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?
$\endgroup$ 34 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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 :
