A wrong mathematical induction Where is the mistake?
Statement: For $n\in \mathbb{N}_{0}$ is $2n=0$.
Bais: Show that the basis holds for $n=0$ : $2*0=0$.
Assumption: The statement is valid for all $k\leq n$ : $2*k=0$ for all $k \leq n$
Inductive Step: For $k = n+1$ is $k=a+b$ for two natural numbers $a,b \leq n$. It is $2(n+1) = 2a + 2b = 0+0=0$.
It is obviously wrong. I have a few ideas but I would like to be sure where the exact mistake is.
The first thing I noticed was that in this equoation $2(n+1) = 2a + 2b = 0+0=0$
the left side $2(n+1)$ is always positive but the right side isn't.
I cant find a mathematical mistake here so I thought the Assumption is wrong. I always thought the Assumption is valid for a specific n. It is is indeed freely selectable but specific. Is that right and is that the problem with this induction? 
If it isn't - maybe it might be the fact that there aren't two natural numbers below 1 that $ a+ b =k$ and thet you cant use the assumption two times?
 A: You use the fact that $2\times1 = 0$, therefore your inductive step does not work for the first  step, since you claim that : $$2\times1 = 2\times0+2\times1 = 0+0=0$$ which is not in your induction hypothesis.
Simply put : $1$ is not the sum of two natural numbers $a+b$ where $a\lt 1$ and $b \lt 1$ thus your indutive step does not hold for $n=0$.
A: 
Inductive Step: For $k = n+1$ is $k=a+b$ for two natural numbers $a,b
 \leq n$. It is $2(n+1) = 2a + 2b = 0+0=0$.

This seems to be a variant of the alternative inductive step
$$
(\forall m \le n: S(m)) \Rightarrow S(m+1)
$$
Which would complete your statement to

Inductive Step: For $k = n+1$ is $k=a+b$ for two natural numbers $a,b
 \leq n$.
  [$2k = 0$ holds for all $k \le n$, therefore it holds for $a$ and $b$ ]
  It is $2(n+1) = 2a + 2b = 0+0=0$.

However only $S(0)$ is true and $S(m)$ is false for $m \in \mathbb{N}$, where
$$ S(m) = ( 2m = 0) $$
so the premise of the implication in square brackets above is wrong and indeed your equation
$$
2a + 2b = 0 + 0
$$
is wrong for $a \in \mathbb{N}$ or $b \in \mathbb{R}$.
