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IMC / 1999 / Problems / Day 2, P7

IMC 1999 · Day 2 · P7

easy
abstract algebraworth 20 pts

Suppose that in a not necessarily commutative ring RR the square of any element is 0. Prove that abc+abc=0abc + abc = 0 for any three elements aa, bb, cc.

Solution (official)

From 0=(a+b)2=a2+b2+ab+ba=ab+ba0 = (a+b)^2 = a^2 + b^2 + ab + ba = ab + ba, we have ab=−(ba)ab = -(ba) for arbitrary aa, bb, which implies abc=a(bc)=−((bc))a=−(b(ca))=(ca)b=c(ab)=−((ab)c)=−abc.abc = a(bc) = -\bigl( (bc) \bigr) a = -\bigl( b(ca) \bigr) = (ca) b = c(ab) = -\bigl( (ab) c \bigr) = -abc.

How the field did

contestants scored
87
average (of 20)
15.06
solved (≥ 80%)
73.6%
near-0 (≤ 10%)
21.8%
discrimination
0.25

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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