A friend of mine claims that in different dimensions, whatever the definition of dimension in any context is, it is possible for 1+1 to not equal 2. Does it matter the dimension one does arithmetic or does 1+1=2 regardless?
Are arithmetic operations independent of dimension?
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0That depends on what you mean by $1$, what you mean by $2$, what you mean by $+$ and what actual context you are attempting to do this "addition" operation in. For example in $\Bbb F_2$ there are only two numbers, zero and one, and you have $1+1=0$ in that specific context. In other contexts, like in $\Bbb R^3$ you have $(1,0,0)+(1,0,0)=(2,0,0)$, but you also have things like $(1,0,0)+(0,1,0)=(1,1,0)$ so it isn't clear what you might mean by $1$ and $2$ in the first place. – 2017-02-26
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0@JMoravitz By numbers I mean objects and by addition I mean putting them together. Regardless of base, symbolic representation, or anything else, if I take a single object and another of the same object and put them in a group what quantity of that object will be there? – 2017-02-26
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0That is a horrible way to define addition. Regardless it still relies on what you mean by "two." I already gave you an example where 1+1=0, where in that context there isn't an entity we call "two" but you could still choose to define two and you would just end up with 2=0. If you choose to define "two" it is usually as S(1), the successor of one, i.e. two is generally *defined* as 1+1. That definition holds whenever it makes sense to add one and one together. – 2017-02-26