Baten80 wrote:
If m and n are nonzero integers, is m^n an integer?
(1) n^m is positve
(2) n^m is an integer.
m and n are non zero integers. So m and n could be positive or negative.
Is m^n an integer?
When will m^n be an integer? When m is an integer (which we are given it is) and n is a positive integer. Also when m = 1/-1 and n is any integer.
So (2)^5, (-5)^3, 1^(-2) etc
Given that m and n are non-zero,
(1) n^m is positve
n^m will be positive when n is positive and m can be any integer.
e.g. 3^(-2), 7^4 etc.
In this case m^n is an integer.
n^m will be positive when n is negative and m is even.
e.g. (-3)^2, (-4)^(-6) etc.
In this case m^n may not be an integer.
Not sufficient.
(2) n^m is an integer.
n^m is an integer when n is any integer and m is positive.
e.g. 4^2, (-2)^3 etc.
In this case, m^n may be an integer (2^4) or may not be an integer (3^(-2)).
We already have two cases so this alone is not sufficient
Using both, n^m is a positive integer.
So n could be positive integer and m may be a positive integer e.g. 3^2, 7^3 etc. Here m^n will be an integer.
or if n is 1, m could be any integer e.g. 1^7. Here m^n will be an integer
or if n is -1, m should be an even integer e.g. (-1)^2, (-1)^-4. Here m^n may not be an integer e.g. (-4)^(-1)
Both statements together are not sufficient.
Answer (E)