Given, $\dfrac{1}{\infty}=0$, then $1=0 \cdot \infty = 0$ (because $0$ times any number or values is $0$ and here that number is infinity). Which gives us $1=0$ i.e, $0=1$. Hence proved....

5$\infty$ is not a number, so you can't divide by it. – May 27 '16 at 00:05

I thought infinity times any number is infinity. Which one would win in this fight? – Kitter Catter May 27 '16 at 00:06

1To add to Zachary Selk's comment, there are some number systems which include $ \infty $ as a number, but these systems must exclude at least one of the rules you used. For example, in the [extended real line](https://en.wikipedia.org/wiki/Extended_real_number_line), the rule that $ 0 $ times any number is $ 0 $ is not always true, since $ 0 \times \infty $ is undefined. – cardboard_box May 27 '16 at 00:17

@KitterCatter You have to put the statement in a technical context. For example in the extended reals we stipulate that the symbol $\frac{a}{\infty} = 0$ for any real number $a$. However we can't then define multiplication by $\infty$ for exactly the reason the OP points out, that it leads to conclusions inconsistent with the rest of math. https://en.wikipedia.org/wiki/Extended_real_number_line – user4894 May 27 '16 at 01:01
2 Answers
$1/\infty=0$ is sometimes taken to mean $1/x$ can be made as close to $0$ as desired by making $x$ big enough. You tell me how close you want $1/x$ to be to $0$; I'll say if you make $x$ at least as big as thusandso, then $1/x$ will be at least as close to $0$ as what you specified.
As for $0$ times $\infty$, consider for example $\dfrac 5 x$ and $x$. If you multiply those, you get $5$. And if you let $x$ approach $\infty$, so that $\dfrac 5 x$ approaches $0$, then the product of those two still approaches $5$. Hence you could argue that $0\cdot\infty=5$. And similarly that it equals $6$. One says that the expression $0\cdot\infty$ is an "indeterminate form", and that means if one thing approaches $0$ and another thing approaches $\infty$, then when you multiply then the product could approach $5$ or $6$ or any other particular number, or $0$ or $\infty$, depending on what those two things are that you're multiplying.
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The reasons why this is false were already explained, but I'll try to word it my way:
First, the usual operations (addition, division, etc...) are defined for numbers, i.e., elements of $\mathbb{R}$. The operations have some properties (associativity, commutativity, existence of inverses for nonzero numbers, etc...) (a field).
$\infty$ is not a number, but rather a quite useful theoretical concept which appears from time to time in some calculations (for example some given by limits), and so it interacts with the numbers in some ways. So it is natural to extend the operations on numbers to operations with $\infty$. However these extended operations do not have the desired properties anymore.
So here's what's happening with your proof: First, some remarks on division: division is not a basic operation, namely we just have addition and multiplication which satisfy the properties of a field. One of the properties is that for each $b$ different from $0$, there exists a unique number $b^{1}$ with $bb^{1}=1$. We then define $a/b=ab^{1}$, whenever $b$ is not zero. In particular a/0$ is not defined.
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