Definition of injective function

[elvis]

I know this definition:
[math]\forall a, b \in \text{Dom}(f) : a \ne b \implies f(a) \ne f(b)[/math]But I was wondering if this definition is correct as well:
[math]a = b \iff f(a) = f(b) \qquad \forall a, b \in \text{Dom}(f)[/math]Thanks in advance

 

[Dr.Peterson]

I know this definition:
[math]\forall a, b \in \text{Dom}(f) : a \ne b \implies f(a) \ne f(b)[/math]But I was wondering if this definition is correct as well:
[math]a = b \iff f(a) = f(b) \qquad \forall a, b \in \text{Dom}(f)[/math]Thanks in advance

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?

 

[elvis]

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?

Ok, but is it correct to say that if a function [math]f(x)[/math] satisfies that condition, then it's injective?

 

[blamocur]

The two definitions look equivalent to me since for every function [imath]f[/imath] we have [imath]a=b \Rightarrow f(a) = f(b)[/imath]

 

[Dr.Peterson]

Ok, but is it correct to say that if a function [math]f(x)[/math] satisfies that condition, then it's injective?

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.

 

[elvis]

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.

Ok, thank you, now it's clear.

 

[topsquark]

Ummmm...
If
[imath]x_1 \neq x_2 \implies f(x_1) \neq f(x_2)[/imath]

then isn't the contrapositive
[imath]f(x_1) = f(x_2) \implies x_1 = x_2[/imath]?

-Dan

 

[Steven G]

A ⟹B is equivalent to ~B⟹~A. That is you don't need double implication--one direction is enough as Dr Peterson stated.

 

[topsquark]

A ⟹B is equivalent to ~B⟹~A. That is you don't need double implication--one direction is enough as Dr Peterson stated.

Oooooh! Okay, now I see what he meant by that. Thank you!

-Dan