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So, here are two ways to say what I interpret as the same statement:

$f_i(x,y)\geq0 \hspace{0.85cm} \forall i \{0,1,2\}$

which implies that $f_0(x,y) \geq0$ and $f_1(x,y)\geq0$ and $f_2(x,y)\geq0$

but doesn't

$f_i(x,y)\geq0 \hspace{0.85cm}\{i\in\mathbb{Z}|i\in[0,2]\}$

imply the same thing?

Is there another reason why these different notations are used, besides the fact that the one consumes less space than the other? Apologies if one of these notations falls into a specific category of mathematics without my knowledge. I am not fully taught (evidently).

Any responses are appreciated.

bof
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1 Answers1

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Actually, in formal mathematics, the condition should come first: $$ \forall i \in \{0,1,2\} \; f_i(x,y) \ge 0 $$

Robert Israel
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  • So you can in fact use them in the same statement? Is this a common occurrence? – joshuaheckroodt Sep 05 '17 at 03:34
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    Yes. Your "$\forall i; {0,1,2}$" is wrong, because there is no indication of the relationship between $i$ and ${0,1,2}$. – Robert Israel Sep 05 '17 at 03:36
  • Thank you for the clarification. – joshuaheckroodt Sep 05 '17 at 03:42
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    This is very interesting. I never knew this—in fact, I always put the condition last! – gen-ℤ ready to perish Sep 05 '17 at 03:43
  • One also sees things like $f_i(x,y)\geq0, i\in{0,1,2},$ but that is a distinctly less formal way of writing the statement. – David K Sep 05 '17 at 03:45
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    @ChaseRyanTaylor Suppose we write "$x^2$ integrated from $0$ to $1$" as $x^2\int_0^1dx$? That probably looks weird to you. Putting $\forall$ to the right of things it's meant to apply to looks weird to me. (And the two notations actually have a lot in common.) – David K Sep 05 '17 at 03:52
  • @DavidK You make perfect sense, and the analogy helps with the understanding. And also, when would you prefer to write in formal mathematical statements rather than informal? – joshuaheckroodt Sep 05 '17 at 03:55
  • The order becomes really important in cases where you have both existential ($\exists$) and universal ($\forall$) quantifiers: $\exists x \forall y ; A(x,y)$ has a very different meaning than $\forall y \exists x ; A(x,y)$. – Robert Israel Sep 05 '17 at 19:48
  • And $\exists x ; A(x,y) \forall y$ would be terribly confusing. – Robert Israel Sep 05 '17 at 19:50