Prove that $a + (b+(c+d)) = (a+b) + (c+d)$

Question

I'm working through Spivak's Calculus and the first section is covering associativity of addition. I am only given $P1: a + (b + c) = (a + b) + c$. I obviously know this intuitively but I cannot make the bridge in logic. It's a trivial proof and the fact I cannot prove something like this means I have a weakness that I'd like to rectify immediately before I dive further into the book.

Answer 1

It's normal to struggle at "intuitive" proofs at the beginning, keep working and you'll get oast it.

Regarding your question, a binary operation $•$ is associative when for any $a,b,c$, $a•(b•c)=(a•b)•c$. Namely the placement of the parentheses doesn't affect the expression.

You can use it as follows define $e=(c+d)$. Then $$ a+(b+(c+d)) = a+(b+e) = (a+b)+e=(a+b)+(c+d) $$

Answer 2

Maybe it will help you to rewrite P1 with different letters: $$ x+(y+z)=(x+y)+z. $$ Now the result you want is just a matter of choosing how to define $x$, $y$, and $z$ in terms of $a$, $b$, and $c$.

Answer 3

You can also approach it as follows: prove associativity within the context of the natural numbers $\textbf{N}$ and then extend such property to $\textbf{Z}$, $\textbf{Q}$ and $\textbf{R}$. Let us prove such claim in the context of $\textbf{N}$ using the Induction axiom on $c\in\textbf{N}$. Clearly, when $c = 0$, the proposed relation holds: \begin{align*} a + (b + c) = a + (b + 0) = a + b = (a + b) + 0 \end{align*} and we are done. Now suppose such relation holds for some $c\in\textbf{N}$ and we shall prove it holds for $s(c)$, where $s:\textbf{N}\to\textbf{N}_{>0}$ is the successor function. Our induction hypothesis is given by: \begin{align*} a + (b + c) = (a + b) + c \end{align*}

Then one has that: \begin{align*} a + (b + s(c)) & = a + s(b + c)\\\\ & = s(a + (b + c))\\\\ & = s((a + b) + c)\\\\ & = (a + b) + s(c) \end{align*}

Hopefully this helps!

EDIT

Apparently, I have misread the title of the question. The corresponding solution is given as follows:

\begin{align*} a + (b + (c + d)) & = a + ((b + c) + d)\\\\ & = (a + (b + c)) + d\\\\ & = ((a + b) + c) + d\\\\ & = (a + b) + (c + d) \end{align*}

Hopefully it contributes!