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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.

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  • $\begingroup$ Actually, you are subtracting $9$ counts of $13$ and additionally subtracting the remainder modulo $10$. $\endgroup$ – abiessu Mar 2 '14 at 1:58
  • $\begingroup$ @abiessu can you elaborate a little on why we subtract by 13 rather than 10? $\endgroup$ – JohnMerlino Mar 2 '14 at 2:13
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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 $a/b = a-\frac {a-a\mod b}b \cdot (b-1)-a\mod b$.

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  • $\begingroup$ I think I could understand this. I just need to clarify a few things. What is that triple equal sign? Also the "⋅" is being used as multiplication? $\endgroup$ – JohnMerlino Mar 2 '14 at 5:30
  • $\begingroup$ "Triple equal sign" means "is congruent to." When $133$ is integer-divided by $10$, there is a remainder of $3$, and this is described as "$133$ is congruent to $3$ modulo $10$" and written $133\equiv 3\mod 10$. $\endgroup$ – abiessu Mar 2 '14 at 16:29
  • $\begingroup$ Also, yes $133\cdot 10$ uses the $\cdot$ to denote multiplication. $\endgroup$ – abiessu Mar 2 '14 at 16:30
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You need to think of division as splitting something up into a number of groups instead of as raw subtraction. You are splitting everything up into ten groups. Imagine you just want to discover the number of items in one of those groups.

In your division problem, you are splitting 133 up into ten groups. The way you explain it is counterintuitive because you rely on subtraction to get you through division. Imagine it more like a partition of objects into many groups. In that example, you are splitting up 133 into ten groups with 13.3 objects each. What you are trying to discover is how many objects are in each group.

Hope it helps.

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