# Negation of $0 = 1$

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 $\langle expr \rangle = \langle expr \rangle$?
• smbc-comics.com/comic/math-puzzles
– 000
Feb 14, 2017 at 23:04
• 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. Feb 15, 2017 at 3:08
• So what was the 'official' solution to the homework problem? Sep 21, 2017 at 8:32
• 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?
– BCLC
Apr 16, 2018 at 16:09

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.

• Great answer! Thank you. The logical equivalence to $(0 > 1) \vee (0 < 1)$ made it a lot easier to understand. Feb 14, 2017 at 19:05
• 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. Feb 15, 2017 at 5:49
• "Finally, note that ¬0 is not well-formed" unless you're programming in C :) 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

• 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! Feb 15, 2017 at 5:52
• @ 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? Feb 15, 2017 at 7:50