Division by infinity

It's a commonly known fact that one does not simply divide by zero. There are many mathematical proofs on the Internet and on Brilliant that are flawed due to dividing by zero, such as proving that 1+1=1. But if we aren't allowed to divide by 0, shouldn't we not be allowed to divide by infinity?

When you type in 0×0\times \infty on wolfram alpha, contrary to popular belief, the answer isn't 0, but rather it's indeterminate.

image image

let me show what I mean

1=0\dfrac{1}{\infty}=0

1=×01=\infty \times 0

Or

2=0\dfrac{2}{\infty}=0

2=×02=\infty \times 0

What... There are two results, but how can this be.

How about this:

1=×01=\infty \times 0

10=\dfrac{1}{0}=\infty

Or

2=×02=\infty \times 0

20=\dfrac{2}{0}=\infty

So my question to you is "why is dividing by \infty allowed but dividing by 0 is not)

#NumberTheory #MathematicalFallacies #NumberFallacy #ThinkAboutIt #FalseProofs

Note by Trevor Arashiro
6 years, 7 months ago

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1 vote

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Comments

There are 7 indeterminant forms namely,

00,,,×0,0,00,1\frac{0}{0}, \frac{\infty}{\infty}, \infty - \infty, \infty \times 0, \infty^{0}, 0^{0}, 1^{\infty}

here 00, 11 and \infty are tending 00,tending \infty and tending 1

But

exact0exact0,tending0exact0,+,×,,±\frac{exact 0}{exact 0}, \frac{tending0}{exact 0}, \infty + \infty, \infty \times \infty, \infty^{\infty}, \pm \infty

Are undefined

Let me explain further

exact1=1exact 1^{\infty} = 1

But

tending1tending 1 ^{\infty} is indeterminant

Another example

exact0tending0=0\frac{exact 0}{tending 0} = 0

But

tending0tending0\frac{tending 0}{tending 0} is indeterminant

exact0×=0\displaystyle exact0 \times \infty = 0

But

tending0×\displaystyle tending0 \times \infty is indeterminant

Ask anything if you don't understand because I copied the comment of mine when I explained same thing to other

Krishna Sharma - 6 years, 7 months ago

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This is actually a great explaintion. If I understand, it's like using limits in a sense.

Trevor Arashiro - 6 years, 7 months ago

One should understand the difference between these two situations:

a.\Large a. When a number is exactly\textit {exactly} equal to some value &

b.\Large b. When it approaches from left or right\textit {approaches from left or right} to some value.

In another way 0×=00×\infty=0 but (something infinitesimally small)×(something infinitely big)=undefined!! \textbf {(something infinitesimally small)×(something infinitely big)=undefined!! }

Sanjeet Raria - 6 years, 7 months ago

Exactly. Using limit sin x 0 value equals 0. Using limit e^x , value equals 1. Using limit e^1/x value equals infinite

Swarupendra Nath Chakraborty - 5 years, 11 months ago
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