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Exercise 12.3 · Q7

Q.If a∗b=a2+b2a*b=\sqrt{a^2+b^2} on the real numbers then ∗* is

(1) commutative but not associative
(2) associative but not commutative
(3) both commutative and associative
(4) neither commutative nor associative
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We check commutativity by symmetry of the formula, and associativity by expanding both groupings and comparing.

Step 1. Commutative. a∗b=a2+b2=b2+a2=b∗aa*b=\sqrt{a^2+b^2}=\sqrt{b^2+a^2}=b*a -- since a2+b2a^2+b^2 is symmetric in a,ba,b. Commutative.

Step 2. Associative -- compute (a∗b)∗c(a*b)*c. (a∗b)∗c=(a∗b)2+c2=(a2+b2)+c2=a2+b2+c2(a*b)*c=\sqrt{(a*b)^2+c^2}=\sqrt{(a^2+b^2)+c^2}=\sqrt{a^2+b^2+c^2} (since (a∗b)2=a2+b2(a*b)^2=a^2+b^2 by definition of ∗*). …

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