I think addition, multiplication and exponents comes naturally in a mental way.
Let's say I have the number 133 in the decimal base 10 system. We know that 1 is in the hundred's place. We know that 3 is in the ten's place. And 3 is in the one's place. So how is this constructed?
1 * 10 * 10 = 10+10+10+10+10+10+10+10+10+10 (hundred's place)
3 * 10 = 10+10+10 (ten's place)
3 = 3 (one's place)
In other words:
1 * 10 * 10 + 3 * 10 + 3 = 133
As we move left, the number of 10's increase by a factor of 1, since the decimal system is base 10.
This can also be written using exponentiation (which is repeated multiplication, just as multiplication is repeated addition):
1*10^2+3*10^1+3 = 133
That's all pretty straightforward and you can visualize that in your head. But undoing all this is what becomes a little unclear.
Division undoes multiplication. So if we want to undo:
1*10*10=100
We can do this:
100 / 10 / 10 = 1
Just as multiplication is repeated addition, division is repeated subtraction:
100 / 10 = 10
100-10-10-10-10-10-10-10-10-10 = 10
When we divide 100 by 10, we get 10 because we are in essence subtracting by 90, which leaves 10 in a base 10 system. And dividing 10 by 10 gives 1, which is the original number we started with.
What I cannot visualize is the following:
133 / 10 = 13 is quotient with a remainder of 3
Now I can easily do long division to solve that, but why do we need to subtract by 120 (that's 12 ten's)? Something's missing that I am not understanding in understanding how this base 10 system is working here.
Best Answer
Consider the number $133/10$ using integer division. There are two subtractions that occur: first, $133\equiv 3\pmod {10}$ and so $3$ is subtracted. Next, dividing by $10$ is the same as subtracting $13\cdot (10-1)=13\cdot 9=117$ from $133-3=130$ leaving $133-3-117=13$.
This is the case primarily because division is the process of scaling the numerator by the amount required to scale the divisor to $1$. This is another way of saying that $a/b = a-\frac {a-a\mod b}b \cdot (b-1)-a\mod b$.