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In calculus by Michael spivak book he say in page 4

It might seem reasonable to regard addition as an operation which can be performed on several numbers at once and consider the sum $a_1+...+a_n$ of n number $a_1,...,a_n$ as a basic concept . it is more convenient , however to consider addition of pairs of numbers only and to define other sums in terms of sums of this types

My questions :
1- How addition will be if we don't define it only on pairs of numbers ?
2- why it is more convenient , however to consider addition of pairs of numbers only and to define other sums in terms of sums of this types?

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    $\begingroup$ To 2: $(((a_1+a_2)+a_3)+a_4)$ and so on is defined on pairs of numbers only. $\endgroup$ Jun 30, 2017 at 15:24
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    $\begingroup$ Thus, having defined binary sum as $+(n,m)$ we can define recursively a $n$-argument sum: $+_n(a_1, \ldots, a_n)=+(+_{n-1}(a_1,\ldots, a_{n-1}), a_n)$. $\endgroup$ Jun 30, 2017 at 15:32
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    $\begingroup$ Reagrding Spivak's seeminngly "weird" assertion, you can consider that we have two concepts in place: definition and computation. In real analysis we define an infinite sum $\Sigma_i a_i$ "all at once": under suitable conditions it is a number. $\endgroup$ Jun 30, 2017 at 15:41

2 Answers 2

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How addition will be if we don't define it only on pairs of numbers ?

The short answer is, it will be summation, as in $\sum_{i = 1}^n a_i$.

The long answer is, Spivak was writing in the 1960s, for people fresh out of high school, who hadn't necessarily ever heard of associative or commutative laws, but were familiar with expressions like "$a + b + c$." He's about to show that that expression can be justified by use of the associative law of addition; he's anticipating objections from readers who don't see why it needs to be justified at all.

why it is more convenient , however to consider addition of pairs of numbers only and to define other sums in terms of sums of this types?

Because then you can define the algebraic properties of the real numbers as consequences of the field laws and order laws, Spivak's properties P1-P12, and these involve only the binary operation.

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  • $\begingroup$ I think Spivak has tried to present an obvious/trivial thing into seemingly non-obvious stuff. Right from primary classes (age 10-11 years) we learn addition is commutative and associative and everyone uses this fact routinely and knowingly. That students know this thing is evident from the fact that they don't try to do the same with subtraction. The main reason for defining addition as a binary operation is that it is much simpler to handle any binary operation compared to an $n$-ary operation $(n>2)$. $\endgroup$
    – Paramanand Singh
    Jul 1, 2017 at 7:01
  • $\begingroup$ @ParamanandSingh Right from primary classes (age 10-11 years) we learn addition is commutative and associative That didn't start happening in the United States until the "New Math" of the 1960s. Spivak's first readers wouldn't have learned maths that way. $\endgroup$
    – A. Burrell
    Jul 1, 2017 at 9:58
  • $\begingroup$ Agree! Things might be different in different countries and hence our differences of opinion. And then I think your answer deserves an upvote. +1 $\endgroup$
    – Paramanand Singh
    Jul 1, 2017 at 14:13
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Unless I am misunderstanding the question, addition this way would be exactly the same. We don't only define addition on pairs of numbers - we define it on any number of numbers by using the associative property. Because we have defined $$a_1+a_2$$ Then we can express a sum of any number of terms $$a_1+a_2+...+a_n$$ as a bunch of sums in the form that we have defined: $$(((a_1+a_2)+a_3)+...+a_n)$$ The reason this is more convenient is because if we were going to define it for any number of addends, we would have to address an infinite number of cases. But if we simply define $a_1+a_2$ and then create the commutative property, we can let it do the heavy lifting for us, since all other cases, using the commutative property, can be reduced to the case $a_1+a_2.$

I suppose you could try to define it for many addends. But if you did, here's what would happen:

Currently, addition is defined, at the most basic level, as an iterative operation. If $S(n)$ is the successor function $$S(n)=n+1$$ then addition can be expressed in terms of the successor function $$a+b=S^b(a)$$ However, trouble arises when you try to do it for three numbers. You would need a "successor function" with multiple arguments, but what would that even represent? Or you could do something like $$a+b+c=S^{b+c}(a)=S^{S^c(b)}(a)$$ or $$a+b+c=S^c(a+b)=S^c(S^b(a))$$ but that is the same as pairing up the addends $$(a+b)+c$$ That's why it is much more convenient just to define it for two addends.

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    $\begingroup$ I think OP wants to know how to define definition for $n$ numbers without defining it for the pair. $\endgroup$ Jun 30, 2017 at 15:28
  • $\begingroup$ @SahibaArora Ah, okay. I'll add that as well. $\endgroup$ Jun 30, 2017 at 15:29
  • $\begingroup$ I'd add that $+$ having two addends is necessary for it to be a group operation ; but OP might wonder why does a group operation requiers two elements of a set and not 3. $\endgroup$
    – krirkrirk
    Jun 30, 2017 at 15:49
  • $\begingroup$ Spivak is writing about addition of real numbers, not natural; and in the part of the book the OP quotes from, he doesn't construct the reals, but characterizes them axiomatically. So succession functions are not his concern. $\endgroup$
    – A. Burrell
    Jun 30, 2017 at 19:57

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