# Is all mathematics based on the concept that $1+1=2$?

Is all mathematics based on the concept that $1+1=2$?

For example, if $1+1\ne2$, then all arithmetic won't work, right?

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No, some more mathematics is based on the concept that $(a+b)^2=a^2 +2ab +b^2$. In fact, the third part of it is based on the craziest concept of all that $(x^2)^{'}_{x}=2x$ –  Alex Feb 10 '13 at 3:03
It might be cheating the question a little, but arithmetic mod 2 has no concept of 2. In fact, $1+1=0 \bmod{2}$. (Also, keep in mind $1$ and $2$ are just symbols! We could just as easily use a different number system, where $\theta + \theta = \upsilon$ and define $\theta$ and $\upsilon$ just as we define $1$ and $2$.) –  George V. Williams Feb 10 '13 at 3:14
This question is pretty much in the makes-no-sense ballpark. Wittgenstein pretty much nailed it. –  Mariano Suárez-Alvarez Feb 10 '13 at 3:21
I think the confusion here is we don't know whether @think123 is asking "does $1+1=2$ hold for all of mathematics" or "does all of mathematics stem from the definition $1+1=2$". –  Alexander Gruber Feb 10 '13 at 3:22
I think that the tagging is a bit off here. –  Asaf Karagila Feb 10 '13 at 3:45

Sometimes $1+1=0$, when you're working in a unital ring with characteristic $2$!

But those are the only possibilities. $2$ is defined as $1+1$. (So actually in the characteristic $2$ case, $1+1=2$ still- it's just that $0=2$.)

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If $1+1=5$ just declare $5$ as $2$ and move on with it. You are mixing the idea of $2$ with its symbol.
We usually define the meaning of the symbol $2$ as a shorthand for $1+1$, but it might as well be any other symbol. It should be noted that if we assume the "usual" axioms of mathematics and we manage to prove that $1+1\neq2$ then our axioms have an inherent contradiction and we need to find it. But so far we're good.