Question 3 Take a look at the operator *, defined as follows: *: ZxZ→Z x*y = x + 3y (a.) Is the operator* commutative? Explain. (b.) Is the operator associative? Explain. (c.) Does the operator * have a unit element e ("identitet" in Swedish)? Explain (d.) Can you find two integers a and b that commute ("kommuterar") for *? Explain.

Algebra and Trigonometry (MindTap Course List)
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Chapter11: Matrices And Determinants
Section11.CT: Chapter Test
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Question 3
Take a look at the operator *, defined as follows:
*: ZxZ→ Z
x+y = x + 3y
(a.) Is the operator ★ commutative? Explain.
(b.) Is the operator associative? Explain.
(c.) Does the operator have a unit element e ("identitet" in Swedish)? Explain.
(d.) Can you find two integers a and b that commute ("kommuterar") for *? Explain.
(Read “Lemurell and Johansson” if you need to know what associative, commutative, unit element is.
Transcribed Image Text:Question 3 Take a look at the operator *, defined as follows: *: ZxZ→ Z x+y = x + 3y (a.) Is the operator ★ commutative? Explain. (b.) Is the operator associative? Explain. (c.) Does the operator have a unit element e ("identitet" in Swedish)? Explain. (d.) Can you find two integers a and b that commute ("kommuterar") for *? Explain. (Read “Lemurell and Johansson” if you need to know what associative, commutative, unit element is.
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2. Almost! But there's nothing that maps to 0. In your notes at the top you've listed ℕ={1,2,3,…}, so you should keep in mind that (at least in computer science-adjacent fields) ℕ includes 0.

3c. It is true that an identity needs to be unique (as in there cannot exist more than one, which can be a fun exercise to prove if you feel so inclined), but I'm guessing you mean either that the left e=-2x is not constant, or that left e ≠ right e. Both are indeed necessary for it to be an identity, but the phrasing is unclear.

3d. That the operator is not commutative doesn't mean there are no particular pairs that commute — for example 2⁴=4²=16. In other words you have to show that x⭑y=y⭑x if *and only if* x=y. To prove this absence, it might be be helpful to consider that a=b ⇔ a-b=0.

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