# Most students asked why is that ${a\over b}\div{c\over d}={ad\over bc}$

Most students asked why is that

$${a\over b}\div{c\over d}={ad\over bc}$$

I just told them: inverse the second fraction and multiply. Why? They ask me. I have no idea.

Any logical answers to them kids?

These day teachers just told students at secondary to memorise and no why is allowed in lessons. I think that is wrong. Putting people of studying maths.

• See mweiss's wonderful answer to an analogous question at MESE. – pjs36 Jun 1 '16 at 20:50
• My experience is that arguments involving commutative, distributive, etc. laws do not work very well for kids. To them you're just replacing one "black box explanation" with some other "black box explanations". In my opinion, this is not something you can take care of quickly, and it requires a slow, steady, and practiced approach. This should be something they got earlier, but . . . For some first steps, see my answer at Teaching fractions: the generalization problem. – Dave L. Renfro Jun 1 '16 at 20:59

By definition, $x \div y$ should be a number $z$ such that $x = y z$.
So you just verify: $$\dfrac{ad}{bc} \times \dfrac{c}{d} = \dfrac{a}{b}$$ Ultimately this works because of the associative and commutative laws of multiplication, but you don't have to tell the students that.

"Why" should always be allowed.

When you divide $x÷y$, it is the same as $\frac{x}{y}=x×\frac{1}{y}$. Students familiar with fractions should be able to understand this. This is true no matter what $x$ or $y$ are, even if they're fractions. Therefore, $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}×\frac{d}{c}=\frac{ad}{bc}$.

I hate the $\div$ symbol, and would ban it if I had the power.

Fractions ARE division. Division IS multiplication by the reciprocal. No need to learn 3 different notations.

I would teach that division is multiplication by the reciprocal first with whole numbers. Then introducing division by fractions, you say that is it exactly the same.

Anyway, I learned the flip it over and multiply as rote process, and I did it, and didn't know why it worked, but I was good at following instructions. It finally clicked when I had:

${\frac ab} \div{\frac cd} = \dfrac {\frac ab}{\frac cd}$

How do you get the fraction out of the denominator? You multiply top and bottom by $d.$ And to get the fraction out of the numerator? You multiply top and bottom by $b.$

$\dfrac {ad}{cb}$

Depending on how much these students know, I'm a fan of this explanation:

$$\frac ab \div \frac cd = \frac{\frac ab}{\frac cd} = \frac{\frac {ad}{b}}{\frac {cd}{d}} = \frac{\frac {ad}{b}}{c} = \frac{\frac{ad}{bc}}{\frac cc} = \frac{ad}{bc} = \frac ab \cdot \frac dc$$

The only thing they have to accept is that we can multiply any fraction by $\frac dd$ or $\frac {\frac{1}{c}}{\frac{1}{c}}$, which should be obvious to them if they're doing fraction multiplication.

The motivation is there, too: we first get rid of the pesky $d$ from the denominator by multiplying by it, and then we get rid of the pesky $c$ by dividing by it: all in all, we've multiplied our $\frac ab$ by $\frac dc$.

I think the merit to this method is that it moves away from the notion of the $\div$ symbol and expresses the result as a "fraction of fractions," which is much more generalizable.

Like the additive inverse of a number is its negative, and subtracting is just adding a negative, the multiplicative inverse of a number is its reciprocal, and therefore dividing is just multiplying by a reciprocal. Assuming they know why the method is what it is for multiplying fractions, they will be able to continue on once they convert dividing to multiplying by a reciprocal.

We formally define addition and multiplication. We then also formally define the additive inverse of an element and the multiplicative inverse of an element (in the case that they exist).

We can then define "subtraction" and "division" in terms of addition and multiplication and inversion of each respectively.

The additive inverse of an element, $b$, we will denote as $(-b)$. The additive inverse of $b$ is the unique element $(-b)$ such that $b+(-b)=0=(-b)+b$.

When we talk about subtraction of two elements, $a-b$, this is really just shorthand for $a$ plus the additive inverse of $b$, i.e. $a-b = a+(-b)$.

In the same way, $a\div b$ is just shorthand for $a\times (b^{-1})$

We also can define fractions in the same way: $\frac{a}{b} = a\div b = a\times (b^{-1})$

One has as a result:

$$\frac{a}{b}\div (\frac{c}{d}) = a\times b^{-1}\times (c\times d^{-1})^{-1} = a\times b^{-1}\times c^{-1}\times d = (a\times d)\times (b^{-1}\times c^{-1})=\frac{ad}{bc}$$

By learning patterns and rules for division and subtraction, one can avoid needing to write everything in terms of multiplications and additions, saving some time.

$\frac{a}{b}=\frac{\frac{ad}{b}}{d}$. So $\frac{ad}{bc}$ is how many times $\frac{c}{d}$ goes into $\frac{\frac{ad}{b}}{d}$.

• "How many times"? It's not in general a positive integer. – Robert Israel Jun 1 '16 at 20:42