# Is 2+2=4 an identity?

I know this seems like a silly question, but someone was trying to debate with me about how $2+2=4$ should be called an identity and not an equation. I mentioned how it has no variables and isn't true for all numbers, but they claimed you could plug in any value for its variables(which don't exist) and it would be true and so is an identity. Which is kind of true in a round about way, but I still don't think anyone would call it an identity in a casual situation.

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It's both an equation and an identity, but not a particular interesting example of either. – Alex Becker Mar 21 '12 at 3:48
To clarify, as I understand it equation refers to the form of the logical statement, while identity refers to the truth of the statement itself. – Alex Becker Mar 21 '12 at 3:54
So they're saying $1 = 1$ is actually $$\forall x,y,z,w,u,v,\ldots (1 = 1)$$ Hm.. – user2468 Mar 21 '12 at 4:02
This is somewhat related. – Asaf Karagila Mar 21 '12 at 18:53

I consider an equation to be a statement that two things (the two sides of the equation, in particular) are equal. In this sense, $2 + 2 = 4$. But perhaps I also know that the set of $b$ in $B$ with some property satisfy the equation $b^4 + b^3 + b = 0$.

I consider an identity to be a relation that's tautologically true. Thus $2 + 2 = 4$ or $\sin^2 + \cos^2 = 1$ are identities. But $b^4 + b^3 + b = 0$ is different from this, and so isn't.

So yes, I consider it an equation and an identity. And as Alex said, it's not particularly interesting as either.

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Essentially, $2+2=4$ by the definition of "$4$". Getting down to basics:

A field is a set that satisfies some axioms that give that set features that we intuitively understand real world numbers to have. (A commenter has pointed out that minimally, all that is needed is the axioms explicitly listed below, which by themselves give a semigroup.) Among the axioms are

• There's a way to "add" two elements together to get a (possibly) new element.
• There's a special element $1$ having some properties that aren't relevant to this question. (But what we need to make use of is that $1$ is being defined as a special element of the set.)
• Associativity of $+$: for all $x$, $y$, $z$ in the set, $(x+y)+z=x+(y+z)$.
• More axioms, not relevant right now.

Once the axioms are established, we can go about defining the digits.

• $2$ is defined as $1+1$. That is, $2=1+1$, and this is not an equation to be solved, nor an identity; it's the definition of "$2$".
• $3$ is defined to be $2+1$.
• $4$ is defined to be $3+1$.

Now as a consequence of these axioms and definitions, and the transitivity of $=$, we find that

\begin{align} 2+2 &=2+(1+1) \\ &=(2+1)+1\\ &=3+1\\ &=4 \end{align}

I would personally consider such an equation (arising as a consequence of axioms and definitions) to be an identity, although it is so close to the definition of "$4$" that it is not an interesting one. (And moderate your argument by pointing out that identities are equations too, since they state that two things are equal to each other, albeit independently of any variables.)

Of course this explains why $2+2=4$ in the context of an abstract algebraic structure. If you accept the numbers that you interact with daily as part of a field, then this is OK. Otherwise the conversation gets philosophical and we have to start asking things like, what is $2$, anyway? And what is $4$? And what is addition?

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The statement $2+2=4$ also makes sense in the context of $\mathbb{Z}$ (a ring and an abelian group), $\mathbb{N}$ (a monoid) and $\mathbb{N}-\{0\}$ (a semigroup) so I'm not sure that the introduction of the field axioms is that enlightening. – Chris Taylor Mar 21 '12 at 8:44
@MDCCXXIX My point is to define digits like 2 and 3 in terms of established defined objects. And a field at least includes 1 in its definition. I've left out irrelevant field axioms and only included the additive identity axiom to illustrate another defined quantity. And while other algebraic structures are fine, my guess is that OP has real numbers in mind. – alex.jordan Mar 21 '12 at 18:30