0
$\begingroup$

A map is one-to-one or injective if $a_1 \neq a_2 $ implies $f(a_1) \neq f(a_2)$. Equivalently, a function is one-to-one if $f(a_1) = f(a_2)$ implies $a_1 = a_2$.

Can I define it in the below mentioned way?

A map is one-to-one or injective if $f(a_1) \neq f(a_2)$ implies $a_1 \neq a_2 $. Equivalently, a function is one-to-one if $a_1 = a_2$ implies$f(a_1) = f(a_2)$ .

$\endgroup$
  • 1
    $\begingroup$ NO; an Injective function "is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain." Your proposed def, is simply the condition of "functionality": the same element has not two difefrent images. $\endgroup$ – Mauro ALLEGRANZA Nov 12 '17 at 13:40
2
$\begingroup$

You cannot use this. Any function satisfies this, because if $a_1 = a_2$, then $f(a_1)$ is the same as $f(a_2)$, because what this notation "means" intuitively, that you evaluate the function $f$ in the point $a_1$. But since $a_1 = a_2$, if you evaluate your function in $a_2$ you are actually evaluating the function in $a_1$, so they will result in the same result.

If you know about relations, what you have is that $a_1 R f(a_1)$, since $a_1 = a_2$ you also have $a_2 R f(a_1)$, and by definition we also have $a_2 R f(a_2)$. But because functions are the relations so that for all points $x$ there is one unique point $f(x)$ such that $x R f(x)$. This means that $f(a_1) = f(a_2)$

$\endgroup$
1
$\begingroup$

No, you definitely can't change the order! If you have two statements $A$ and $B$ then $$ (A\Rightarrow B) \Leftrightarrow (\neg B\Rightarrow \neg A). $$ But $A\Rightarrow B$ is not equivalent to $B\Rightarrow A$. You can never do it for any statements $A$ and $B$! Never!

In your case you can see it on the not injective function $f(x)=x^2$. It isn't injective since $f(1)=1=f(-1)$ and $1\neq -1$. But for $f(x)\neq f(y)$ you can conclude $x\neq y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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