I look at $(a+b)+c = a+(b+c)$ for $a,b,c \in \mathbb{R}$ and think this tells me if I see three addends and two of them are in parentheses I can shift them without changing the sum. It obvious, at least with simple numbers, that I can group addends however I want without making an error.

But how does the associative property justify $(a+b+c)+d = a+(b+c+d)$ or even $a+(b+c+d) = (a+b)+c+d$? I mean, the way I see it is the associative property makes a statement about what you are allowed do if you have exactly three numbers and two parentheses.

Or is $(a+b)+c = a+(b+c)$ for $a,b,c \in \mathbb{R}$ just the mathematical way to say if you have a bunch of addends you can put parentheses wherever you want without changing the result?

  • $\begingroup$ Related: math.stackexchange.com/questions/2459697/… $\endgroup$ – Theo Bendit May 27 at 7:55
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    $\begingroup$ The expression $a + b + c$ does not even make sense without some form of associativity (even if it's a left or right-associativity understood by convention). Is it equal to $(a + b) + c$, or $a + (b + c)$? $\endgroup$ – Theo Bendit May 27 at 7:57
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    $\begingroup$ Building on Theo's comment, think about "$a\div b\div c$." Really the takeaway from associativity should be that we can write an expression like "$a_1+a_2+...+a_n$" and not worry about needing to put in parentheses to make it meaningful; and the proof of this (once precisely formulated) is by "basic" associativity + induction. $\endgroup$ – Noah Schweber May 27 at 8:03

Or is $(a+b)+c=a+(b+c)$ for $a,b,c∈ℝ$ just the mathematical way to say if you have a bunch of addends you can put parentheses wherever you want without changing the result?

You are right that is the meaning basically.

To be exact addition is a binary operation and thus defined for exactly two operands. For any binary operator $\circ$ you need some explanation what expressions like $a\circ b\circ c$ mean.

For example for the subtraction we have the convention $a-b-c := (a-b)-c$. (To be continued recursively).

For addition and any other associative operation the brackets do not matter.

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    $\begingroup$ +1 This convention (for subtraction) is known as "left-associativity". It's not something you prove; it's just a convention for interpreting $a \circ b \circ c$ for non-associative $\circ$. On the other hand, we adopt right-associativity for exponentiation. That is, we interpret $a^{b^c}$ as $a^{\left(b^c\right)}$. $\endgroup$ – Theo Bendit May 27 at 11:23

To prove that $(a+b+c)+d = a+(b+c+d)$, let $x = b+c$. Then the equation becomes $(a+x)+d = a+(x+d)$, which is immediately implied by the associative property. Similarly, to prove $a+(b+c+d) = (a+b)+c+d$, we let $c+d = x$, and the equation becomes $a+(b+x) = (a+b)+x$, again implied by the associative property.

In general, it is possible to prove that placing parentheses in any position can be moved around purely from manipulating the associative property.


We have a theorem in group theory (part of abstract algebra that deals with these types of fundamental properties) that shows that the "simpler" statement that you give, regarding $a,b,c \in \mathbb{R}$ actually is the same as saying that every possible way to add parentheses around $a_1 \cdot a_2, ...\cdot\ a_n $ is equivalent. This is called the generalized associative law. This proof can be done using induction.


This is just a development of the answer given by " auscrypt".

I think the question that is behind your question is : how to practice substitutions using an algebraic formula ( and conjointly, what do variables stand for exactly)?

The associative property is often stated like this :

(a+b)+c = a+(b+c).

But when it is rigorouly stated, it is phrased like this :

" for all numbers a, b and c : (a+b)+c = a+(b+c) " .

So, anything that is a number can be represented by the variables a, b, or c.

Even "complex" expressions such as : (2+d) / 4² could be substituted for a, b , or c (assuming d is a number).

Even the expression : a+b ( a and b being numbers) could be substituted for a, because a ( in the formula defining associativity) represents any number and (a+b) is a number.

That would give :

[(a+b)+c] + d = (a+b) + [c+d],

and this equality is perfectly true.

Personnaly, I think it is usefull to state formulas with capital letters as variables, in order to remember that anything can be substituted for them, even "big"/ " complex" expressions.

So I would state it like this:

For all A, B, C : (A+B)+C = A+(B+C).

Using this idea, we can show that :

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