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Question Number 83965 by Rio Michael last updated on 08/Mar/20
proveordisprove(withcounter−example)thata)Foralltwodimensionalvectorsa,b,c,a.b=a.c⇒b=c.b)Forallpositiverealnumbersa,b.a+b2⩾ab
Commented by mr W last updated on 08/Mar/20
(a)statementiswrong!a⋅b=∣a∣∣b∣cosβa⋅c=∣a∣∣c∣cosγa⋅b=a⋅c⇒∣b∣cosβ=∣c∣cosγ⇏b=cexamplea=(1,0)b=(1,1)c=(1,2)a⋅b=a⋅c=1butb≠c
(b)fora,b>0:(a−b)2⩾0⇒a−2ab+b⩾0⇒a+b2⩾ab
Commented by Rio Michael last updated on 08/Mar/20
perfectsir,andthanksforthecorrection
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