Is the set of irrational numbers closed under the operation of exponentiation? That is, for all irrational numbers a and b, is a^b irrational? Why?
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Is the set of irrational numbers closed under the operation of exponentiation? That is, for all irrational numbers a and b, is a^b irrational? Why?
i do not know ifis rational, but if it is not, then certainly
is.
http://www.math.sc.edu/~filaseta/gra...h785Notes8.pdfQuote:
Originally Posted by thamathkid1729
Consider. if
is a rational number than
is irrational. So here you have an example of an irrational number raised to an irrational number yielding a rational number result.