# How do I convince someone that $1+1=2$ may not necessarily be true?

Me and my friend were arguing over this "fact" that we all know and hold dear. However, I do know that $1+1=2$ is an axiom. That is why I beg to differ. Neither of us have the required mathematical knowledge to convince each other.

And that is why, we decided to turn to stackexchange for help.

What would be stack's opinion?

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Related: math.stackexchange.com/questions/95069/… (possibly a duplicate) –  Asaf Karagila Nov 23 '12 at 9:22
Thank you. Maybe we could delete my thread. I will leave it up to you. I couldn't find. Similar question, –  Aces12345 Nov 23 '12 at 9:25
@Asaf: definitely related. Unfortunately the two question are worded in opposite senses (one asks for a positive proof, one asks for a negative intuition), so I would not advice closing and merging. –  Willie Wong Nov 23 '12 at 9:45
This question was popularised by Einstein .This was discussed while developing a formal logical language for computers by Bertrnd Russel ,Whitehead and others.The Truth table used in Computer logic -> [ 0+0 =0 ;1 +0 =1 ;0+1 =1 ; 1+1 =1 ] , uses 1 +1 =1 When a circuit is switched ON ,it has state ' one' --when it is switched OFF , it has a state ' zero ' . In Genetic Algorithm , the symbols ' one ' and ' zero ' are used . But the rules of permuattion and combination are different. Arithmatics is one form of Symbolic Logic , where 1+1 =2 Depends on what you mean by the Symbols ' one' and 'plus ' –  B.Sahu Nov 28 '12 at 7:19
Teach your friend binary. –  Parth Kohli Dec 17 '12 at 10:19

It seems that you and your friend lack the mathematical knowledge to handle this delicate point. What is a proof? What is an axiom? What are $1,+,2,=$?

Well, let me try and be concise about things.

• A proof is a short sequence of deductions from axioms and assumptions, where at every step we deduce information from our axioms, our assumptions and previously deduced sentences.

• An axiom is simply an assumption.

• $1,+,2,=$ are just letters and symbols. We usually associate $=$ with equality, that is two things are equal if and only if they are the same thing. As for $1,2,+$ we have a natural understanding of what they are but it is important to remember those are just letters which can be used elsewhere (and they are used elsewhere, often).

You want to prove to your friend that $1+1=2$, where those symbols are interpreted as they are naturally perceived. $1$ is the amount of hands attached to a healthy arm of a human being; $2$ is the number of arms attached to a healthy human being; and $+$ is the natural sense of addition.

From the above, what you want to show, mathematically, is that if you are a healthy human being then you have exactly two hands.

But in mathematics we don't talk about hands and arms. We talk about mathematical objects. We need a suitable framework, and we need axioms to define the properties of these objects. For the sake of the natural numbers which include $1,2,+$ and so on, we can use the Peano Axioms (PA). These axioms are commonly accepted as the definition of the natural numbers in mathematics, so it makes sense to choose them.

I don't want to give a full exposition of PA, so I will only use the part I need from the axioms, the one discussing addition. We have three primary symbols in the language: $0, S, +$. And our axioms are:

1. For every $x$ and for every $y$, $S(x)=S(y)$ if and only if $x=y$.
2. For every $x$ either $x=0$ or there is some $y$ such that $x=S(y)$.
3. There is no $x$ such that $S(x)=0$.
4. For every $x$ and for every $y$, $x+y=y+x$.
5. For every $x$, $x+0=x$.
6. For every $x$ and for every $y$, $x+S(y)=S(x+y)$.

This axioms tell us that $S(x)$ is to be thought as $x+1$ (the successor of $x$), and it tells us that addition is commutative and what relations it bears with the successor function.

Now we need to define what are $1$ and $2$. Well, $1$ is a shorthand for $S(0)$ and $2$ is a shorthand for $S(1)$, or $S(S(0))$.

