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I'm looking at basic proofs in Lang and in Downing. Here's the proof in question in Downing:

$m$ and $n$ are natural numbers $\ne 0$.

$$s = (2n) + (2m + 1)$$ $$s = 2(m + n) + 1$$ Thus the answer is odd by an earlier definition of odd as $2n + 1$. This seems to me like just selectively applying the rules to get an answer you want. Why can't I just rewrite this as $s = 2(m + n + 1)$ by associativity, thus producing an even number? I suspect I'm misunderstanding something, but I can't figure it out.

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  • $\begingroup$ $2(m+n+1)=2m+2n+2$ $\endgroup$
    – Tito Eliatron
    Dec 2, 2020 at 11:17
  • $\begingroup$ $(2n)+(2m+1)\neq2(m+n+1)$ instead $(2n)+(2m+1)=2(m+n)+1$ $\endgroup$
    – user799688
    Dec 2, 2020 at 11:17
  • $\begingroup$ Ahhh i see, thanks Tito and Vlad!! $\endgroup$
    – Beegs
    Dec 2, 2020 at 11:23

1 Answer 1

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Associativity does not mean "move the parentheses wherever you want".

Note that associativity is the axiom that $(a+b)+c = a+(b+c)$.

If you tried rewriting $2(m+n)+1$ in the way you suggest, you would first have to reduce $2(m+n)+1$ to $(2m + 2n)+1$ by the distributive property,

which would stop you from writing the expression as $2(m + n + 1)$.

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