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Is $\; \infty = \frac{1}{0}$? My teacher says no but he wouldn't explain it. My question is why $\; \infty \neq \frac{1}{0}\;?$

My thinking:

Let $\frac{1}{x}=p$

Now as $x$ becomes smaller $p$ gets bigger. Ultimately when $x$ is smallest in magnitude then $p$ is largest which is what infinity is. Can anyone help me out?

PS: I have just started Calculus, therefore please try to give answers according to the level of my understanding

amWhy
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Mathslove
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4 Answers4

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I'd like to say why I suspect your teacher refused to address this issue.

In mathematics you often can't understand an individual thing without understanding the larger system of which it is part. For example you can't really understand what the number $17$ is in isolation. You really have to have a good idea of what numbers are in general before understanding a specific number like $17$.

To understand the symbol $\infty$ you need to understand the system in which it is playing a role. In particular, when the symbol $\infty$ (as opposed to other kinds of infinity like $\omega$ or $\aleph_0$) is used it's part of a special notation used for talking about limits. Until you have a good grasp of limits, it doesn't make sense to talk about this $\infty$. When you do grasp how it's being used, it'll become clear that the proposition $1/0=\infty$ isn't actually part of the standard framework in which the symbol $\infty$ is used.

There are ways to make sense of a statement like $1/0=\infty$. But none of them in standard in any way. If a textbook were to say $1/0=\infty$ you can be sure that earlier in the book they took care to say exactly what they meant because there isn't a standard.

(For example it might make sense to say $1/0=\infty$ if you're working on the Riemann sphere. But you probably ought to finish calculus before looking at that. And this use of $\infty$ is geometrical and is probably quite different to what you're currently imagining.)

Dan Piponi
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Approach it from both sides

You need to approach 0 from both sides in order to realize that 1/0 is undefined. If we approach 0 from the positive side ($\lim_{x\rightarrow 0^+}$), then we approach $+\infty$. If we approach it from the left (negative side) ($\lim_{x\rightarrow 0^-}$), then we approach towards $-\infty$.

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No, it's impossible. Doesn't exist any n s.t. $n*0=1$.

What is true is that $\lim_{n\rightarrow 0}1/n=\infty$

Jody
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  • Your final claim is incorrect. That limit itself does not exist, because if you approach zero from $n<0$, then you get a different result. – rajb245 Dec 03 '14 at 16:43
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You are both right and wrong, so to say. You are correct that, as $x$ gets closer to zero, $1/x$ gets larger and larger (towards infinity). We write this as

$$\lim_{x \to 0+} \frac{1}{x} = \infty$$

However, it does not make sense to simply plug in $x=0$ and say that $1/0 = \infty$.

You can argue this in many different ways, but it is perhaps enough to realise that this implies $1 = \infty \cdot 0$. But since it, by the same logic, would be true that $2/0 = \infty$ and hence $2 = 0 \cdot \infty$, we are forced to conclude that $1 = 2$, which is nonsense.

naslundx
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  • Why are we forced to conclude that? $\frac{0}{0}$ is undefined. isn't it? – Mathslove Jul 06 '14 at 17:19
  • @Mathslove Yes, that's true (just like 1/0, 2/0 and so on are undefined), but I did not mention 0/0? – naslundx Jul 06 '14 at 17:22
  • But $0 \times \infty$ is also undefined. Isn't it? Thats what you make use of to prove $1=2$ – Mathslove Jul 06 '14 at 17:42
  • @Mathslove Yes, $0 \cdot \infty$ is undefined. Hence, we must say that $1 / 0 = \infty$ is incorrect, because otherwise, by multiplying by 0, we get $0 \cdot \infty = 1$, which is a contradiction. – naslundx Jul 06 '14 at 18:54