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This question was featured on Saturday Morning Breakfast Cereal and I haven't been able to find a proof. Can anyone help?

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  • $\begingroup$ It might surprise you but $2+2$ is sometimes $0$. $\endgroup$ – Berci Jan 4 '14 at 22:22
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    $\begingroup$ @Berci But in those cases $4=1+1+1+1=0$. $\endgroup$ – Josué Tonelli-Cueto Jan 4 '14 at 22:30
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    $\begingroup$ "Mom, if you don't believe me, don't use my result!" $\endgroup$ – Apprentice Queue Jan 4 '14 at 22:32
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    $\begingroup$ What is your definition of $2$, $4$, and $+$? $\endgroup$ – chubakueno Jan 4 '14 at 22:53
  • $\begingroup$ And sometimes $2+2=1$. $\endgroup$ – Christopher Carl Heckman Apr 26 '16 at 5:48
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When proving theorems in mathematics, one starts from a set of axioms, statements that are accepted as true without argument. You might ask, "but what if an axiom isn't true?", and the answer is that we would be dealing with different mathematics. For example, Euclid included the parallel postulate as an axiom in his elements. For years mathematicians tried to prove that the parallel postulate could be derived from the other axioms. It turns out that if you don't accept the parallel postulate, you end up with different types of geometry that we now call non-euclidean. Einsteins theory of general relativity depends on these geometries.

To come up with a proof of such a seemingly simple fact as $2 + 2 = 4$, we need a set of axioms to start with, and we need precise definitions of all the terms we are using. Depending on what set of axioms you start with, proving that $2 + 2 = 4$, and that no other natural number can equal $2+2$ may be either very simple or surprisingly difficult. For example in Russell and Whitehead's Principia, it famously took over 300 pages of work before they could prove that $1+1=2$. They started with a very sparse set of axioms though.

The most common set of axioms for the natural numbers are the Peano Axioms.

They are

  1. $0$ is a natural number.
  2. For every natural number $x$, $x=x$.
  3. For all natural numbers $x$ and $y$, if $x = y$, then $y = x$.
  4. For all natural numbers $x$, $y$, and $z$, if $x = y$ and $y = z$, then $x = z$.
  5. For all $a$ and $b$, if $a$ is a natural number, and $a = b$, then $b$ is a natural number.
  6. For every natural number $n$, $S(n)$ is a natural number.
  7. For every natural number $n$, $S(n) = 0$ is false.
  8. For all natural numbers $m$ and $n$, if $S(m) = S(n)$ then $m = n$.
  9. If $K$ is a set such that $0 \in K$, and for every natural number $n$, $n \in K$ implies that $S(n) \in K$, then $K$ contains all natural numbers.

Here $S$ is the successor function, it takes each natural number to its successor. This might seem like a complicated mess compared to the simplicity of natural numbers, but we need to be precise. We need to carefully construct the axioms so that no contradiction can be derived from them, and so they encapsulate what we understand to be the natural numbers. We want to be able to prove interesting statements about the natural numbers from them. Note that the axioms contain undefined terms. The axioms don't need to state what the terms mean, only what they do.

The following definitions are commonly used within this axiomatization. They are the definitions from Peano's original paper (An English translation is available in the book From Frege to Gödel), modified to start at $0$ instead of $1$.

$1$ is defined as $S(0)$, $2$ is defined as $S(1)$, $3$ is defined as $S(2)$, and $4$ is defined as $S(3)$. Addition is defined recursively as follows.

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

Thus

$$2 + 2 = 2 + S(1) = S(2 + 1) = S(2 + S(0)) = S(S(2 + 0)) = S(S(2)) = S(3) = 4$$

proving that $2+2 = 4$.

This is the unique value of $2+2$ by axiom 4.

If $x = 2+2$ and $2+2 = 4$, then $x = 4$.

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    $\begingroup$ You ran in your own when you defined the numbers themselves, and then defined the addition, that is not in any of the $9$ axioms of Peano arithmetic. $\endgroup$ – chubakueno Jan 4 '14 at 23:33
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    $\begingroup$ @chubakueno. To prove this result, one needs not only the axioms, but definitions of the terms used. The definitions I used are perfectly standard within Peano Arithmetic. They are not my own. $\endgroup$ – ajs3 Jan 4 '14 at 23:35
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    $\begingroup$ Of course we need the definitions of the operations and the symbols we were using, that is why I asked for them in the comment. My point was that you just defined them as you wished to. With the edit, my only suggestion now is to cite your source of the definitions. +1. $\endgroup$ – chubakueno Jan 4 '14 at 23:42
  • $\begingroup$ For every x, x=x? So this is why NaN does not equal NaN in many programming languages? $\endgroup$ – Loupax Apr 19 '16 at 10:47
  • $\begingroup$ @Loupax NaN is generally a property of floating-point numbers (real numbers), which are not the natural numbers (non-negative integers). The axioms listed are for natural numbers. $\endgroup$ – merthsoft May 8 '16 at 3:20
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From the Peano axioms.

