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How to prove "v" is not a commutative operation?
Give the operation "v" on the integers such that: avb = a|a-b|
*those are absolute value symbols*
And V does not stand in for any standard operation, it just IS. How do you prove the operation is not communtative?
An explanation with steps would also be greatly appreciated =)
1 Answer
- Ron WLv 71 decade agoFavorite Answer
avb = a|a-b|
bva = b|b-a|
Proof by contradiction:
Suppose v is commutative. Then avb = bva for all integers a and b. Let integers a and b, with a > b, be given. Then |a-b| = a-b and |b-a| = -(b-a) = a-b. If v is commutative, then
a|a-b| = b|b-a|
But we just determined that |a-b| = |b-a| = a-b so
a(a-b) = b(a-b)
Since a > b. a-b ≠ 0 so we can divide both sides by (a-b), giving us
a = b
But this contradicts the assumption that a > b. So the initial supposition, that v is commutative, is false. Therefore, v is not commutative.
Actually, you don't need to resort to such a general proof to prove non-commutativity. A single counterexample will serve as well. Choose any two distinct integers and show that they do not satisfy a|a-b| = b|b-a|.
