# Intuition for Division by fraction / Can you only divide by (non-zero) integers?

Goal: I need help to intuitively understand division by a fraction.

Background: I've read this and this. I understand that division is "even distribution" so, with integers, $\frac 6 2 = 3$ could mean that 6 friends get 2 chocolates each.

Observation: When the denominator is >0, the solution is less than the numerator (e.g., $\frac 3 4$ and $\frac 4 3$), but when 0 < denominator < 1, the solution is larger than the numerator (e.g., $\frac {1} {\frac 1 2}$ = 2). Graph of y=$\frac 1 x$.

Problem: "Dividing" (as understood by "into smaller parts") by a fraction doesn't seem possible: you end up multiplying, never dividing--and yes, I understand that division by a fraction is multiplication by that fraction's multiplicative inverse (Khan Academy).

Question: Since 3--2 (three minus negative two) = 3+(-(-2))--i.e., addition not subtraction (source)--is division by a fraction similar? In other words, is division by a fraction not division at all, but multiplication?

The definition of division is: $a \div b$ is $c$ where $a = b \times c$. In the case of positive integers $b$ you can interpret this as even distribution of an amount $a$ into $b$ equal parts. This is because if $b$ is a positive integer, $b \times c$ is the result of adding together $b$ copies of $c$.

If $b$ is a fraction $m/n$, where $m$ and $n$ are positive integers, you can think of $b \times c$ as the result of adding together $m$ copies of $c/n$, i.e. you split $c$ into $n$ equal pieces, and you add together $m$ of those pieces (or pieces equal to one of those if $m$ is greater than $n$). Thus $a/b$ would be an amount $c$ that if you divide into $n$ equal pieces and add $m$ of those pieces together you get $a$.

However, when you go beyond fractions (e.g. to real numbers), even this doesn't make sense. Again, the definition is in terms of multiplication, the interpretation is secondary.

• Thank you! Just to clarify (at least for me), the second paragraph refers to re-organizing a/(m/n) = c as m * (1/n) * c = a Mar 10, 2017 at 15:38

I don't have a math research job. I don't know everything with certainty. This is just my guesswork that the following is a simplifying approximation of reality.

There's no universal law that a binary operation whose steps include only multiplication and not division of integers cannot be called division. A rational number can be defined as an equivalence class of ordered pairs of integers $$(a, b)$$ where $$b \neq 0$$. From now on when I talk about how operations are defined on ordered pairs of integers, I really mean that that's how it's defined on ordered pairs of integers whose second coordinate is nonzero.

For $$(a, b)$$ is said to be equivalent to $$(c, d)$$ when ever $$ad = bc$$. We can do that because it can be shown that that is an equivalence relation. Addition is defined such that the sum of the equivalnce class containing $$(a, b)$$ and the equivalence class containing $$(c, d)$$ is the equivalence class containing $$(ad + bc, bd)$$ and multiplication is defined such that the product of the equivalcne class containing $$(a, b)$$ and the equivalence class containing $$(c, d)$$ is the equivalence class containing $$(ac, bd)$$. It can also be shown that there is a way to define it that way.

Technically, a field is not a set but an ordered triplet of a set and operations on that set. You can denote a set with operations how ever you want and $$\mathbb{Q}$$ is just one possible variable you could use to denote it. Most mathematicians agree to use the notation $$(\mathbb{Q}, +, \times)$$ to refer to any field that is isomorphic to the desired structure of the rational numbers with those operations and call it the rational numbers with addition and multiplication. I believe the mathematical community also agreed that once you have a field and you denote the second coordinate for that field as $$+$$ and the third coordinate of it as $$\times$$, $$a - b$$ is defined to mean $$a$$ plus the additive inverse of $$b$$ and $$a \div b$$ is defined to mean $$a$$ times the multiplicative inverse of $$b$$ if $$b$$ is not the additive identity.

The rational numbers the way I defined them with addition and multiplication the way I defined them can be shown to be a field. Let * denote the operation of taking the equivalence class an ordered pair of integers is in. It can be shown that when ever $$a$$ is an integer and $$b$$ is a nonzero integer, $$*(a, b) = *(a, 1) \div *(b, 1)$$. However, this is true only when $$a$$ is an integer and $$b$$ is a nonzero integer. Since technically no rational numbers are integers according to this definition, it's in fact the case that $$\forall a \in \mathbb{Q}\forall b \in \mathbb{Q}(a, b) \notin \mathbb{Q}$$ so it is not correct to say $$*(a, b) = *(a, 1) \div *(b, 1)$$ or $$*(a, b) = a \div b$$. Despite that, we can still define division as $$*(a, b) \div *(c, d) = *(ad, bc)$$ when ever $$c$$ is nonzero.

Also once $$(\mathbb{Q}, +, \times)$$ has been defined, $$(\mathbb{Z}, +, \times)$$ can be redefined in such a way that $$\mathbb{Z}$$ is a subset of $$\mathbb{Q}$$; the operations in $$(\mathbb{Z}, +, \times)$$ are the exact same operations as the ones in $$(\mathbb{Q}, +, \times)$$ in that order; and $$(\mathbb{Z}, +, \times)$$ has the exact same structure as it was defined to have when using that to construct $$(\mathbb{Q}, +, \times)$$. Most authors use the fractional notation synonymously with division so when $$a$$ an $$b$$ are elements of set that was originally called $$\mathbb{Z}$$ with $$b$$ nonzero, we cannot use $$\frac{a}{b}$$ to denote $$*(a, b)$$ but when $$a$$ and $$b$$ are elements of $$\mathbb{Q}$$ with $$b$$ nonzero, we can use $$\frac{a}{b}$$ to denote $$a \div b$$. In some books, the fractional notation is used to denote division when describing the rule for differentiating a quotient.

First of all, a fraction is any real number of the form $$\frac{a}{b}$$, where $$a, b$$ are integers and $$b \neq 0$$. So to answer your second question in the title, yes---you can only divide by nonzero integers; for otherwise, to say that this is possible:

$$\frac{a}{0}$$

is to say that $$0$$ divides into $$a$$ a certain number of times, say $$x$$; that is, $$x$$ is that number by which

$$0 x = a;$$

But as we can see, no such number $$x$$; and so, division by $$0$$ is not permitted.

Now, getting to the first part of what you ask in the title---

Any rational number is expressible as a fraction; e.g., $$7$$ is a rational number, and so, can be written exactly for example, as

$$\frac{7}{1};$$

We can also express $$7$$ as

$$\frac{1}{\frac{1}{7}}$$

although in this form it is not a fraction, but a rational number equivalent to $$7$$.

To see this, I now want to multiple the last expression by a number equivalent to $$1$$ but expresses such that when I multiply $$\frac{1}{1/7}$$ by it, the denominator becomes $$1$$.

The number I look for is $$\frac{7}{7}$$; and so,

$$\frac{1}{\frac{1}{7}} \cdot \frac{7}{7} = \frac{(1)(7)}{\left(\frac{1}{7}\right)(7)} = \frac{7}{1} = 7.$$

We usually think of division as being an arithmetic operation separate from multiplication; but this is not true. Just as subtraction is a form of addition, so is division a form of multiplication.

Again, take the fraction $$1/7$$. We think we are dividing $$7$$ into $$1$$ (which we are), but equivalently, we are multiplying $$1$$ by the reciprocal of the denominator; that is,

$$\frac{1}{7} = 1 \left(\frac{1}{7}\right).$$

I hope this helps a little.