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 $\langle expr \rangle = \langle expr \rangle$?
  • 4
    $\begingroup$ smbc-comics.com/comic/math-puzzles $\endgroup$
    – 000
    Feb 14, 2017 at 23:04
  • $\begingroup$ Interestingly; while only things with truth values can be negated: questions about things without truth values can be negated. So if you were to define 0 as "does 0 exist?" and 1 as "does 1 exist?", (where a thing is defined to exist when it has a truth value of T). You would end up with the statement: Is the answer to the question; "does 1 exist at the same time as 0" true or not? And the negation of this will be does 1 exist when 0 does not and does 0 exist when 1 does not? This isn't an answer to the OP but I thought you might like to know about it. $\endgroup$
    – user400188
    Feb 15, 2017 at 3:08
  • $\begingroup$ So what was the 'official' solution to the homework problem? $\endgroup$
    – lalala
    Sep 21, 2017 at 8:32
  • $\begingroup$ Even if you did $A=B$ with $A=\{0\}$ and $B=\{1\}$, you still wouldn't do $A^C \ne B^C$. You're thinking of things like $(A \cup B)^C = A^C \cap B^C$...I guess? $\endgroup$
    – BCLC
    Apr 16, 2018 at 16:09

3 Answers 3


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.

  • 4
    $\begingroup$ Great answer! Thank you. The logical equivalence to $(0 > 1) \vee (0 < 1)$ made it a lot easier to understand. $\endgroup$
    – m0meni
    Feb 14, 2017 at 19:05
  • 6
    $\begingroup$ The part about $0>1$ or $1>0$ is true for the real numbers, but in other contexts there might not be an order, and even if there is an order it might not be linear. $\endgroup$
    – Asaf Karagila
    Feb 15, 2017 at 5:49
  • $\begingroup$ "Finally, note that ¬0 is not well-formed" unless you're programming in C :) $\endgroup$ Oct 31, 2022 at 4:10

$0\neq 1$ 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.


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

  • 1
    $\begingroup$ No matter how numbers are introduced $1\neq0$ is the way do go. Your suggestion of what 1=0 might be written as makes no sense, as far as I can tell! $\endgroup$ Feb 15, 2017 at 5:52
  • 1
    $\begingroup$ @ Mariano Suárez-Álvarez: In ZF the numbers are defined as sets. So equality of two numbers is set equality. For the definition of the empty set, I used the one from Dieudonne. For the equality check, I skipped the trivial part (empty set subset of any set). Which part of this doesnt make sense? Could you clarify? $\endgroup$
    – lalala
    Feb 15, 2017 at 7:50

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