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I want to prove that some functions are not one-to-one.

Are both these proofs written correct?

  1. $\exists\,x_1=1,\,x_2=-1\;[1\neq -1 \wedge f(1)=f(-1)]$

  2. $\exists\,x_1=1,\,x_2=-1\;[x_1\neq x_2 \wedge f(x_1)=f(x_2)]$

Angelo
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DScounterGO
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  • The use of notation has variations. They could be ok or not, depending on the context. Both look understandable, and expressing the existence of one example proving that a function is not one-to-one, from the usual ways the symbols in them are commonly used. – plop Jan 10 '23 at 17:01
  • I haven't seen either of these notations, but the notation is clear. So it depends really on the notation of your class/book/professor. – Thomas Andrews Jan 10 '23 at 17:05
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    I would normally write $f(1)=f(-1)$ and $1\neq -1,$ so $f$ is not one-to-one. Basically, $\exists$ is a formal symbol, and your notation is not a formal usage of that symbol. – Thomas Andrews Jan 10 '23 at 17:07
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    Correct? Yes. Clear? Maybe not. – Sean Roberson Jan 10 '23 at 17:07
  • Context? Missing. I suppose $f\colon \Bbb R\rightarrow \Bbb R$ is a function. Or is the domain different? Over $\Bbb F_2$, say, $1=-1$ is true. – Dietrich Burde Jan 10 '23 at 17:08
  • I think it’d be better to just write in normal sentences. – littleO Jan 10 '23 at 17:14
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    Thank all, I have to use that notation. My teacher use first one but I was curious will she give me points for the second one i used on the test. – DScounterGO Jan 10 '23 at 17:24

1 Answers1

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I have never seen the notation $\exists_{x_1=1}$. It is clear what you mean, but unless you are following the notation of your instructor it would be best to avoid this.

I would personally suggest just saying

Since $f(-1) = f(1)$ we can see that $f$ is not one-to-one.

If you wanted to be extremely formal you could say something like:

A function $f:X \to Y$is said to be one to one if $$\forall x_1 \in X \forall x_2 \in X f(x_1) = f(x_2) \implies x_1 = x_2$$ We want to show that our particular $f$ is not one-to-one. Assume $f$ is one-to-one. Notice that $f(-1) = f(1)$ for our particular $f$. So if $f$ were one-to-one we would have that $-1=1$, which is absurd. Thus $f$ is not one-to-one.

Steven Gubkin
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  • Yup. Using both "there exists" and exhibiting a specific value is redundant, kinda confusing, and makes it harder for your readers to understand your intention. If your prof requires it, then I guess you have to use it, but I encourage you to drop it as soon as you get out of that class – JonathanZ Jan 10 '23 at 19:14