Subhotosh Khan said:
However,
(nothing)[sup:7eq9bm3a]0[/sup:7eq9bm3a] = indeterminate
Yes,
for instance, let k be a nonzero real number.
\(\displaystyle 0^{k - k} \ = \ \frac{0^k}{0^k} \ = \ \frac{0}{0}, \ which \ is \ indeterminate.\)
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Loci: Convergence
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What is 0^0?
by Michael Huber and V. Frederick Rickey
Today's Algebra Books
Pick up a high school mathematics textbook today and you will see that 0^0 is treated as
an indeterminate form. For example, the following is taken from a current New York Regents text [6]:
We recall the rule for dividing powers with like bases:
x^a/x^b = x^(a-b) (x not equal to 0) (1)
If we do not require a > b, then a may be equal to b. When a = b:
x^a/x^b = x^a/x^a = x^(a-a) = x^0
(2)
but
x^a / x^a = 1 (3)
Therefore, in order for x^0 to be meaningful, we must make the following definition:
x^0 = 1 (x not equal to 0) (4)
Since the definition x^0 = 1 is based upon division, and division by 0 is not possible,
we have stated that x is not equal to 0. Actually, the expression 0^0 (0 to the zero power)
is one of several indeterminate expressions in mathematics. It is not possible to assign a
value to an indeterminate expression.
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http://www.math.utah.edu/~pa/math/0to0.html
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http://www.homeschoolmath.net/teaching/ ... -proof.php
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http://hotmath.com/hotmath_help/topics/ ... -zero.html
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http://www.mathsteacher.com.au/year8/ch ... w/zero.htm
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http://www.mathsteacher.com.au/year8/ch ... w/zero.htm
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http://oakroadsystems.com/math/expolaws.htm
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Also, Subhotosh Khan, it is correct to type this without an asterisk,
because the parentheses are for multiplication, as well as for grouping:
\(\displaystyle 5^{2x} - 5^x = 5^x(5^x - 1)\)
