Generating Numbers Proof We can do the following actions: multiply by $10$ (add $0$ at the end of number) , multiply by  $10$ and add $4$ (add $4$ at the end of number) and divide by $2$. I need to prove that we can get every natural number from $4$ using these actions.
Is it easier to prove that from any number I can get $4$ by multiplying it $2$ , dividing by $10$ and extract it's last $4$, or it is the same problem?

This is a trickier problem than I expected, so this is only a partial answer for now.  
Working backwards, it's clear that you can get from any number that ends in a $0$ or a $4$ to a smaller number in a single step, simply by deleting the final digit (i.e., divide by $10$, in the first case, or subtract $4$ then divide by $10$, in the second case).  Similarly, if $N$ ends in a $2$, $5$, or $7$, you get to a smaller number in two steps: first multiply by $2$ and then remove the resulting $0$ or $4$ in the ones digit.  Likewise, if $N$ ends in a $1$ or a $6$, three steps will suffice to arrive at a smaller number, while if it ends in a $3$ or $8$, four steps produce a smaller number -- that is, three doublings followed by a reduction by at least a factor of $10$.
The sticking point is if $N$ ends in a $9$.  In that case it takes four doublings before you can remove a $4$, but at that point you're left with a number that's larger than what you started with.  For example, let's look at $N=49$:
$$49\to98\to196\to392\to784\to78$$
Even continuing for another round leaves us something bigger than $49$:
$$78\to156\to312\to624\to62$$
It's only by going one more round that we wind up with something smaller than $49$:
$$62\to124\to12$$
It feels like it shouldn't ever take more than three rounds to get from $N$ to a number smaller than $N$ (at which point one can say "strong induction" and call it a day), but it'll take a closer look than I've given it here to be sure.  The problem has enough of the flavor of the classic $3x+1$ problem for me to be sure there isn't some lingering loophole.  On the other hand, maybe there's some elementary argument that I'm overlooking.  If I have a chance I'll give the problem some more thought, but I'd be happy if someone else posted a complete answer.
Update (April 7, 2016):  I was wrong about it never needing more than three rounds to get to a smaller number (where each "round" ends in the removal of trailing $4$'s and $0$'s).  Here's an example that takes four rounds:
$$749\to1498\to2996\to5992\to11984\to1198\\
1198\to2396\to4792\to9584\to958\\
958\to1916\to3832\to7664\to766\\
766\to1532\to3064\to306$$
