One to one function

Question

Following is the way of checking the one to one function

or

Can i use bi implication in between these?If not,then why?

Answer 1

I think the second one is the correct representation of the one to one function.

i.e for all of x and y that belongs to x if f(a)=f(b) then a=b because what one to one or injection function is that each and every value in the domain should only maps to one value in codomain .

Thus this statement actually says that : if the value of function for two different values are same it implies that they are equal.
Now why we cant use bi implication :

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The answer lies in the reasoning provided above that we only need to check the implication when the value generated by the function is same.
If they are not same why we need to check the other side ?
Thus if the statement F(a) is not equal to F(b) we dont need to check the other part of the implication because we are checking whether its one to one or not and it is !

Hope you got it?

Comments

Actually both are true .statement  1 is contra positive of another so both are true.

Answer 2

both are true

∀ a,b ∈ x,,, if f(a)=f(b) ⇒ a=b

and its contrapositive is

∀ a,b ∈ x,,, if a ≠ b ⇒  f(a) ≠ f(b)

and u know direct and contrapositive are equivalent (ie. A⇒B ⇔ ! B ⇒ ! A)

and  inverse and converse are equivalent

also u can think for 1st one

for ONE-ONE whenever a ≠ b then f(a) and f(b) also be different(if (a) and f(b) are same then it will be many - one)

Comments

no,,, it is converse

how can u say A⇒B is equivalent to B⇒A

and u know A⇒B is !A+B

B⇒A is !B+A

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if  f(a)!=f(b) => a!=b.

Can i use this to check one to one?

I know that it is converse of what is being mentioned in question.Just to clarify some doubt can you give me a small example that will satisfy my above constion mentioned in comment but voilate one to one ?

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yes u can use it

but it will be simple

if u use

f(a)=f(b) ⇒ a=b

Answer 3

Let  p = a not equal to b

        q = f(a) not equal to f(b)

Now your first statement is p->q . Bi-implication and implication only differs when p is false and q is true.

Here both your statements mean the same as one is contrapositive of another.

Now if u add a bi implication between the two then it will be

(P->Q) <->(~Q -> ~P)

If it is a one to one function then both will be true and hence it is true.

If it's not a one to one function , suppose when P is true and Q is false, then it will be

F <-> F which is true .

This should not be the case as answer should be false because is not a one to one function .

I don't know if I have understood your question properly. Do let me know if you have any queries.

Comments

Let f be a function whose domain is a set X. The function f is said to be injective provided that for all a and b in X, whenever f(a) = f(b), then a = b; that is, f(a) = f(b) implies a = b.  Equivalently, if a ≠ b, then f(a) ≠ f(b).

Symbolically,

which is logically equivalent to the contrapositive,

both statements are correct