# Reduction of fractions

Last night I was fiddling with some equations and admittedly, I made a careless mistake because I was exhausted. However, in doing so, I began to question the process of what I used to understand as "cancelling" in fractions. The equation was:

(6 ÷ 2)/2

I cancelled the divison by 2 with the 2 in the denominator.

After some research, I came to realise that a better term is reduction to 1 as it is a better indication of what we are actually doing when simplifying fractions.

However, I would like to know the reason why we can't do what I did because if a student asks, I would like to see if there is an explanation beyond BODMAS.

Also, I do believe that we can only cancel factors but if a student were to say ((6 ÷ (2)(1))/2, we now have factors that we seem to be able to cancel, but because of the division in the numerator, we cannot. Is there a more articulate and mathematically correct way of putting this?

• The answer to the problem you posed is 1.5 (1.5 = 3/2). There are two divisions by 2. They do not cancel one another. Using the fraction bar only, as suggested in the answer, will help you see your mistake. Aug 11 at 16:10
• I think the question "why can't we cancel in this case?" has already fallen prey to a false frame, which is "we can generally cancel when we see common terms". That's not how cancellation works: rather, in very specific situations, where we see common terms related by specific operations, we can cancel the common terms. So the default assumption should be that cancellation is not valid, and any time we want to cancel, we should ask ourselves "why can we cancel in this case?". Aug 12 at 15:59
• Who would use both ÷ and / in the same equation? They both mean the same thing. Most people would stick with one symbol or the other and use it consistently unless it was deliberately designed to confuse people. Aug 12 at 16:31

Your first example is equivalent to $$(6\div2)\div2$$. Writing it this way, the erroneous cancelation can be seen as treating division as associative, that is, as writing $$(6\div2)\div2=6\div(2\div2)=6\div1=6.$$ The mistake in this calculation is that division, like subtraction, is not associative. Compare the similar erroneous calculation $$(6-2)-2=6-(2-2)=6-0=6.$$

Turning to your second example: by $$((6 ÷ (2)(1))/2$$ do you mean $$(6\div(2\times1))/2$$ or $$((6\div2)\times1)/2$$? By the precedence rules I am accustomed to it should mean the latter, but this kind of expression is sufficiently confusing to most people that it is best to be more explicit about what you mean by putting in additional parentheses.

The reason I ask is that you refer to the $$2$$ in the numerator as a "factor", which I can only make sense of if you are using the first interpretation (so that $$2$$ is a factor in the product $$2\times1$$). Assuming that's the interpretation you had in mind, the reason the cancelation of $$2$$s is not allowed is that it would violate a general principle of cancelation in fractions. That general principle is that cancelation is only possible when both numerator and denominator are products and, in addition, those two products have a factor in common. To emphasize, the numerator and denominator must be products; it's not enough for them merely to contain products as subexpressions. So just as you can't cancel the $$2$$s in $$(1+2\times3)/(2\times4)$$, you can't cancel the $$2$$s in $$(6\div(2\times1))/2$$. In neither case is $$2$$ a factor of both the numerator and the denominator.

Returning to your first example, you can write $$(6\div2)/2$$ as $$\left(6\times\frac{1}{2}\right)/2$$. Writing it this way, it is clear that a factor of $$\frac{1}{2}$$ does not cancel a factor of $$2$$. On the other hand, consider $$(6\div2)/(3\div2)$$. This can be written as $$\left(6\times\frac{1}{2}\right)/\left(3\times\frac{1}{2}\right)$$, and cancelation of the factors $$\frac{1}{2}$$ is legitimate.

• this really helped. The last 3 sentences in the 4th paragraph are exactly what I was looking for but couldn't articulate. Thank you so much! Aug 11 at 22:50

Cancelling is a shorthand for:

• The rule for multiplying fractions, used in reverse
• Any number (except 0) divided by itself is 1
• The rule for multiplying by 1

$$\frac{ac}{bc} = \frac{a}{b} \cdot \frac{c}{c} = \frac{a}{b} \cdot 1 = \frac{a}{b}$$

• I don't know if you'd agree, but I think the usage of fraction lines instead of $/$ or $\div$ (which is implicit in your answer) makes one less prone to errors. Aug 11 at 10:42
• @MichałMiśkiewicz Your comment is spot-on: it really did take me multiple parses to figure out that cancelling out the '2' in $(6 ÷ 2)/2$ is a mistake, whereas it would've been immediate from $\dfrac{6\div2}2$ (despite the '÷' still in the numerator). $\quad$ It also doesn't help that the expression uses differing symbols to represent the same operation. $\quad$ Thirdly, Will's point about division not being associative. $\quad$ All points considered, $(6 ÷ 2)/2$ is simply more taxing than $\dfrac{6\div2}2$ to correctly process. Aug 12 at 4:09

I think the best defense against this sort of error (and explanation for students) is to know what the process of cancelling is actually doing and why it works.

If you think of cancelling as just spotting the same number (or a multiple thereof) above and below the line and getting rid of them, then the sort of mistake you made is easy. To guard against it you need to remember a bunch of auxiliary rules like "don't cancel something that's a right operand of a division".

However it actually gets much simpler if you remember that the process of cancelling is not "spot two of the same number and get rid of them" but rather dividing the numerator and denominator by the same number.1

In this case dividing the denominator by 2 gives us $$2\div2$$, which results in 1 as we wanted. But $$(6\div2)\div2$$ simply isn't $$6$$ by the ordinary rules for division (which the students hopefully already understand); it isn't anything to do with special conditions on cancelling. The cancelling process was done when we added $$\div2$$ to both sides; the rest is ordinary simplification.

1 Which is based on the observation that if we multiply or divide both numerator and denominator the same number (other than zero) it's equivalent to multiplying by $$\frac{n}{n}$$ for some $$n \ne 0$$, which is multiplying by $$1$$, which must produce an equal fraction.

• Clean and simple-to-understand answer. Welcome to this site! Aug 12 at 3:39
• This was a very good explanation. Thank you for your help! Aug 14 at 12:19