Exercise 5.4.18. Give a similar proof that 1 is the multiplicative identity for Zn when n > 1. What is the multiplicative identity for Zn when n = 1?

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Chapter2: Second-order Linear Odes
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Please do exercise 5.4.18 and please show step by step

Proposition 5.4.17. 0 e Z„ is the additive identity of Zn.
PROOF. Given any a e Zn, then a 0 is computed by taking the remainder
of a+0 mod n. Since a+0 = a, and 0< a < n, it follows that the remainder
of a is still a. Hence a e0 = a. Similarly we can show 0 e a = a. Thus 0
satisfies the definition of identity for Zn.
Exercise 5.4.18. Give a similar proof that 1 is the multiplicative identity
for Zn when n > 1. What is the multiplicative identity for Zn when n = 1?
Transcribed Image Text:Proposition 5.4.17. 0 e Z„ is the additive identity of Zn. PROOF. Given any a e Zn, then a 0 is computed by taking the remainder of a+0 mod n. Since a+0 = a, and 0< a < n, it follows that the remainder of a is still a. Hence a e0 = a. Similarly we can show 0 e a = a. Thus 0 satisfies the definition of identity for Zn. Exercise 5.4.18. Give a similar proof that 1 is the multiplicative identity for Zn when n > 1. What is the multiplicative identity for Zn when n = 1?
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To show 1 is the multiplicative identity of  n for n>1 and 0 is the multiplicative identity of  0 .

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