Properties of integer halving by powers of 2. I'm dividing integers with a method where the remainder is removed after division. 
Example: 17/2 = 8 , 11/2 = 5 , 8/2 = 4 , 2/2 =  1
Is this property true for every integer; dividing by 4 yields the same result as dividing two times by 2 or dividing by 8 yields the same result as dividing three times by 2 and so on.
 A: Yes, this property is true.
Actually, dividing $a$ by $b_1$ then by $b_2$ yields the same quotient as dividing by $b_1b_2$.
To see why, write the euclidian divisions where $q_i$ denotes the quotient and $r_i$ the remainder:
$$a=b_1 q_1 + r_1$$
Then
$$q_1=b_2 q_2 + r_2$$
With $0 \leq r_1 < b_1$ and $0 \leq r_2 < b_2$
And your final result is $q_2$, with
$$a=b_1b_2 q_2+b_1 r_2 + r_1$$
And $$0 \leq b_1r_2+r_1 \leq b_1 (b_2 -1)+(b_1-1) < b_1b_2$$
Thus $b_1r_2+r_1$ is the remainder of the division of $a$ by $b_1b_2$, and the quotient is $q_2$.
A: This process is called integer division. Your quesion is then:
For $n \in \mathbb N$ is it true that $\def\b{\backslash}$
$$n \b 4 = (n \b 2) \b 2$$
Where $\b$ denotes integer division. To prove this, we consider only $n = 4k + i, n\b 4=k$ where $i\in \{1,2,3\}$, since the remainig case is trivial.
1. $i = 1$:
$$((4k + 1) \b 2) \b 2 = 2k \b 2 = k$$
2. $i = 2$:
$$((4k + 2) \b 2) \b 2 = (2k + 1)\b2 = k$$
3. $i=3$:
$$((4k + 3) \b 2) \b 2 = (2k + 1) \b 2 = k$$
This concludes the proof.
A: In addition to the other answers, you can represent what you are doing by means of the floor function, and then use this property:
$$\lfloor\frac{\lfloor\frac{x}{n}\rfloor}{m}\rfloor = \lfloor\frac{x}{mn}\rfloor$$
This is listed here on Wikipedia and a proof can be found in this MSE question.
