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I am looking at this question from Hardy's book, A Course of Pure Mathematics and have no idea where to begin.

I was wondering, what is the first step to deriving the conditions?


What are the conditions that $ax+by+cz=0$ for all values of $x,y,z$?

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1 Answer

up vote 1 down vote accepted

It needs to be an identity which implies $a=0,b=0,c=0$

If it were anything other than $(0,0,0)$, then equation $ax+by+cz=0$ would represent a plane that doesn't cover all $x,y,z\in\Bbb R$

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@ Avatar I see that as a condition that will definitely work. Is this condition unique, or is there another set of conditions that will also satisfy the equality? –  Jordan Mahar Jan 19 '13 at 4:38
well, it depends on what you mean by all values of $x,y,z$. you need to specify the sets to be they belong. –  Aang Jan 19 '13 at 4:47
Alright, sadly Hardy does not provide the set they belong to. If the assumption is made that $x,y,z \in \mathbb R$ would we be able to prove uniqueness? –  Jordan Mahar Jan 19 '13 at 4:50
Surely, choose set of values $(\alpha,0,0)$($\alpha\neq 0$). putting in the equation gives $a\alpha=0\implies a=0$. –  Aang Jan 19 '13 at 4:55
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