Why does this show that $\frac{0}{0}= 1$?

Question

So I was learning how to divide polynomials and I was making this one...

$$\frac{x^2+3x+2}{x+1}=x+2$$

When I graph the equation in Desmos, it looks like when $x = -1$ we have undefined in one part of the equation and $1$ in the other, when we solve $x = -1$ the left side ends up being $\frac{0}{0}$ while the right side becomes $1$... Both functions are by the way exactly the same with the exception that for $x=-1$ one has a blank point and the other doesn't.

Does that means then that, it is at least possible to divide $\frac{0}{0}$ or that $x+2$ is NOT equal to the polynomial division?

Thanks!

Answer 1

From $\frac{2(x^2+3x+2)}{x+1} = 2(x+2)$, we discover by exactly the same method that $0/0 = 2$. In fact, from $\frac{k(x^2+3x+2)}{x+1} = k(x+2)$, we discover that $0/0 = k$ for any $k$ you like.

But if $a = b$ and $a = c$, we must have $b = c$. This means $1 = 0/0 = 2$, which is clearly nonsense. (In fact, we have $k = 1$ for any $k$ you like, which is even more clearly nonsense.)

The difficulty is you have read that equals sign as promising more than it actually does. In the form $f = g$ where $f$ and $g$ are functions, the equality only promises that $f$ and $g$ give the same output values for a given input value. This claim is only possible when that input value is in both of their domains. Equality of functions only happens at points where both functions are defined.

The resolution to your question is "$x-2$ is equal to the polynomial division but may or may not be the same function as $x-2$ depending on what domain you assign to that polynomial." Remember every function has a domain. Two functions given by the same recipe, "$x-2$" but with different domains are different functions. The maximal domain of the rational function $\frac{x^2+3x+2}{x+1}$ is $(-\infty, -1) \cup (-1,\infty)$. (I say "maximal domain" because the convention is that a function has its maximal domain unless specified otherwise.) On that domain the quotient is the same as $x-2$, but now we have an ambiguity. Do we mean

• $x-2$ with domain $(-\infty, -1) \cup (-1,\infty)$ or
• $x-2$ with domain $(-\infty, \infty)$?

We can't tell which is meant from what is written. The quotient is equal to both of these functions. These two functions are equal (recall that this is because equality only inspects the functions where both are defined).

The quotient is exactly the same function as "$x-2$ with domain $(-\infty, -1) \cup (-1,\infty)$" (some values and same domains). The quotient is equal to "$x-2$ with domain $(-\infty, \infty)$" (same values on their common domain). When you wrote "$x-2$", which of these two functions did you mean?

Answer 2

Eric's answer deals well with the domains of functions, but I wanted to add an answer more specifically about $0/0$. $0/0$ is undefined, not because it is nothing, but because it could be anything. Remember, division is the reverse of multiplication. So, when we say, for instance, $8 / 2$, what we are asking for is the number, when multiplied by $2$, will yield $8$.

So, for $0/0$, we are asking for the number, when multiplied by zero, will be zero. Well, that's any number. That is why it is undefined. Not that there is necessarily no answer, but because, when you have a zero multiplier, it destroys all the information that used to be there, so you can't reverse the process. In fact, this was one of the reasons why zero was not originally a number. Since you could not divide by it, they thought it couldn't actually be a number.

Anyway, even though $0/0$ is generally nonsense (because it could mean anything), you actually can use some tools to figure out what it is supposed to mean in some particular case. In calculus, we call these "limits". What you can do is, rather than put in $-1$ for $x$, put in a number that is infinitely close to $-1$. Let us use $\epsilon$ as a value that is infinitely small, an infinitesimal. Therefore, let us substitute in $(-1 + \epsilon)$ in for $x$ and see what jumps out:

$$ \frac{x^2 + 3x + 2}{x + 1} \\ =\frac{(-1 + \epsilon)^2 + 3(-1 + \epsilon) + 2}{(-1 + \epsilon) + 1} \\ =\frac{1 + -2\epsilon + \epsilon^2 + -3 + 3\epsilon + 2}{\epsilon} \\ =\frac{\epsilon^2 + \epsilon}{\epsilon} \\ =\frac{\epsilon(\epsilon + 1)}{\epsilon} \\ =\epsilon + 1 $$

So, what is $\epsilon + 1$? Since $\epsilon$ is an infinitely small value, $\epsilon + 1$ is infinitely close to $1$. In other words, when $x$ is infinitely close to $-1$, the result is infinitely close to $1$. A technical way of saying this is "the limit as $x$ approaches $-1$ is $1$." This technique works well for smooth, continuous functions.

So, in summary, the reason why $0/0$ is undefined is that it could be any value - it is not possible to tell what the value should be just using the formula as-is. However, by pushing the formula an infinitely small amount one way or another, we can see what the value should be in our particular version of $0/0$.