Finally! We can write a proof that $1+1=2$:

1. $S(0)+S(0)=S(S(0)+0)$ (by axiom 6).
2. $S(0)+0 = S(0)$ (by axiom 5).
3. $S(S(0)+0) = S(S(0))$ (by the second deduction and axiom 1).
4. $S(0)+S(0) = S(S(0))$ (from the first and third deductions).

And that is what we wanted to prove.

Note that the context is quite important. We are free to define the symbols to mean whatever it is we want them to mean. We can easily define a new context, and a new framework in which $1+1\neq 2$. Much like we can invent a whole new language in which Bye is a word for greeting people when you meet them, and Hi is a word for greeting people as they leave.

To see that $1+1\neq2$ in some context, simply define the following axioms:

1. $1\neq 2$
2. For every $x$ and for every $y$, $x+y=x$.

Now we can write a proof that $1+1\neq 2$:

1. $1+1=1$ (axiom 2 applied for $x=1$).
2. $1\neq 2$ (axiom 1).
3. $1+1\neq 2$ (from the first and second deductions).

If you read this far, you might also be interested to read these:

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This amazing answer is definetely what i am looking for. –  Aces12345 Nov 23 '12 at 10:11
Just a note: you can find examples in the nature where (in some sense) $1+1=2$ fails, for example: counting humans as ones, Husband+Wife$=2$ for first, but it may be well possible that Husband+Wife$=3$ or even more as time passes. –  Berci Nov 23 '12 at 11:14
@MJD: Well, I had thought about doing that. I felt like the post was already long and that the explanation that two things are equal if and only if they are the same sufficed. –  Asaf Karagila Nov 23 '12 at 13:39
@Thomas: If one wishes to have a ring (or field) structure on a Boolean algebra, then one needs to use exclusive or. However as a structure on its own accord Boolean algebra's $+$ is bitwise or, or union of sets. –  Asaf Karagila Nov 24 '12 at 0:03
@KCd: I agree, this is not a very natural context. However this is exactly what I'd written in the answer, and in the previous comments: there is no natural context in which $2$ is not defined as $1+1$, simply because we naturally define $2$ as $1+1$. So any context in which this fails has to be artificial. I prefer writing distinct axioms that will give out a clear proof rather than "mixing labels" on a familiar context which is confusing. This is an issue throughout mathematics where counterexamples might be abundant, but they are all pathological. Like non-measurable sets. :-) –  Asaf Karagila Nov 24 '12 at 17:33

I personally would define the symbol $2$ as $1+1$.

However, depending on what you assume, this does not ensure $2\ne 0$. What do people usually assume? Often, mathematicians work with the notion of a commutative ring. Think of the integer numbers in the following definition: A commutative ring is a mathematical structure with an addition operator $+$ and a multiplication operator $\times$, containing elements $1$ and $0$ subject to the conditions that

• $a + b = b + a$ and $a\times b = b\times a$
• $a + (b + c) = (a + b) + c$ and $a \times (b \times c) = (a \times b)\times c$
• $a + 0 = a$ and $a\times 1 = a$
• For each $a$ there exists $b$ with $a+b=0$, we write $b=-a$
• We have $a\times(b+c)=(a\times b) + (a\times c)$

Now the integer numbers clearly satisfy these conditions, and we all have a good idea what $2$ means there. However, consider $\{0,1\}=:\mathbb F_2$. It also satisfies the conditions, if we perform addition and subtraction as usual, but decree that $1+1=0$. Now you have $2=0$ inside $\mathbb F_2$.

However, note that nothing stops you from limiting yourself to mathematical structures where $1+1\ne 0$. In fact, quite often, one limits himself to structures where $1+\cdots+1\ne 0$ no matter how many times you add $1$ to itself.

In the end, it's all just a matter of definitions. But truly, I would always say $2=1+1$ simply because the symbol $2$ should reasonably be defined as that.

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Those interested in pushing this question back further than Asaf Karagila did (well past logic and into the morass of philosophy) may be interested in the following comments that were written in 1860 (full reference below). Also, although Asaf's treatment here avoids this, there are certain issues when defining addition of natural numbers in terms of the successor operation that are often overlooked. See my 22 November 2011 and 28 November 2011 posts in the Math Forum group math-teach.

Consider this case. There is a world in which, whenever two pairs of things are either placed in proximity or are contemplated together, a fifth thing is immediately created and brought within the contemplation of the mind engaged in putting two and two together. This is surely neither inconceivable, for we can readily conceive the result by thinking of common puzzle tricks, nor can it be said to be beyond the power of Omnipotence. Yet in such a world surely two and two would make five. That is, the result to the mind of contemplating two two’s would be to count five. This shows that it is not inconceivable that two and two might make five: but, on the other hand, it is perfectly easy to see why in this world we are absolutely certain that two and two make four. There is probably not an instant of our lives in which we are not experiencing the fact. We see it whenever we count four books, four tables or chairs, four men in the street, or the four corners of a paving stone, and we feel more sure of it than of the rising of the sun to-morrow, because our experience upon the subject is so much wider and applies to such an infinitely greater number of cases.

The above passage comes from:

James Fitzjames Stephen (1829-1894), Review of Henry Longueville Mansel (1820-1871), Metaphysics; or, the Philosophy of Consciousness, Phenomenal and Real (1860), The Saturday Review 9 #244 (30 June 1860), pp. 840-842. [see page 842]

Stephen’s review of Mansel's book is reprinted on pp. 320-335 of Stephen's 1862 book Essays, where the quote above can be found on page 333.

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Here's an argument to consider. Take the following three propositions:

(A) 1 + 1 = 2

(B) If there is exactly one $F$ and exactly one $G$ (and no $F$ is $G$), then there are exactly two things which are $F$-or-$G$.

(C) $(\exists x)(Fx \land \forall y(Fy \to y = x)$, $(\exists x)(Gx \land \forall y(Gy \to y = x) \vdash$ $\quad\quad\quad(\exists x)(\exists y)(\neg x = y \land (Fx \lor Gx) \land (Fy \lor Gy) \land (\forall z)(Fz \lor Gz \to (z = x \lor z = y))$

Now, it is a vexed question what exactly the relation is between (A) and (B). But it is very plausible to say that if (A) is indeed interpreted as a statement of ordinary schoolroom arithmetic (and not as a mere de-intepreted string of formal symbols), then so understood, (A) is true iff, for any $F$, $G$, (B) is true. But (B), with numerical quantifiers, can be regimented as (C), with ordinary quantifiers. And (C) is a straight logical theorem of first-order logic.

So (C) is true as a matter of logical necessity, hence (B) is. And it looks a priori that (A)'s truth necessarily goes with the truth of (B) for any $F$, $G$. So it follows that (A), understood the ordinary way, is indeed necessarily true. too.

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My I "elaborate" a little bit your answer ? If I apply Deduction Th, (C) becomes $\vdash A \rightarrow B$. You are saying : this is a valid sentence of FOL, so it is "logically" true. But what does it means ? Can we translate it as : "In every 'universe of discourse' having (exactly) one thing white and (exactly) one thing balck, we have exactly two different things that are black-or-white". Is this a support for Frege's thesis (his form of logicism) that logic is simple the collection of laws that applies to everything and that arithmetic is only "logic in disguise" ? –  Mauro ALLEGRANZA Dec 19 '13 at 16:54
One of the issue with Frege's project (but also in Dedekind's logicism) was how to derive the infinity of natural numbers. W&R in PM assumed Infinity Axiom, and this was seen as a irreparable flaw in their development of logicism. But your method can be iterated; what happens with $\omega$ ? If your argument supports the thesis that (finite) numbers are a priori, what is missing in order to circumvent PM's difficulty ? –  Mauro ALLEGRANZA Dec 19 '13 at 17:03