How is dividing basically "subtracting several times"?

Question

I keep hearing that dividing is basically "subtracting by a number several times" but I don't see how. ie multiplying 5 5 times: $$5+5+5+5+5 = 25 = 5\times5$$

But if you want to divide 25 by 5 you can't subtract 5 five times or you get zero, not 5.

Answer 1

Notice that for addition, you started off with $0$. When you added $5$ to the number $0$ five times, you attained $25$. Division is the inverse process. It asks how many times one must subtract $5$ to come back to $0$.

Answer 2

To multiply $5$ by $5$ using addition:

$$\begin{align*} &0+5=5\qquad\text{count }1\\ &5+5=10\qquad\text{count }2\\ &10+5=15\qquad\text{count }3\\ &15+5=20\qquad\text{count }4\\ &20+5=25\qquad\;\text{ count }5\; \end{align*}$$ Here, the number $5$ has been repeatedly added $5$ times. We can say that the number $5$ has been added $5$ times from $0$. So, we can write this addition as $5 \times 5 = 25$.

To divide $25$ by $5$ using subtraction:

$$\begin{align*} &25-5=20\qquad\text{count }1\\ &20-5=15\qquad\text{count }2\\ &15-5=10\qquad\text{count }3\\ &10-5=5\qquad\text{count }4\\ &5-5=0\qquad\;\text{ count }5\; \end{align*}$$

Here, the number $5$ has been repeatedly subtracted $5$ times. We can say that the number $5$ has been subtracted $5$ times from $25$. So, we can write this subtraction as $25 \div 5 = 5$.

Answer 3

I think the fact that 25 is a perfect square is confusing you because the divisor and quotient are the same when divided by 5. If you take the number 30 and subtract the number “5” 6 times, you get 0. Thus, 30/5 = 6.

Answer 4

I remember from years ago a question about the Euclidean algorithm for the greatest common divisor (GCD). Someone else wrote something to the effect that the Euclidean algorithm does not actually use division.

That might be true from a philosophical point of view, though in practice you'll probably want to implement it with division.

For example, calculate $\gcd(5, 28)$. You already know the answer is 1, but please play along for a minute. Dividing, we find that $$\frac{28}{5} = 5.6.$$ But for the Euclidean algorithm we care more about the remainder, so $$\frac{28}{5} - \left\lfloor \frac{28}{5} \right\rfloor = 3,$$ and $0.6 \times 5 = 3,$ so this checks out.

But we can also get there by repeated subtraction: $28 - 5 = 23$, $23 - 5 = 18$, $18 - 5 = 13$, $13 - 5 = 8$ and $8 - 5 = 3$. I'd hate to have to do this for $\gcd(7028, 5)$.

So $28 = 5m_1 + 3$ (we don't really care what $m_1$ is), and then $5 = 3m_2 + 2$, and $3 = 2m_3 + 1$ and there's our answer.

Now, the analogy doesn't quite hold up when 0 is involved. For example, what's $\gcd(0, 43)$? Don't you try to divide 43 by 0, your calculator will melt... just kidding. But... $43 - 0 = 43$... tells us nothing, it does not move the algorithm along.

The point is that division is "basically" repeated subtraction, but it is not exactly that.