If it were not assumed that f is a (well-defined) function, this would not be equivalent.I know this definition:
∀a,b∈Dom(f):a=b⟹f(a)=f(b)But I was wondering if this definition is correct as well:
a=b⟺f(a)=f(b)∀a,b∈Dom(f)Thanks in advance![]()
Ok, but is it correct to say that if a function f(x) satisfies that condition, then it's injective?If it were not assumed that f is a (well-defined) function, this would not be equivalent.
Your equivalence corresponds to two implications. The definition you start with has an implication in only one direction. Do you see the difference?
So in proving that a function satisfies your alternative "definition", you would be proving two different things. What are they?
As a result, I would call it "overkill"; it requires more than the minimum that needs to be said. Can you rewrite your version as a single implication?
Yes.Ok, but is it correct to say that if a function f(x) satisfies that condition, then it's injective?
Ok, thank you, now it's clear.Yes.
What I said is that your condition is "overkill": it is more than sufficient, not less. It shows both that f is a function and that it is one-to-one. And sometimes that's what you need to do. Do you see that? That's what I asked.
Oooooh! Okay, now I see what he meant by that. Thank you!A ⟹B is equivalent to ~B⟹~A. That is you don't need double implication--one direction is enough as Dr Peterson stated.