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The question is in the title. I am very appreciative of any time and concern put into belaboring this relatively little problem.

Asaf Karagila
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Jeremy Chavez
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    Can we assume that $1+1 = 2$? – Ben Grossmann Apr 02 '15 at 23:38
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    Before this gets downvoted into oblivion, can you elaborate on your question? Is this more of a philosophical question or ... ? (You need to add context; otherwise, your question will be closed) – Daniel W. Farlow Apr 02 '15 at 23:39
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    You may read some classic text like Landau's Foundations of Analysis. – Salomo Apr 02 '15 at 23:39
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    What is your definition of $4$? If your definition is that it's $2+2$, then you're done. If your definition is that it's $3+1$, then we go further: what is your definition of $1$? Of $2$ and $3$? Of $+$? When you ask such basic questions, you need to go this deep down into the matter in order to prove anything. – Arthur Apr 02 '15 at 23:41
  • 2+2=4 only when the Party says it does. – Jonny Apr 02 '15 at 23:41
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    $2+2\equiv 1 \bmod 3$. $2+2=1$ within $\mathbb Z_3$. Definition and context are important. – Mark Bennet Apr 02 '15 at 23:42
  • http://mathworld.wolfram.com/PeanosAxioms.html – Tae Hyung Kim Apr 02 '15 at 23:44
  • @MarkBennnet But at the same time, $2+2\equiv 4\pmod 3$, so the premise isn't wrong in that case either. – Arthur Apr 02 '15 at 23:44
  • I've heard engineers say that 2+2=5 for extremely large values of 2. – Jonny Apr 02 '15 at 23:45
  • @MarkBennet Good point. In any monoid you can choose any element to call "1" and then 2 is simply defined to be $1+1$ and 3 is defined to be $1+1+1$ etc. In every case $4=2+2$. – Gregory Grant Apr 02 '15 at 23:47

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The definition of 2 is $1+1$. The definition of 4 is $1+1+1+1$. In general any $n$ is defined as $\sum_{i=1}^n1$. You can think of that as a purely abstract symbol. And the sum $n+m$ is defined to be $\sum_{i=1}^{n+m}1$. It is easy to show that this set and addition make this a well defined monoid. Then you must simply show that $(1+1)+(1+1)=1+1+1+1$. This is true because addition, as defined, is associative.

Gregory Grant
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    Hmmm. This seems like a bit of a reductive answer. It also glosses over the fact that we are presumably computing sums in a base-10 system. I think it would be wise/appropriate to wait until OP clarifies the question. – Daniel W. Farlow Apr 02 '15 at 23:41
  • @crash: the base isn't relevant, as long as by "$4$" you mean the number usually represented by that symbol and not the symbol itself (which you should). – A.P. Apr 02 '15 at 23:46
  • @A.P. The base is certainly relevant. What is $2+2$ in base-3? It certainly isn't $4$. – Daniel W. Farlow Apr 02 '15 at 23:47
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    @crash Where is the base ten? Anyways, base is irrelevant to the math itself, $10$ is just two squiggles and we have agreed that when they appear together, they should mean $1+1+1+1+1+1+1+1+1+1$, to save space and improve readability. The fact that the first squiggle resembles $1$ is purely coincidental. – Arthur Apr 02 '15 at 23:48
  • In any monoid you can choose any element to call "1" and then 2 is simply defined to be $1+1$ and 3 is defined to be $1+1+1$ etc. In every case $4=2+2$. That's total generality, true in any monoid. Not necessarily true in a non-associative algebra. – Gregory Grant Apr 02 '15 at 23:48
  • All of you are proving my point made in my first comment--we should, all of us, wait until the OP provides context for the question. Otherwise, answers may be completely off-base (no pun intended but enjoyed). – Daniel W. Farlow Apr 02 '15 at 23:50
  • @GregoryGrant how do you define $1+1+1+1$, though? Usually addition is defined as a binary operation first and then extended by, say, left associativity, in which case $1+1+1+1 := ((1+1)+1)+1$. – A.P. Apr 02 '15 at 23:52
  • @A.P. Originally I was just thinking of it as an abstract symbol and defining addition on abstract symbols. In the context of a monoid however, we are given associativity, so there is no ambiguity. – Gregory Grant Apr 02 '15 at 23:56