\[\infty\times0=\]

- UnkleRhaukus

\[\infty\times0=\]

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- anonymous

0!!!!

- UnkleRhaukus

@lgbasallote

- anonymous

0

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## More answers

- anonymous

definitely \(\pi\)

- anonymous

@satellite73 IKR

- UnkleRhaukus

what is your evidence @satellite73

- anonymous

oh wait, i am wrong it is \(e^2\) sorry

- lgbasallote

you can rewrite \[\infty \times 0\] as \[\infty \times \frac 1\infty\]
right?
then it becomes l'hospital...

- lgbasallote

assuming of course these are functions...

- UnkleRhaukus

there is another way

- lgbasallote

really? hmm

- lgbasallote

change infinity to 1/0 to make it 0/0?

- lgbasallote

either way...it's indeterminate....

- anonymous

if this is a serious question, presumably it is about limits, i.e. if \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to \infty}g(x)=0\) then what is
\[\lim_{x\to \infty}f(x)g(x)\] the answer is it could be anything, it depends on \(f\) and \(g\)
the form is not determined

- anonymous

\[\large a \times b = c\]
\[\large a=\frac{c}{b}\]
\[\large b=\frac{c}{a}\]
Let a = ∞, b=0
\[\large ∞=\frac{c}{0}\]
\[\large 0=\frac{c}{∞}\]
But c/0 is undefined and so is c/∞, right?
What if c were positive? What if c were negative?

- anonymous

so it could be \(\pi\) or it could be \(e^2\) or it could be anything

- lgbasallote

an interesting question though is... \[\frac 1 \infty = 0\]
therefore... \[0 \times \infty = 1\]
that should be right?

- lgbasallote

just expressing how weird math is

- anonymous

not at all

- anonymous

infinity is not a number, and so \(\frac{1}{\infty}\) is not a number either

- anonymous

Agree with @satellite73
To say that 1/∞ = 0, you have to take the limit of 1/x as x-->∞
and then 0 × (x is really really big) still equals 0.

- UnkleRhaukus

|dw:1349968485543:dw|

- lgbasallote

math is still weird....

- anonymous

(I'm going to have to go get coffee, then come back for this.
Unkle is about to get all Twilight Zone on us, I can feel it.)

- anonymous

you are making short cut statements about limits
namely if \[\lim_{x\to\infty}f(x)=\infty\]
then \[\lim_{x\to \infty}\frac{1}{f(x)}=0\]

- lgbasallote

i never really understood the function of limits...

- lgbasallote

math is too ambiguous for me

- anonymous

\(\infty\) is not a number
\(\infty\times 0\) is not a number
\[\frac{5}{\infty}\] is not a number

- anonymous

(aside: Anybody hear the full treatment of 'Hilbert's Hotel?')

- anonymous

Is the slope of the vertical line +∞ or -∞ @UnkleRhaukus ?

- UnkleRhaukus

i dont know what 'Hilbert's Hotel?' is

- anonymous

Check it out some time. David Deutsch gives a good telling of it in his book, 'The Beginning of Infinity.' (Fantastic book on the philosophy of science; I recommend it to everyone.)

- UnkleRhaukus

the product of the slopes of perpendicular lines is

- anonymous

Yes, but what is the slope of the vertical line?

- UnkleRhaukus

\[\pm\infty\]

- helder_edwin

this might help

##### 1 Attachment

- anonymous

infinity is stupid. \[\infty + 1 = \infty \]
\[1 = \infty - \infty = 0\]

- UnkleRhaukus

\[\infty\times 0=-1\]

- anonymous

all the even numbers = \[\infty\] which is smaller than all the numbers = \[\infty\]
so \[\infty < \infty\]

- anonymous

\[\large 2 \times ∞ \times 0 = -2?\]

- anonymous

@JamesWolf there are different cardinalities of infinity. Research "Aleph Numbers" for more info.
The infinity of integers is less than the infinity of real numbers, etc.

- anonymous

The bastardization of mathematics that has taken place in this thread is like salt rubbed into a fresh wound of my soul. Is there a legitimate question, or are you guys just playing around? :)

- anonymous

this is definitely an undefined case because any number multiplied with zero become zero and any number multiplied with infinity become infinity that's why both of these cases are possible here therefore we cannot consider one these case separately. thus this is an undefined case.

- anonymous

@Jemurray3 I don't think there's anything serious going on.
Unkle came up with a cute link to slopes of parallel lines, but made the error in saying that vertical lines have a slope, m=∞, which, of course, is false.

- anonymous

if the question is something like lim x->inf of 0*x then it is 0
otherwise we should be more specific about the problem i guess

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