Negation of $0 = 1$

Solution 1:

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.

Solution 2:

$0\neq 1$ 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.

Solution 3:

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