That’s if you add two of the same odd number. The more general proof is basically the same though: let n and m be integers, then 2n+1 and 2m+1 are odd. (2n+1) + (2m+1) = 2n + 2m + 2 = 2(n+m+1) which will be even.
Or you can just like, understand that an odd number is one more than an even number so if you add them together it’s two more than an even number, hence even.
That’s if you add two of the same odd number. The more general proof is basically the same though: let
nandmbe integers, then2n+1and2m+1are odd.(2n+1) + (2m+1) = 2n + 2m + 2 = 2(n+m+1)which will be even.Or you can just like, understand that an odd number is one more than an even number so if you add them together it’s two more than an even number, hence even.
Definitely, that’s how I’d explain it in words
This is why I failed at uni. I’m struggling so hard to make sense of such proofs, even if I understand the underlying concepts… :(
It helped me to lean on the different principals as an example.
The easiest being Principal of Induction. Substitute
mandnwith 1,3,5,7,9…After going through a few iterations you can see if it holds up enough to keep testing with other principals. (Super simplified).