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Suppose we had a function defined over the complex numbers:

$ f(x)= \begin{cases} 1&\text{if } x\in\mathbb{R}\\ 0&\text{if } x\not\in\mathbb{R} \end{cases} $

And we are asked to prove that $f(x)=1$ iff $x$ is real.

Obviously, if $x$ is real, then $f(x)=1$ because that is given in the function's definition. Is it enough to use the following to justify the converse and thus complete the iff statement?

Because a complex number is either real or non-real, the function will produce some output for all complex numbers. For all real numbers, the function produces an output of $1$; and for all non-real numbers, the function produces an output of $0$. The function will only produce an output of $1$ if $x$ is real; otherwise, it would produce an output of $0$. Thus, if $f(x)=1,$ then $x$ is real.

I think that is enough to justify the iff statement, but it feels too wordy, and I don't have much experience writing proofs like this. Is what I wrote sufficient to complete the proof, and is there a good way to make it less wordy?

Arcturus
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  • You're right that it's a bit wordy. When people are first learning proofs, they often explain (unnecessarily) the definitions in the beginning of their proofs (which is what you did here). For example, you don't need to explain what a function is or that it always produces an output (the definition and domain of the function are enough for that). – Michael Burr Jun 30 '15 at 01:14
  • Hint: Use the contrapositive for the second direction in the iff (it'll be easier). – Michael Burr Jun 30 '15 at 01:14

1 Answers1

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Here's a simpler way to do it: Use the fact that $p \Leftrightarrow q \equiv (p \Rightarrow q) \wedge (\neg p \Rightarrow \neg q)$

So, if $x$ is real, then $f(x) = 1$.

If $x$ is not real, then $f(x) = 0$, i.e. $f(x) \neq 1$.

This proves the statement.

MT_
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