Tim Brauch wrote: > email@example.com (Beston Kader) wrote in news://vdfob6vb0f6a@legacy: > > >>answer to your question: >> >>infinity = 1/0 ---> 1=0*infinity > > > A slight wave of the hand here, no? For surely you are not saying > 0*infinity is 1, I could understand saying it is zero, I could understand > saying it is infinity. But surely not one. > > How about this then: > > infinity = 2/0 ---> 2 = 0*infinity = 1 ---> 2 = 1 ---> 0 = 1 > > Uh oh. > > >>infinity - 1 = infinity-0*infinity = infinity(1-0) = infinity >> >> >> ---> >> >>infinity - 1 = infinity >> >> > > > Perhaps this will work better... > > 1 + 2 + 3 + 4 + 5 + 6 +.... tends towards infinity > > Subtracting 1 gives > > 1 - 1 + 2 + 3 + 4 + 5 + 6 + .... = 2 + 3 + 4 + 5 + 6 + ... > > which also tends towards infinity. By this logic, infinity less any finite > number is still infinite. And infinity - infinity is not defined. (You > know why?) > > - Tim >
Is this because a + (-a) = 0 is defined on the set of reals, and infinity is nonreal?
This question was posed to my calc teacher in high school when discussing indeterminate forms, and she replied (in essence) that it was a magical thing that didn't have a reason, while I disagreed.
Similarly, if my reasoning is correct, this would mean that 1*infinity is undefined, since that is defined as the multiplicative identity property of the reals. Is this correct?