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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?

N. F. Taussig
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Ham Sandwich
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  • You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample. – lEm Sep 09 '16 at 02:48
  • The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations. –  Sep 09 '16 at 02:58
  • It is however true that the function $$g : [0,\infty)\to [0,\infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $\sqrt{\bullet}$. –  Sep 09 '16 at 02:59
  • Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right? – Ham Sandwich Sep 09 '16 at 03:11
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    You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=\sqrt4$. –  Sep 09 '16 at 03:17
  • "For $x = 2, y = 4.$ So, $f(x) = 4,$ but $f(y) = 2$" No! $f(4) = 16$! I think what you are trying to say is $f(2) = f(-2) = 4$ hence $f$ is not an injection. Next paragraph, the contrapositive of the contrapositve? Do you mean the positive? If the positive is true then the contrapositive is true. – Doug M Sep 09 '16 at 03:32
  • I was tired. I was confusing myself. It was a stupid question. – Ham Sandwich Sep 09 '16 at 21:55

4 Answers4

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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.

Ross Millikan
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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.

Arji
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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.

  • First sentence: "no". You need to prove that $f(x_1)=f(x_2)\implies x_1=x_2$ –  Sep 09 '16 at 03:36
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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 $$ x\neq y\:\text{ but }\:f(x)=f(y)=4. $$
Daniele Tampieri
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Jacob Kraus
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  • Some books, professors usually keep referring 'y' as the output of the function i.e. y=f(x), the op may have become confused with that. But yeah for sure 'y' is any dummy point. – Aayush Karna Aug 16 '22 at 14:24