Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x,y,z$?

share|cite|improve this question
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 $x,y,z\in\Bbb R$

share|cite|improve this answer
@ 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? – GovEcon 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? – GovEcon 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.