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m_a_m_a [10]
3 years ago
8

Explain why rational exponenets are not defined when the denominator of the exponent in lowest terms is even and the base is neg

ative.
Mathematics
2 answers:
ch4aika [34]3 years ago
6 0
Assume x > 0 then (-x)^(1/2) = sqrt(-x) and we cant take sqaure roots of negative numbers.............yet

so (-3)^(1/4) is<span><span>−3</span><span>−−−</span>√4</span>which is not ok but<span><span>−3</span><span>−−−</span>√3</span>is fine
Leona [35]3 years ago
4 0
Such a square root of a negative base wouldn't be possible. Because two (or just even numbers in general) negative numbers will always equal a positive number when multiplied, the rational exponent would not be defined because it isn't possible to conclude with a negative answer. However, if the denominator of the rational exponent happened to be an odd number, the problem would be solvable. When the denominator is an odd number it allows for the base to be negative because three negatives multiplied together give a negative answer. If the denominator of the exponent is even and the base is negative, the rational exponent will not be defined because it isn't possible.
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