Why does $x$ divided by zero not equal $x$?

Why does $x$ divided by zero not equal $x$? After all, $x$ is not being divided by anything.

Solution 1:

Think of division as an "inverse" operation to multiplication.

For instance, let us look at $\frac{a}{b}$ means.

Let $x = \frac{a}{b}$. What it means is when $x$ is multiplied by $b$ it gives $a$.

So, if you want to make sense of $\frac{a}{0}$.

Say $\frac{a}{0}$ is $y$.

So, we want $y$ such that $y \times 0$ gives us $a$.

But we know that $y \times 0 = 0$, $\forall y$.

So there is a logical inconsistency here.

Solution 2:

I just want to mention that in certain situations it is perfectly valid to think of something nonzero divided by something zero as "infinity." For example, the slope of the line between the points $(a, b)$ and $(c, d)$ is $\frac{d - b}{c - a}$, unless $c - a = 0$ and $d - b \neq 0$, where this expression stops making sense. However, the corresponding line is vertical, and it is perfectly reasonable to think of its slope as being "infinity." More precisely, the space of slopes is a gadget called the projective line $\mathbb{P}^1(\mathbb{R})$ and if all one cares about is the geometric structure of this space (that is, one is not too concerned with adding and multiplying) it is perfectly valid to think about "infinity" in the same way as any other element, since they're all taken to each other by projective transformations. These ideas are very important in certain branches of mathematics, such as algebraic geometry.

A more concrete application is that the tangent function can be thought of as taking values in $\mathbb{P}^1(\mathbb{R})$ instead of in $\mathbb{R}$, and then it no longer has "singularities"; in fact, it defines a homeomorphism from the circle to $\mathbb{P}^1(\mathbb{R})$. Doing the same thing over $\mathbb{C}$ allows you to think of meromorphic functions as functions to the Riemann sphere $\mathbb{P}^1(\mathbb{C})$, and this idea also has many fruitful applications.

Solution 3:

I would say that if you're dividing by 1, then you're "not really dividing", which is why $X/1=X$. Dividing by $0$, however, is undefined. Here's a way to see this based on basic properties of arithmetic. Suppose we could define, say $3/0$ as a real number $r$. Then multiplying both sides of the equation $3/0=r$ by $0$ yields $3=0\cdot r$. But $0\cdot r=0$ for all real numbers $r$, so this is impossible.

Informally, dividing $X$ by $n$ tells you how big to make the pieces of pie if you have to serve them to $n$ people. This means that $n$ pieces of size $X/n$ give a total amount of $X$. But if $n=0$, then no matter the size of each piece, $0$ of them can never give a total amount of $X$. (If $X=0$, then the reasoning becomes more intricate.)

(This reminds me of another confusion involving $0$ and language. I have seen a temptation to replace "the solution is 0" with "there is no solution," and vice versa among precalculus or beginning calculus students.)

Added:

Perhaps this problem of contrasting dividing by $0$ and "not dividing by anything" can be further clarified by comparison with what happens when adding and subtracting. When you add $0$, you're "not adding anything", and sure enough $X+0=X$ holds for all $X$. Similarly for subtraction: $X-0=X$, and here the interpretation "not subtracting anything" works out. This works because $0$ is a neutral element with respect to addition (by definition). Changing operations to multiplication, $0$ is no longer the neutral element; $1$ is. Multiplying by $1$ is what translates to "don't change anything": $X\cdot 1=X$. Thus "undoing" multiplication by $1$, naming dividing by $1$, also doesn't change anything. But since $0$ no longer has this innocuous behavior with respect to multiplication, things similarly change for division. It changes drastically, because multiplication by $0$ becomes something that can't be undone.

Solution 4:

The best way to think of it is check out what happens when you divide by the numbers near zero. For example $$\frac{1}{1} \;\;\text{and} \;\; \frac{1}{-1}$$ Then get closer to zero on either side. For example $$\frac{1}{.5} \;\;\text{and} \;\; \frac{1}{-.5}$$. Then even closer. $$\frac{1}{.1} \;\;\text{and} \;\; \frac{1}{-.1}$$. Do you see how the numbers are getting further and further apart? This is why we have no clue what dividing by zero might mean.