I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).
However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.
Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.
This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($\sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.
However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.
Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?
$\endgroup$ 74 Answers
$\begingroup$The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x \neq y$. The contrapositive fails as well because you have $x \neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.
$\endgroup$ $\begingroup$to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 \Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 \Rightarrow f(x_1)=f(x_2)$ yet $x_1 \neq x_2$, making $f(x)$ not injective.
$\endgroup$ 1 $\begingroup$Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.
If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.
The other definition is just the other way around. Like $A \Rightarrow B$ is equal to $\neg B \Rightarrow \neg A$.
It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.
Let me take an example. Lets show that $f(x) = x^3$ is injective.
We take general $x,y \in \mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.
The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $\sqrt{y} = x$ but that not enough. You also get $\sqrt{y} = - x$. For example $\sqrt{y = 1} = \pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.
$\endgroup$ $\begingroup$I think that the syntax of the definition from your class is the point of confusion. The class definition depends on functions being defined as $$ f(x)=(\text{expression of }x) $$ rather than as $$ y=(\text{expression of }x). $$ Consequently, the "$y$" in "$f(y)$" is just some dummy variable that gets input into $f$ and is unrelated to $x$. So, for your example, $f(x)=x^2$, $f(x)= 4$ for $x=2$ and $f(y)=f(4)=16$ for $y=4$. Again, note that $y$ is unrelated to $x$.
Finally, in order to show that $f(x)=x^2$ isn't injective, you can start with the definition or with its contrapositive, as you stated:
- Starting with the definition, $$ f(-2)=f(2)\:\text{ but }-2\:\text{ isn't equal to }2.$$
- Starting with the contrapositive, let's consider $x=2$ and $y=-2$: then $$ x\neq y\:\text{ but }\:f(x)=f(y)=4. $$
