(2n + 1) + (2n) is odd?

Question

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.

$(2n)+(2m+1)\neq2(m+n+1)$ instead $(2n)+(2m+1)=2(m+n)+1$ – Dec 02 '20 at 11:17 — , Dec 02 '20 at 11:17
Ahhh i see, thanks Tito and Vlad!! – Beegs Dec 02 '20 at 11:23 — Beegs, Dec 02 '20 at 11:23

Derek Luna · Accepted Answer · 2020-12-02T11:32:00.147

3

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 $2(m + n + 1)$.

edited Dec 02 '20 at 11:32

answered Dec 02 '20 at 11:26

Derek Luna

2,732
8
19

(2n + 1) + (2n) is odd?

1 Answers1