Remainder and long division

Question

I have been thinking about this. Lets say we take 7 divided by 3, we know the remainder is 1. However, if we let x=7 and x-4=3, and we take x/(x-4), after performing long division the remainder is 4. Why is it not 1?

Answer 1

Because the quotients are different:

$ 7 = 2 \cdot 3 + 1 $

$ x = 1 \cdot (x-4) + 4 $

When $x=7$ we get

$ 7 = 1 \cdot 3 + 4 $

which is correct, of course.

Answer 2

If someone handed you $x$ and $x-4$, why would you assume that they are $7$ and $3$?

In general, we know that $\displaystyle \frac{x}{x-4}=1+\frac{4}{x-4}$.

For $x=7$, it is the same that $\displaystyle \frac{7}{7-4}=1+\frac{4}{7-4}$, just like any arbitrary $x$.

However, it just so happens that $\displaystyle 1+\frac{4}{7-4}$ can be rewritten as $\displaystyle 2+\frac{1}{7-4}$.

The polynomial remainder is still $4$.

Answer 3

Notice:

$7=4*3+(-5) $

$7=3*3+(-2) $

$7=2*3+1$

$7=1*3+4$

$7=0*3+7$

$7=-1*3+10$

Etc. So which one qualifies as "the" remainder? Why is it $1$? Whis it not $4$, $10$ or $-2$?

Well, because the remainder is defined to be at least $0$ but strictly less than the divisor.

The 1 is "the" remainder. The rest are congruences. And there are an infinite number of them.

So $\frac x {x-4}= 1 + \frac 4 {x-4} $. Or in other words $x=1*(x-4)+4$.

But is $4$ "the" remainder? Or is $4$ just a congruence?

The depends on whether $4 < x-4$ or not. If $x=7$ then $4> x-4$ and $4$ is not "the" remainder. It is a congruence.

In particular this is $7=1*3+4$.

... which is $7=1*3+3+1$ which is $7=(1+1)*3+1$ which is $7=2*3+1$

Note: we also have: $x = 2*(x-4)+(4-(x-4))=2*(x-4)+(8-x) $.

If we use that expression, then the remainder is $8-x$ which is $1$ for $x=7$.