(√2) since 3+√2 can be re-written in the form
7
a+b√2 using the following steps:
7
7
3-5√2 21-35√2
21-35√2 21 - 35√2 21 35
== + √2
3+5√2 3+5√2 3-5√2 9 - 25(2) -41
41 41
=
*
Your task: Show that the system (Q(√2), +, *) is a field, where + is the
ordinary addition operation and * is the ordinary multiplication operation.
So, you need to show that:
a. (Q(√2), +, *) is a commutative ring, and
b. (Q(√2) - {0}, *) is a group.
Transcribed Image Text:Another Field We'll let Q(√2) represent the set of numbers that can be written in the form a+b√2, where 'a' and 'b' and rational numbers (i.e., 'a' and 'b' are from 13 2 the set Q). For example, +¹3√2, 7.2-6√2, and 15 (= 15 + 0√2) are three numbers that would be in the set Q(√2), but √√3+ 4√2 or 3 + √2 would not be in the set Q(√2) (neither √√3 nor are rational numbers). A number such 3+5√2 would be in the set Q(√2) since 3+√2 can be re-written in the form 7 a+b√2 using the following steps: 7 7 3-5√2 21-35√2 21-35√2 21 - 35√2 21 35 == + √2 3+5√2 3+5√2 3-5√2 9 - 25(2) -41 41 41 = * Your task: Show that the system (Q(√2), +, *) is a field, where + is the ordinary addition operation and * is the ordinary multiplication operation. So, you need to show that: a. (Q(√2), +, *) is a commutative ring, and b. (Q(√2) - {0}, *) is a group.
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