I'm currently reading "Knowing and Teaching Elementary Mathematics" by Liping Ma. For those unfamiliar with it, it's a comparison of the teaching approaching of US versus Chinese elementary teachers in Mathematics from around the late 1990s. In Chapter 3, which discusses division by fractions, they are looking at alternate approaches to a problem and proofs of why each approach works. I have no problem understanding or answering the problem, but I find in one approach I don't understand the logic behind one of the steps in the proof. Actually, that's not strictly true, I understand why it's valid, I just don't understand what the underlying "theorem" is that allows you to apply the specific transformation "straight off" in that context. Unfortunately there's only one example so I can't compare/contrast to try and spot what I'm missing. The problem/proof is:

$$ \begin{align} 1\frac{3}{4} \div \frac{1}{2} & = \frac{7}{4} \div \frac{1}{2} \\ & = (7 \div 4) \div (1 \div 2) \\ & = 7 \div 4 \div 1 \times 2 \\ & = 7 \div 1 \div 4 \times 2 \\ & = (7 \div 1) \div (4 \div 2) \\ & = \frac{7 \div 1}{4 \div 2} \end{align} $$

I can happily prove why the transformation from the improper fraction to the final step is correct. The step I have a problem with between the 2nd and 3rd steps where the parentheses are removed and the final $\div$ is transformed to a $\times$. Apparently this whole proof is based on basic principles such as the order of operations, and the equivalence between a fraction and a division expression, so I guess I'm missing some of those :-).

If we have $a \div (b \div c)$ where $a$ substitutes for the first bracket then we want to transform the divisor in the bracket to a 1 - i.e:

$$ \begin{align} a \div (c \div d) & = \frac{a}{\frac{b}{c}} \\ & = \frac{a}{\frac{b}{c}} \times 1 \\ & = \frac{a}{\frac{b}{c}} \times \frac{\frac{c}{b}}{\frac{c}{b}} \\ & = \frac{ac}{b} \end{align} $$

Using that we can say:

$$ \begin{align} (7 \div 4) \div (1 \div 2) & = ((7 \div 4) \times 2) \div 1 \\ & = 7 \div 4 \times 2 \div 1 \\ & = 7 \div 4 \div 1 \times 2 \end{align} $$

So I can justify the step, I just don't understand the rationale behind doing it in one step. Can anyone see what I'm missing please?

  • $\begingroup$ Without parenthesis, they are applying division/multiplication from left to right in lines 3 and 4. The whole sequence is ghastly though. Awful. $\endgroup$ – Paul Mar 5 '18 at 12:15
  • $\begingroup$ I'm OK with the reordering of the operations since division is equivalent to multiplication by reciprocal so associative/commutative laws apply. $\endgroup$ – Windscale Mar 5 '18 at 13:29

In this context, multiplication and division work similarly to addition and subtraction. What is being done is the transforming $\;(a-b)\!-\!(c-d)\;$ to $\;(a-c)\!-\!(b-d)\;$ but using $\div$ instead of $-$.

You can think of the numbers $\;a,b,c,d\;$ as being on the corners of a square. The top has $a$ and $b$ and the bottom has $c$ and $d$. If we add all of the numbers together we get $\;S:=a+b+c+d.\;$ Suppose that we can add only two numbers at a time, so we add pairs of numbers and then add the resulting sums in three ways: $\;S = (a+b)+(c+d) = (a+c)+(b+d) = (a+d)+(b+c)\;.$

Now do the same thing but with using $-b$ and $-c$. In this case $\;S := a-b-c+d.\;$ Adding only two numbers at a time, then $\;S = (a-b)+(d-c) = (a-c)+(d-b) = (a+d)-(b+c),\;$ where we used the results: $\;x+(-y)=(-y)+x=x-y\;$ and $\;(-x-y)=-(x+y).\;$ Moerover, by using the result $\;+(x-y)=-(y-x)\;$ twice we now have the equalities $$S = (a+d)-(b+c) = (a-b)-(c-d) = (a-c)-(b-d)$$ and you can see how $b$ and $c$ can be switched because addition is commutative.

  • $\begingroup$ Thanks. I'm trying to get my head round this :-). $\endgroup$ – Windscale Mar 5 '18 at 14:30
  • $\begingroup$ I can understand your first bit: (a - c) - (b - d) = (a - b) - (c -d) = a + (-b) - (c + (-d)) = a + (-b) - c - (-d) = a - c + (-b) - (-d) = (a - c) - (b - d). And I can see that you can do the same with division except instead of turning the subtractions into the addition of negatives, you turn the divisions into multiplications by the reciprocal. $\endgroup$ – Windscale Mar 6 '18 at 5:52
  • $\begingroup$ But I don't understand your square argument. I can see you might consider a, b, c, d as being the values of vertices of a graph with edges a <-> b, a <-> c, c <-> d, b <-> d. But I don't understand why the three equalities you've mentioned would hold? If I start with the first one, expand the bracket the rearrange I get: (a + d) - (b + c) = a + d + (-b) + (-c) = a + (-b) + (-c) + d = (a - b) + (d - c) = (a - b) - (c - d) so that one holds. $\endgroup$ – Windscale Mar 6 '18 at 6:09
  • $\begingroup$ And also: (a + d) - (b + c) = a + d + (-b) + (-c) = a + (-c) + (-b) + d = (a - c) + (d - b) = (a - c) - (b - d). So clearly all three are equal. So I think I don't understand where the square idea comes from and why it works? $\endgroup$ – Windscale Mar 6 '18 at 6:16

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