Answer:
The proof itself
Step-by-step explanation:
We can define the set of all even numbers as
[tex]E = \{ a \in \mathbb{Z} \setminus a = 2.k , k\in \mathbb{N}\}[/tex]
This is, we can define all even numbers as the set of all the multiples of [tex]2[/tex]
As for the odd numbers, we can always take every even number and sum one to each one. This is
[tex]O = \{ a\in \mathbb{Z} \setminus a=2.k+1,k\in\mathbb{N}_{0}\}[/tex]
Note that [tex]k\in\mathbb{N}_{0}[/tex](the set of all natural numbers adding the zero) so that for [tex]k=0[/tex] then [tex]a=1[/tex]
Now, given 2 odd numbers [tex]a[/tex] and [tex]b[/tex] we can write each one as follows:
[tex]a = 2k+1\\b = 2l+1\\k,l \in\mathbb{N}_{0}[/tex]
And then if we multiply them with each other we obtain:
[tex]a.b = (2k+1).(2l+1)\\= 4kl+2k+2l+1\\= 2(2kl+k+l) + 1\\= 2k'+1 \\where\ k'=2kl+k+l[/tex]
Then we have that [tex]a.b[/tex] is also an odd number as we defined them.