Exercise 14.4.13. Fill in the blanks with the appropriate variables in the following proof of the proposition. (*Hint*) Proof: For any integer { we may write { = ak + b, where be Z <1> · It follows that =7>k+<> = (7<4>k)p<5> = (7k)<>> = (id)<>, = 7 <10>. = id. However, we know that Therefore r = id if and only if < 12 > < k, and we also know that < 13 > is the smallest positive integer such that 714> = id. Hence it must be the case that b = < 15 > , which is the same thing as saying that mod(e, < 17 > ) = 0.
Exercise 14.4.13. Fill in the blanks with the appropriate variables in the following proof of the proposition. (*Hint*) Proof: For any integer { we may write { = ak + b, where be Z <1> · It follows that =7>k+<> = (7<4>k)p<5> = (7k)<>> = (id)<>, = 7 <10>. = id. However, we know that Therefore r = id if and only if < 12 > < k, and we also know that < 13 > is the smallest positive integer such that 714> = id. Hence it must be the case that b = < 15 > , which is the same thing as saying that mod(e, < 17 > ) = 0.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please do exercise 14.4.13 please show step by step and explain.
The hint states "The first blank should be replaced by k"
![Definition 14.4.12. The order of a permutation 7 is the smallest positive
integer k such that * = id. As before, the order of T is denoted by the
notation |T|.
A
Proposition: Let 7 be a permutation, and let k = |7|. Then r = id if and
only if mod(e, k) = 0.
Exercise 14.4.13. Fill in the blanks with the appropriate variables in the
following proof of the proposition. (*Hint*)
Proof: For any integer l we may write l = ak + b, where b e Z <1> . It
follows that
=7>k+<> = (7<4>-k)p<5> = (7k)<>>Z> = (id)<>,> =7 10>.
Therefore 7
< 12 > < k, and we also know that < 13 > is the smallest positive
integer such that 714> = id. Hence it must be the case that b = < 15 >,
which is the same thing as saying that mod(e, < 17 > ) = 0.
= id if and only if r11>
= id. However, we know that](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2Fed2c280a-6c51-4de5-95a7-ef5dca60e1d0%2Fy5vtome_processed.png&w=3840&q=75)
Transcribed Image Text:Definition 14.4.12. The order of a permutation 7 is the smallest positive
integer k such that * = id. As before, the order of T is denoted by the
notation |T|.
A
Proposition: Let 7 be a permutation, and let k = |7|. Then r = id if and
only if mod(e, k) = 0.
Exercise 14.4.13. Fill in the blanks with the appropriate variables in the
following proof of the proposition. (*Hint*)
Proof: For any integer l we may write l = ak + b, where b e Z <1> . It
follows that
=7>k+<> = (7<4>-k)p<5> = (7k)<>>Z> = (id)<>,> =7 10>.
Therefore 7
< 12 > < k, and we also know that < 13 > is the smallest positive
integer such that 714> = id. Hence it must be the case that b = < 15 >,
which is the same thing as saying that mod(e, < 17 > ) = 0.
= id if and only if r11>
= id. However, we know that
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