## Exploding Dots

### 6.2 Division in Any Base

Lesson materials located below the video overview.

Check out Goldfish & Robin and Friend’s “Where Young Minds Collide” video on this topic: Kids Explain Math For Kids.

Here’s the division problem $$276 \div 12$$ we did earlier in a $$1 \leftarrow 10$$ machine. We see the answer $$23$$. Stare at this picture for a moment – it will soon sneak back up on us.

Let’s now do the same division problem in another base. But the only tricky part is that I am not going to tell you which machine we are in! We could be in a $$1 \leftarrow 10$$ machine again, I am just not going to say. Maybe it will be in a $$1 \leftarrow 2$$ machine, or a  $$1 \leftarrow 4$$ machine or a $$1 \leftarrow 13$$ machine. You just won’t know as I am not telling. It’s the mood I am in!

Now, in high school algebra there seems to be a favorite letter of the alphabet to use for a quantity whose value you do not know. It’s the letter $$x$$.

So let’s work with an $$1 \leftarrow x$$ machine with the letter $$x$$ representing some number whose actual value we do not know.

In a $$1 \leftarrow 10$$ machine the place values of the boxes are the powers of ten: $$1$$, $$10$$, $$100$$, $$1000$$, … .

In a $$1 \leftarrow 2$$ machine the place values of the boxes are the powers of two: $$1$$, $$2$$, $$4$$, $$8$$, $$16$$, ….

And so on.

Thus, in an $$1 \leftarrow x$$ machine, the place values of the boxes will be the powers of $$x$$.

As a check, if I do tell you that $$x$$ actually is $$10$$ in my mind, then the powers $$1$$, $$x$$, $$x^{2}$$, $$x^{3}$$, … match the numbers $$1$$, $$10$$, $$100$$, $$1000$$, … , which is correct for a $$1 \leftarrow 10$$ machine. If, instead, I tell you $$x$$ is really $$2$$ in my mind, then the powers $$1$$, $$x$$, $$x^{2}$$, $$x^{3}$$, … match the numbers $$1$$, $$2$$, $$4$$, $$8$$, $$16$$, …, which is correct for a $$1 \leftarrow 2$$ machine.

This $$1 \leftarrow x$$ machine really is representing all machines all at once!

Okay. Out of the blue! Here’s an advanced high school algebra problem.

 Compute $$\left(x^2+7x+6\right)\div\left(x+2\right)$$.

Can you figure out what this means on an $$1 \leftarrow x$$  machine? Try playing with this before reading on.

Here’s what $$\left(x^2+7x+6\right)$$ looks like in an $$1 \leftarrow x$$ machine. It’s two $$x^2$$s, seven $$x$$s, and six ones.

And here’s what $$x+2$$ looks like.

The division problem $$\left(x^2+7x+6\right)\div\left(x+2\right)$$ is asking us to find copies of $$x+2$$ in the picture of $$2x^2+7x+6$$.

I see two copies of $$x+2$$ at the $$x$$ level and three copies at the $$1$$ level. The answer is $$2x+3$$.

 Stare at the picture for $$\left(x^2+7x+6\right)\div\left(x+2\right)=2x+3$$. Does it look familiar?

We’ve just done a high school algebra problem as though it is a grade school arithmetic problem!

What’s going on?

Suppose I told you that $$x$$ really was $$10$$ in my head all along. Then $$2x^2+7x+6$$ is the number $$2\times 100 + 7\times 10 +6$$, which is $$276$$. And $$x+2$$ is the number $$10+2$$, that is, $$12$$. And so we computed $$276 \div 12$$. We got the answer $$2x+3$$, which is $$2\times10 +3=23$$, if I am indeed now telling you that $$x$$ is $$10$$.

So we did just repeat a grade-school arithmetic problem!

Aside: By the way, if I tell you that $$x$$ was instead $$2$$, then

$$2x^2+7x+6=2\times4+ 7\times 2 +6$$, which is $$28$$,

$$x+2=2+2$$, which is $$4$$,

and

$$2x+3=2\times2+3$$, which is $$7$$.

We just computed $$28 \div 4 = 7$$, which is correct!

Doing division in an $$1 \leftarrow x$$ machine is really doing an infinite number of division problems all in one hit. Whoa!

Try computing $$\left(2x^{3}+5x^2+5x+6\right)\div \left(x+2\right)$$ in an $$1 \leftarrow x$$ machine to get the answer $$2x^2+x+3$$. (And if I tell you $$x$$ is $$10$$ in my mind, can you see that this matches $$2556 \div 12 = 213$$?)

In high school, numbers expressed in an $$1 \leftarrow x$$ machine are usually called polynomials. They are just like numbers expressed in base $$10$$, except now they are “numbers” expressed in base $$x$$. (And if someone tells you $$x$$ is actually $$10$$, then they really are base-ten  numbers!)

Keeping this in mind makes so much of high school algebra so straightforward: it is a repeat of grade school base-ten arithmetic.

Here are some practice problems for you to try, it you like. My answers to them appear in the final section of this chapter.

1. a) Compute $$\left(2x^{4}+3x^{3}+5x^{2}+4x+1\right)\div\left(2x+1\right)$$.

b) Compute $$\left(x^{4}+3x^{3}+6x^{2}+5x+3\right)\div\left(x^2+x+1\right)$$.

If I tell you $$x$$ is actually $$10$$ in both these problems what two division problems in ordinary arithmetic have you just computed?

2. Here’s a polynomial division problem written in fraction notation. Can you do it? (Is there a slight difficulty to watch out for?)

$$\dfrac{x^{4}+2x^{3}+4x^{2}+6x+3}{x^2+3}$$

3. Show that $$\left(x^{4}+4x^{3}+6x^{2}+4x+1\right)\div\left(x+1\right)$$ equals $$x^{3}+3x^{2}+3x+1$$.

a) What is this saying for $$x=10$$?

b) What is this saying for $$x=2$$?

c) What is this saying for $$x$$ equal to each of 3, 4, 5, 6, 7, 8, 9, and 11?

d) What is it saying for $$x=0$$?

e) What is it saying for $$x=-1$$?

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