First recall that $2$ is really a shorthand for $S(S(0))$ and $4$ is a shorthand for $S(S(S(S(0))))$. Now we have the axioms for addition, $x+0=x$ and $x+S(y)=S(x+y)$. We calculate:

$$2+2=S(S(0))+S(S(0))=S(S(S(0))+S(0))=S(S(S(S(0))+0))=S(S(S(S(0)))=4$$

To prove uniqueness we use the axioms that $0\neq S(x)$, $x\neq0\rightarrow \exists y(x=S(y))$ and $S(x)=S(y)\rightarrow x=y$:

$$S(S(S(S(0))))=S(S(0))+S(S(0))=x\implies\exists y(x=S(y))\implies S(S(S(0))=y\implies\exists z(S(z)=y)\implies z=S(S(0))=2$$

Therefore $z=2$ and so $y=3$ and $x=4$ as wanted.

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The question doesn't make sense; the definition of $2 + 2$ doesn't involve an indeterminate, so there's no "solution" to be had, just the fact that $2 + 2 = 4$.

Contrast this with $\sqrt{4}$, which could be interpreted as "a solution to $x^2 = 2$", in which case there are two solutions, $2$ and $-2$.

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    $\begingroup$ By definition, $\sqrt {x^2}=|x|$. $\endgroup$ – John Bentin Jan 5 '14 at 0:35
  • $\begingroup$ Then by definition, $\sqrt{(3 + 4i)^2} = 5$, which is a definition that I think a few of us would prefer not to use. $\endgroup$ – DanielV Jan 5 '14 at 3:36
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    $\begingroup$ @DanielV: The context here is the real numbers. The square root isn't defined for negative real or complex numbers. $\endgroup$ – John Bentin Jan 5 '14 at 10:51
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The cartoon is a joke which doesn't bear too much analysis. The "mathematician parent" is the cartoonist's parody of such a parent. A real mathematician would not ask for a proof that the "solution is unique" in the case of addition of two specific integers, because such a request would make little sense. If we must credit the request with some sense, then perhaps an appropriate response would be at the level of "because addition is unique, Stupid!".

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$+$ is an operation, which means it is a function. The rest follow from definition of a function.

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    $\begingroup$ $\sin$ is a function, but $\sin(\pi)=0$ is not a unique solution $\endgroup$ – Tim Jan 5 '14 at 0:16
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    $\begingroup$ @Tim:Yes it is: $0$ is the unique solution to $\sin(\pi)=x$. $\endgroup$ – Gina Jan 5 '14 at 0:50
  • $\begingroup$ I see Gina's point: if $f$ is a function, then $f(x)$ has one value only. However, I don't see why operations are necessarily functions. $\endgroup$ – Ian Mateus Jan 5 '14 at 0:56
  • $\begingroup$ @IanMateus: a binary operation on the set $X$ is a function $f:X\times X\rightarrow X$. So in the above case, $+$ is a function where $+((2,2))=4$ (the double bracketing is intended, one is for function argument, one is for 2-tuple). Of course, such notation is cube sum compared to $2+2=4$. $\endgroup$ – Gina Jan 5 '14 at 1:03
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You need a mathematical proof by contradiction.

You assume 2 plus 2 is not equal 4.

2+2 <> 4

You substract 2 from the two sides of the equation.

2+2-2 <> 4-2

2 <> 2

2 is not equal to 2? CONTRADICTION. Thus, proof.

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    $\begingroup$ You make the assumption that $2 + 2 = 4$ when you say that $4 - 2 = 2$. $\endgroup$ – okarin Jan 5 '14 at 0:32
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Prove it geometrically...

  1. Fist define your "anchor/base" unit -> a line with a defined length is #1.

  2. Next define #2... just join 2 of your #1 units... and postulate it's #2, the double of the "base"... you can measure it... then you can prove it...

  3. draw 2 of your #2 units, parallel in the geometric space, like in a vector space...

  4. Draw the next sequence -> Move the beginning of the the 2nd #2 to the end of the 1st #2 and voilá... you have #4

*you can prove all the relations -> #1 = 1x #1 | #2 = 2x #1 | #4 = (4x #1 or 2x #2) *if at a superior academic level you have to define and prove the operations also '+' and 'x'... at least the "neutral element" of operation 'x'... Good luck...

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