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1/0=+/-infinity


Andrew26

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It can be shown that as x approaches 0 from the right side, 1/x will approach 1/0 and apporach positive infinity. And it can be shown that as -x approaches 0 from the left side, -1/x will approach negative infinity and apporach -1/0, which is equal to (-1/-1)*-1/0 equals 1/0, since 0 is neither positive or negative.

Showing that 1/0 is not even a number, and does not represent a distinct point on the real number line.

 

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8 hours ago, Andrew26 said:

The question here is what is b/0. 

 

the problem is you are dividing by zero which has undefined elements to it.

0 = 0/4 = 0/-2579 = 0/323i = 0/π = ...

From a simple seeming 1/0 you can forever pull any number(excluding zero) from below, preventing you from even saying it equals anything in particular even if it heads that way.

1/0 = 1/((((0/.33)/9i)/-196)/5)

consider 0a and a-a for added fun.

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14 hours ago, Endy0816 said:

 

the problem is you are dividing by zero which has undefined elements to it.

0 = 0/4 = 0/-2579 = 0/323i = 0/π = ...

From a simple seeming 1/0 you can forever pull any number(excluding zero) from below, preventing you from even saying it equals anything in particular even if it heads that way.

1/0 = 1/((((0/.33)/9i)/-196)/5)

consider 0a and a-a for added fun.

As stated in the OP, 1/0 equals +/-infinity as a limit. 

Your post is impertinent here 

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