lim x->0 1/x = +inf from the right, - inf from the left.
But it's still a limit, not the actual result.

>>7978612
No, because 0 times infinity does not equal 1.

>>7978612
1/0=0

No. 1/0 does not equal anything.
Suppose 1/0 did equal something, let it be a.
a = 1/0. Then 0*a = 1. Clearly that does not make sense.

learn how to spell "acquaintance" first and then we'll talk about numbers.

Usually when I see this it has to do with complex manifolds such as the Riemann sphere that map the extend complex plane [math]\mathbb{C} \cup \left \{ \infty \right \}[\math]. These people say division by zero results in 'complex infinity' [math]\bar{\mathbb{C}}[\math]. Which is true in cases where division by zero results in a vertical asymptope.

Then what happens is stupid people who don't know fuck all about complex analysis or algebraic geometry hear this and say "1/0=infinity" - this is false. The concept of infinity and complex infinity are two completely different things.

>>7978755

fucking dumbass, so 0 * 0 = 1?

>>7979231
>stupid people
you mean people who didn't hear about the Riemann sphere ? Do you mean to say intelligence is EXACTLY equal to trivial facts?
because that's retarded. Faggot.

>>7979261
You're right, I'm sorry for my poor choice in words. A more adequate label would've been ignorant.
I didn't really put too much thought into that post before I sent it out (as demonstrated by [\math] tags), I know it's not an excuse; I just thought you might want an explanation. When I look back on what I said my heart is filled with shame. I'm deeply remorseful if I hurt you with my comment.

>>7978612
If I have a cake and I divide it into 0 parts, I am left with one whole. The cake is made up of a finite amount of parts.

>>7978612
In the lim true->false he is right.

>>7978762
saw something similar in spivak,

bc a*a^-1 = 1

0*0^-1=1

but

0*a =0

contradiction

>>7978612
>(1/0)=infinity.
>implying 0 * infinity = 1

>>7978612
Not quite, as >>7978627 has said, the limit is +/- infinity BUT actual division by zero is undefined, and there are no fancy ways around that. The answer to your question is no, there is no answer, and any answer you do get will be to a modification of your question.

>>7978762
>>7979435
"There is clearly no square root value for negative integers. Assume [math]\sqrt{-1}=a[/math] then [math]a^2 =-1[/math] this clearly violates the properties of multiplication."

>Assuming a is in the field [math]\mathbb{R}[/math]
Go back to 1156 old man