Respuesta :

Nehuen

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.

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