5.2 Practice Problems

Problem 5.1. State whether each of the following is a binary operation on \(\mathbb {Z}\):

(a).: \(a*b=a-2b\)
(b).: \(a*b=\sqrt {a+b}\)
(c).: \(a*b=(a-b)^{2}\)
(d).: \(a*b=a^{2}-b^{2}\)

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Solution. The question in each case is closure: does the rule give an answer, and does that answer land back in \(\mathbb {Z}\)?

(a). Yes. If \(a,b\in \mathbb {Z}\) then \(2b\in \mathbb {Z}\) and \(a-2b\in \mathbb {Z}\).

(b). No. Take \(a=1\), \(b=1\): then \(\sqrt {2}\) is not an integer. Worse, with \(a=1\), \(b=-3\) the expression \(\sqrt {-2}\) is not even real, so the rule fails to give an answer at all for some pairs.

(c). Yes. A difference of integers is an integer and its square is an integer.

(d). Yes, for the same reason.

Problem 5.2. Let \(*\) be defined on \(\mathbb {R}\) by \(a*b=(a-b)^{2}-3ab\).

(a).: Is the operation commutative? Is it associative?
(b).: Evaluate \(-2*(3*4)\) and \((-4*3)*2\).

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Solution. (a). Commutative. Compute \(b*a\) and compare: \[b*a=(b-a)^{2}-3ba=\left (-(a-b)\right )^{2}-3ab=(a-b)^{2}-3ab=a*b,\] using \((-x)^{2}=x^{2}\) and the commutativity of ordinary multiplication. So the operation is commutative.

Associative. It is not, and one counterexample settles it. Using the values from part (b), \(3*4=(3-4)^{2}-3(12)=1-36=-35\), so \[-2*(3*4)=-2*(-35)=(-2+35)^{2}-3(-2)(-35)=33^{2}-210=1089-210=879 .\] Going the other way, \(-2*3=(-2-3)^{2}-3(-6)=25+18=43\), so \[(-2*3)*4=43*4=(43-4)^{2}-3(172)=1521-516=1005 .\] Since \(879\neq 1005\), the operation is not associative.

(b). The first was found above: \(-2*(3*4)=879\). For the second, \(-4*3=(-4-3)^{2}-3(-12)=49+36=85\), and \[(-4*3)*2=85*2=(85-2)^{2}-3(170)=6889-510=6379 .\]

Note 5.19. Notice the asymmetry in what the two parts required. To prove commutativity we had to argue for arbitrary \(a\) and \(b\) — checking a few numerical cases would prove nothing. To disprove associativity, one specific triple was enough.

That is the general shape of such questions: a universal claim needs a proof, its denial needs only an example. When asked whether a property holds, it is usually quickest to try two or three numbers first — if they disagree you are finished, and if they agree you have some evidence that a proof is worth attempting.

Problem 5.3. The operation \(*\) is defined on \(\mathbb {R}\) by \(a*b=3(a-b)^{2}\). Show that it is commutative.

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Solution. \[b*a=3(b-a)^{2}=3\left (-(a-b)\right )^{2}=3(-1)^{2}(a-b)^{2}=3(a-b)^{2}=a*b .\] The squaring destroys the order, which is exactly why the operation cannot tell \(a-b\) from \(b-a\).

Problem 5.4. Let \(*\) be defined on \(\mathbb {Z}\) by \(a*b=(a-b)^{2}-ab\).

(a).: Determine whether \(*\) is commutative.
(b).: Evaluate \(2*(-3*5)\).

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Solution. (a). \(b*a=(b-a)^{2}-ba=\left (-(a-b)\right )^{2}-ab=(a-b)^{2}-ab=a*b\), so the operation is commutative.

(b). Work outwards from the bracket: \[-3*5=(-3-5)^{2}-(-3)(5)=64+15=79,\] \[2*79=(2-79)^{2}-2(79)=(-77)^{2}-158=5929-158=5771 .\]

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