Indeterminate form

burt

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Is [MATH]\infty\infty[/MATH] in indeterminate form? I would think not because it is just making an even bigger infinity. How about [MATH]\infty+\infty[/MATH]? What about 0fraction with zero denominato\frac0{fraction\ with\ zero\ denominato}?
 
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One way to think about [MATH]\infty[/MATH] is the projectively extended reals.

In that system, [MATH]\infty[/MATH] is defined as [MATH]\infty = \dfrac{1}{0}.[/MATH]
All operations involving real numbers except 0/0 are determinate (including r/0 if r is not zero). Most arithmetic operations involving [MATH]\infty[/MATH] and a real number are also determinate, but operations that involve only infinity are indeterminate.

 
what about the others though?
Here are mine.
One way to think about [MATH]\infty[/MATH] is the projectively extended reals.
In that system, [MATH]\infty[/MATH] is defined as [MATH]\infty = \dfrac{1}{0}.[/MATH]
If that is the case for real numbers. then 0=1\displaystyle 0\cdot \infty=1 Does anyone really want that?
There is a perfectly worked out, logically consistent, system known as non-standard analysis.
Here is a completely free textbook available for download. Elementary Calculus: An Infinitesimal Approach In Elementary Calculus: An Infinitesimal Approach by Jerome Keisler in chapter 3, there is good proof of this. The chapters and whole book is a free down-load at http://www.math.wisc.edu/~keisler/.
Down load that text. See how enlarging the real number system can accomplish what you seek in a completely logical way.
 
If that is the case for real numbers. then 0=1\displaystyle 0\cdot \infty=1 Does anyone really want that?
Actually it does not.

[MATH]0 \cdot \infty[/MATH] is indeterminate in that system.
 
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