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

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.

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

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 $$2(m + n + 1)$$.