How to define the amount of additions. E.g. $1+2+3+4+5+6+7+8+9$
Are there $9$ additions, because of the nine numbers that are added together. Or can you also say that there are $8$ additions, because there are only $8$ '$+$' signs.
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Addition is an operation and hence we are performing 8 additions as we repeat the operation of 'adding' 8 times. In contrast, I would say that we are adding 9 numbers.
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$$ \sum_{i=1}^n a_i = a_1 + \cdots + a_n $$ There are $n$ terms in the addition but what if there are minuses? Hence counting terms are kinda less ambiguous. Also we can do: $$ \sum_{i=1}^n a_i = 0+ a_1 + \cdots + a_n $$ and we have $n$ additions now.
Tony Stark
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So, in my case there would be 8 additions? – user8005 May 30 '13 at 01:36
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@user8005 we can say there are 9 additions as well: 0+1+2+...+9 – Tony Stark May 30 '13 at 02:01
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Well if you define it that way, you can say there are 100 additions or 1000 additions as well, if you keep adding zeros. – user8005 May 30 '13 at 02:09
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@user8005 that's why we take minimum additions so it will 8.(no.of operands-1) – iostream007 Jun 09 '13 at 14:42
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Addition is a binary operator that uses two operands to execute.so if there are $n$ number of elements then number of addition will be $$\;n-1$$.In other answers people uses "addition of $0$",because $0$ is addition identity,it will give $n$ addition but it is only in case of certain operator like $+,-$ but there are other binary operator also like $\times ,/$ with these operator we can not use $0$ there.For these operator we have multiplicative identity $1$.But regardless we can say there are minimum $$n-1$$ operations.
iostream007
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