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I am trying to prove that

$$m \times a + n \times a = (m + n) \times a$$

while $m, n \in N$ and $a,b \in (Z, +, \times)$.

This is what I have got:

$a \times 0 = 0$ $\Rightarrow$ $a \times 0 + a \times 0 = a \times 0$, because $0 + 0 = 0$

assumption: $\underbrace{a + a + ... + a}_{\text{m}} = m \times a$

induction: $\underbrace{a + a + ... + a}_{\text{m}} + a = (m + 1) \times a$

therefore: $m \times a + a = (m + 1) \times a$ $\Rightarrow$ $m \times a + 1 \times a = (m + 1) \times a$, because $a \times 1 = a$ $\Rightarrow$ $m \times a + n \times a = (m + n) \times a$

is this right ?

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The problem seems to be missing the definition of $n\times a$, which is - I assume - inductive.

Namely:
$0\times a=0$
$(n+1)\times a=n\times a+a$.

Once we know the definition, it is not difficult to show this using induction on $n$.

$1^\circ$ For $n=0$ we have $$m\times a+0\times a\overset{(*)}=m\times a=(m+0)\times a$$ where the equality $(*)$ follows from $0\times a=0$, which is part of the definition.

$2^\circ$ Inductive step.
You know that $m\times a+n\times a=(m+n)\times a$.
What can you say - using the definition - about
$m\times a + (n+1)\times a=?$
$(m+n+1)\times a=?$

$m\times a + (n+1)\times a= m\times a + n\times a + a \overset{(\triangle)}= (m+n)\times a + a$
$(m+n+1)\times a= (m+n)\times a +a$
The equality marked as $(\triangle)$ is the place where we are using the inductive hypothesis. The first equality in both cases follows from the definition.

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    $\begingroup$ m x a + (n + 1) x a = (m + n + 1) x a => (using the definition) m x a + n x a + a = (m + n + 1) x a => (using the assumption) (m + n) x a + a = (m + n + 1) x a => (using the definition again) (m + n + 1) x a = (m + n + 1) x a is this right ? $\endgroup$ – Filip Sulik Nov 15 '15 at 20:53
  • $\begingroup$ That seems about right. I understand what you mean in those steps. One problem with the way it is phrased is that the first thing written there is what you want to prove. So it looks like you are starting with the thing you need to prove. Perhaps writing it as $m\times a + (n+1)\times a \overset{?}= (m+n+1)\times a$ would make clearer to the reader of your comment that you are working with the LHS and RHS separately. (If I understand your comment correctly, you are trying to do basically the same thing as I wrote above in the spoiler, but you wrote it as a series of equalities.) $\endgroup$ – Martin Sleziak Nov 15 '15 at 20:59
  • $\begingroup$ Yes, thank you very much. :) $\endgroup$ – Filip Sulik Nov 15 '15 at 21:03

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