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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?

Arnaud D.
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  • Related: https://math.stackexchange.com/questions/2459697/prove-generalized-associative-law/2459739#2459739 – Theo Bendit May 27 '19 at 07:55
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    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 $a + (b + c)$? – Theo Bendit May 27 '19 at 07:57
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    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. – Noah Schweber May 27 '19 at 08:03

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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.

auscrypt
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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.

pH 74
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    +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 $a^{\left(b^c\right)}$. – Theo Bendit May 27 '19 at 11:23
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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.

rb612
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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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