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Is it possible to prove that $1+1 = 2$? Or rather, how would one prove this algebraically or mathematically?

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    $\begingroup$ In fact, this occasionally useful proposition was proved by Russell and Whitehead. $\endgroup$ – Srivatsan Dec 29 '11 at 23:19
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    $\begingroup$ You should go read up on set theory, particularly the Peano Axioms. This is a consequence of how the natural numbers and addition on them is defined. $\endgroup$ – Potato Dec 29 '11 at 23:19
  • $\begingroup$ Google "Russel and Whitehead's proof that 1+1=2"! Good Luck. $\endgroup$ – draks ... Dec 29 '11 at 23:20
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    $\begingroup$ In many formal developments, "$1+1=2$" is very close to being the definition of $2$ -- so don't expect any very exciting proof. $\endgroup$ – Henning Makholm Dec 29 '11 at 23:24
  • $\begingroup$ I tried to retag, since [proof-theory] was completely wrong. I hope the tag is less wrong now, but if someone thinks it's still not fully fitting - please retag. $\endgroup$ – Asaf Karagila Dec 29 '11 at 23:26
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In a very "raw" sense the symbol $2$ is just a shorthand for $1+1$. There is really not much to prove there.

If we want to talk about proof we need axioms to derive the wanted conclusion from. Let us take the "usual" axioms of the natural numbers here, namely Peano Axioms. These axioms give a very basic rules which describe the natural numbers, and from them we will derive $1+1=2$.

In those axioms the numbers $1$ and $2$ don't exist. We have $0$ and we have $S(n)$, which can be thought of as a "successor function" which generates the next number, so to speak. In this system $1$ is a symbol for $S(0)$ and $2$ is a symbol for $S(S(0))$.

Addition is defined inductively, that is $x+0=x$, $x+S(y)=S(x+y)$. From this we can derive:

$$1+1 = 1+S(0) = S(1+0) = S(1)$$ Now replace $1$ with its "full form" of $S(0)$ and we have: $$S(0)+S(0) = S(S(0)+0) = S(S(0))$$

Which is what we wanted.


In a more general setting, one needs to remember that $0,1,2,3,\ldots$ are just symbols. They are devoid of meaning until we give them such, and when we write $1$ we often think of the multiplicative identity. However, as I wrote in the first part, this is often dependent on the axioms - our "ground rules".

If we consider, instead of the natural numbers, the binary digits $0,1$ with addition $\bmod 2$, then we have that $1+1=0$. Now you can argue that of course that $0\neq 2$, however in this set of axioms (which I have not expressed explicitly here) we can prove that $0=2$, where $2$ is the shorthand for $1+1$ and $0$ is the additive neutral element.

Actually, just writing $1+1=0$ is a proof of that.

I can't really stress that enough, because this is a very important part of mathematics. We often use some natural notion, such as the natural numbers, before we define it. Later we define it "to work as we want it to work" and only then we have a formal framework to work with.

These axioms, these frameworks, those often remain "in the shadows" and if you don't know where to look for them then you are less likely to find them.

This is why the question "Why $1+1=2$?" is nearly meaningless - since you don't have a formal framework, and the interpretation (while assumed to be the natural one) is ill-defined.

On that same note, this question is also very important when starting with mathematics. It helps to you understand what there is to prove, and how to do it. Of course this too lack of context because one would have to define what is a proof, and all the other things first.

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    $\begingroup$ That is a really good answer to a annoying question that comes up a lot. I can't count the amount of people who says you can't prove 1+1=2. $\endgroup$ – simplicity Dec 30 '11 at 0:38
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    $\begingroup$ I'm always confused why people find it annoying - maths at hs was boring until I tried to prove 1+1=2 one da. Being a naive 17 year old I was soon swepped towards set theory and then ended up studying the subject at University and have been offered placings after my exams with my current uni for further study if things go like I want them to. All in all I would say that question changed the course of my adult life - I never really took work seriously until then! Not that I do now, I guess I should use the word respect, though that's another tale altogether ... $\endgroup$ – Adam Dec 30 '11 at 3:20
  • $\begingroup$ This seems like a pretty good answer. If it still doesn't solve the author's problem, what can you do? You can only do your best to explain it. You have to start somewhere. The natural numbers can be thought of as the numbers that can be constructed from 0 and the successor operation. We could decide that when we write 1 + 1, we really mean S(0) + S(0) and when we write 2, we really mean S(S(0)) so the real task is to show that S(0) + S(0) = S(S(0)). Now addition can be defined inductively to derive S(0) + S(0) = S(S(0)). $\endgroup$ – Timothy Oct 19 at 3:17
  • $\begingroup$ @simplicity I think we should keep them to show that some people have a real different way of thinking which I respect. If they want to figure out how to further break down the proof into one they accept, I respect them for that. Maybe they were looking for an answer like this answer. Some people only accept proofs they find very intuitive. Maybe it's a sign that they're a slow and careful person who thinks about a problem slowly and deeply. Maybe as a result of being a slow deep thinker, they will reject the axiom of choice and reject proofs in Naive set theory. Try adding 1000 + 20 + 30 + $\endgroup$ – Timothy Oct 19 at 3:25
  • $\begingroup$ 1000 + 1030 + 1000 + 20 and see what you get. Did you get it right? Are you sure? If you got the answer 5000, I can tell you that is not the right answer. I didn't invent the problem. I got it somewhere. $\endgroup$ – Timothy Oct 19 at 3:28
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1 is the convention name of 0++, 2 is the convention name of (0++)++, so what you need to prove is 0++ + 0++ = (0++)++.

$+$ is defined as:

  • $0+m := m$
  • $(n++)+m := (n+m)++$

Just apply this definition to the left of the equation and you will get the right. You don't even need to know what '0' or '++' is, all you do is shuffling some sequences of symbols according to some rules.

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  • $\begingroup$ Why is this getting no upvotes? $\endgroup$ – Pacerier May 15 '17 at 17:01
  • $\begingroup$ @Pacerier: Possibly because (a) it came five years after Asaf's answer that already makes the same substantial point, and (b) uses a strange idiosyncratic notation for the successor function. The notation may be an attempt at analogy from programming languages in the C tradition, but that doesn't even work so well: In C, a++ evaluates to the original value of a and as a side effect changes what a means subsequently to the successor of that original value. $\endgroup$ – Henning Makholm May 15 '17 at 20:07
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One standard set of axioms says that $0$ is a natural number, and for every natural number $n$ the successor of that number, $S(n)$ is a unique natural number.

Then there are these two axioms of addition (amongst other axioms):

$$a+0=a\\a+S(b)=S(a+b)$$

Now, if we define $1$ as $S(0)$ and $2$ as $2=S(S(0))=S(1)$ then:

$$1+1=S(0)+S(0)=S(S(0)+0)=S(S(0))=2$$

However, this is just abstract nonsense.

You can also have other axiom systems with no successor function, and then usually $2$ is just defined as $1+1$.


A lambda-calculus proof might start with definitions:

$$\begin{align}1&=\lambda\, f.f\\ 2&=\lambda\, fx.f(f(x))\\ +&=\lambda\,mnfx.(mf)(nfx) \end{align}$$

Then you can prove that $+11fx=2fx$ for any $f,x$.

In lambda calculus, this will mean that there is an equivalence between $+11$ and $2$, but they are not necessarily equal. (Consider lambda calculus to describe programs for computation - two programs are equivalent if they compute the same thing, but they might not be identical as programs.)

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