this proof has to show that adding an even number and an odd number give an odd number. So let $a$ be an even number and $b$ is an odd number $c$ is an even number and $c=b-1$ so:
$$a+b=d$$ substitute $b=c+1$ so, $$a+c+1=d$$ use the fact that $even+even=even$ so $a+c$ is even and $even+1=odd$ so $d$ is odd. Hence $$even+odd=odd$$ Is this proof valid?
On
You proof seems unclear, although partially correct. I'll try to propose a more precise version of the same proof.
Suppose that $a$ is even, and $b$ is odd. This means that there exist $k$ and $h$ such that:
$$a = 2k,$$
and
$$b = 2h + 1.$$
Then:
$$d = a + b = 2k + 2h + 1 = 2(k+h) + 1.$$
Since $d$ is expressed as $$d = 2z+ 1,$$
where $z = k+h$, then we can conclude that $d$ is odd.
On
First of all: if you are trying to prove that $o(dd)+e(ven)=o$ and you use the fact that $e+e=e$ you have a little hole in the logic. This can be easily proven: $$a+b=2c+2d=2(c+d)=2k$$ hence $e+e=e$
The other part is that $e+1=o$. You should also prove this:
Let $a_n$ be the $n^{th}$ even number: $a_n=2n$ the $(n+1)^{th}$ element is $2(n+1)=2n+2$ hence $a_n<a_n+1<a_{n+1}$ hence $e+1$ is odd.
I want to point out that those are just things I think you can improve to make the proof better, you don't have to use them, the original proof is good.
Yes, well, mostly. You have shown that $d$ is the addition of $1$ to the sum of two even numbers (from the premises), and claim by some facts that infers that $d$ is an odd number.
Your argument is valid if you are given those facts. Were they provided as part of the problem? Otherwise you will have to demonstrate them from the definitions of even and odd numbers.