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?


3 Answers 3


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.

  • $\begingroup$ 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 $\endgroup$ 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.

Source: https://en.wikipedia.org/wiki/Rational_number


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:


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


We can also express $7$ as


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.


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