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definitely \(\pi\)
what is your evidence @satellite73
oh wait, i am wrong it is \(e^2\) sorry
you can rewrite \[\infty \times 0\] as \[\infty \times \frac 1\infty\] right? then it becomes l'hospital...
assuming of course these are functions...
there is another way
really? hmm
change infinity to 1/0 to make it 0/0?
either way...it's indeterminate....
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
\[\large a \times b = c\] \[\large a=\frac{c}{b}\] \[\large b=\frac{c}{a}\] 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?
so it could be \(\pi\) or it could be \(e^2\) or it could be anything
an interesting question though is... \[\frac 1 \infty = 0\] therefore... \[0 \times \infty = 1\] that should be right?
just expressing how weird math is
not at all
infinity is not a number, and so \(\frac{1}{\infty}\) is not a number either
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.
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math is still weird....
(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.)
you are making short cut statements about limits namely if \[\lim_{x\to\infty}f(x)=\infty\] then \[\lim_{x\to \infty}\frac{1}{f(x)}=0\]
i never really understood the function of limits...
math is too ambiguous for me
\(\infty\) is not a number \(\infty\times 0\) is not a number \[\frac{5}{\infty}\] is not a number
(aside: Anybody hear the full treatment of 'Hilbert's Hotel?')
Is the slope of the vertical line +∞ or -∞ @UnkleRhaukus ?
i dont know what 'Hilbert's Hotel?' is
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.)
the product of the slopes of perpendicular lines is
Yes, but what is the slope of the vertical line?
\[\pm\infty\]
this might help
1 Attachment
infinity is stupid. \[\infty + 1 = \infty \] \[1 = \infty - \infty = 0\]
\[\infty\times 0=-1\]
all the even numbers = \[\infty\] which is smaller than all the numbers = \[\infty\] so \[\infty < \infty\]
\[\large 2 \times ∞ \times 0 = -2?\]
@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.
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? :)
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.
@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.
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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