7
$\begingroup$

All positive and negative numbers including zero are called integers. So in the form $a=bq$, since $0 = 0ㆍq$ is true for any integer $q$, $0$ can have $0$ as a divisor of itself as well as a multiple of itself by the definition expressed by $a=bq$.

But why it's said "We cannot divide by $0$"? It's understood as "$0$ cannot be a divisor" to me.

"Definition: An integer a is called a multiple of an integer $b$ if $a=bq$ for some integer $q$. In this case we also say that b is a divisor of $a$, and we use the notation $b | a$ . . . On the other hand, for any integer $a$, we have $0 = aㆍ0$ and thus $0$ is a multiple of any integer."

Source: Abstract Algebra: Third Edition, John A. Beachy, William D. Blair, p.4.

"Rule Division by $0$ is undefined. Any expression with a divisor of $0$ is undefined. We cannot divide by $0$"

Source: Prealgebra: A Text/Workbook, Charles McKeague, p.61.

"Observe that division by the integer $0$ is not defined, since for $n≠0$ there is no integer $x$ such that $0ㆍx=n$ and since for $n =0$ every integer $x$ satisfies $0ㆍx=0$"

Source: Introduction to Mathematical Proofs, Second Edition, Charles Roberts, p.99.

[Now I understand my question more after reading number theory chapter of a book]

$0=d\cdot 0 $
Thus, 0 is a multiple of every integer except 0.

enter image description here

$\endgroup$
8
  • $\begingroup$ Is there a single unique value that zero divided by zero would be? $\endgroup$
    – JB King
    Dec 28, 2015 at 15:49
  • $\begingroup$ @JB King Should there be a single value of zero divided by zero so that zero can be a divisor? $\endgroup$
    – buzzee
    Dec 28, 2015 at 17:10
  • $\begingroup$ @buzzee, for all other integer divisions the answer $q$ is a unique value. To have a multi-valued solution does present some challenges compared to other operations. $\endgroup$
    – JB King
    Dec 28, 2015 at 17:31
  • 4
    $\begingroup$ You are mixing terminology from three different books with very different focuses. One of them is pre-algebra - intended for students who may be 12 years old. Another is abstract algebra for college students. The underlying phenomenon is always the same, but the terminology that the books use to describe it varies depending on their audience and on the author's taste. $\endgroup$
    – Carl Mummert
    Dec 29, 2015 at 14:51
  • 1
    $\begingroup$ What do you mean by "$0\cdot q=0$ is not allowed"? We certainly are allowed to write $0\cdot q=0$ for any integer $q$, and this is how we know that $0$ is a multiple of $q$ for any integer $q$. It is also how we know that $0$ is a multiple of $0$ and that $0$ is a divisor of $0$. $\endgroup$
    – David K
    Jan 4, 2016 at 5:47

1 Answer 1

Reset to default
8
$\begingroup$

Every integer divides zero, including zero itself; however, the only integer that zero divides is itself. That is, $b \mid 0$ for all integers $b$; but if $a$ is an integer and $0 \mid a$, then $a = 0$.

When it is said that "you can't divide by zero", what is meant is that, given an integer $a$, there is not a unique quotient upon division by zero.

Specifically, given integers $a$ and $b$, with $b \ne 0$, if $b \mid a$ then there is a unique integer $q$ such that $a = qb$, namely $q=\frac{a}{b}$. However, if we allow the case when $b=0$, then we lose the uniqueness. Indeed, as already mentioned, $0 = q \cdot 0$ for all integers $q$, so it makes no sense to assign a value to the expression $\frac{0}{0}$. And if $a \ne 0$ then there is no integer $q$ such that $a = q \cdot 0$, so it also doesn't make any sense to assign a value to the expression $\frac{a}{0}$.

$\endgroup$
8
  • $\begingroup$ So is the statement "We cannot divide by 0" wrong? $\endgroup$
    – buzzee
    Dec 28, 2015 at 17:07
  • 2
    $\begingroup$ @buzzee It's not wrong. It's just that "$a$ is divisor of $b$" and "$b$ can be divided by $a$" mean two different things. $\endgroup$
    – Wojowu
    Dec 28, 2015 at 17:39
  • $\begingroup$ @Wojowu Can I get a more detailed explanation of how two statements are different? I thought they all can be expressed by "a=bq", where a is a multiple of b and, b is a divisor of a when we let a, b, q are integers. $\endgroup$
    – buzzee
    Dec 28, 2015 at 17:56
  • 1
    $\begingroup$ @buzzee We define $a/b$ to be the unique number $q$ such that $a=bq$, if it exists. For $b=0$, either existence or uniqueness fails. $\endgroup$
    – Wojowu
    Dec 28, 2015 at 18:03
  • 1
    $\begingroup$ @buzzee It is. The problem is with other direction. If we know that $a$ is divisor of $b$, then we can't conclude $a/b$ exists. For example, take $a=b=0$. $0$ is a divisor of itself, but there is no unique number $q$ giving $0=0q$. Hence $0/0$ doesn't exist. $\endgroup$
    – Wojowu
    Dec 28, 2015 at 18:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .