I'm taking my first proof-heavy class (real analysis), and one practice problem on the first homework is to write the negation of
$$0 = 1$$
My immediate thought was that it would simply be
$$0 \neq 1$$
but I'm not 100% certain of that answer. I was wondering if there's more to it than just inverting the $=$ sign, and perhaps you'd distribute the negation like
$$\neg 0 \neq \neg1$$
but logically that doesn't make sense to me. I've tried looking this up, but a statement as simple as $0 = 1$ has given me a hard time finding any good search results.
Basically to break down my questions:
- Is $0 \neq 1$ right?
- if so, do I prove it somehow?
- if not, how do you negate expressions like $\langle expr \rangle = \langle expr \rangle$?
3 Answers
$\begingroup$Your negation is correct. Note also that
$$\neg(0=1) \equiv 0 \neq 1 \equiv (0 > 1) \vee (0 < 1).$$
($\equiv$ means logical equivalence and $\vee$ stands for inclusive "or".)
Finally, note that $\neg 0$ is not well-formed. Only sentences (things with truth-values) can be negated, and $0$ is not a sentence; it's a numeral.
$\endgroup$ 2 $\begingroup$$0\neq 1$ is correct. $\neg 0=\neg 1$ is hard to interpret; what does $\neg 0$ even mean?
There's not much to prove here, I think you're just being asked to demonstrate understanding that $\neg (a=b)$ means $a\neq b$. In fact, that usually how $\neq$ is defined.
$\endgroup$ 0 $\begingroup$It depends a bit on your class. Were the natural numbers defined as sets? In Zermelo-Fraenkel 0 would be the empty set, and $1$ the set containing the empty set. So $0=1$ can be written as: for all $x$ in $1$ : $x \neq x$ Negation would be: there exists an x in 1 : $ x=x$. It depends on the definitions used in class. Although if your class didn't introduce the numbers, probably the obvious answer $ 0\neq 1$ is requested.
Edit: an introduction to the set theoretic construction of the natural numbers can be found in the highest rated answer here: Set theoretic construction of the natural numbers
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