If a and b are nonzero integers, is a^b an integer?

If a and b are nonzero integers, is a^b an integer?
Constraint: a and b ain’t zero
Asked : a^b = Integer ,Yes /No?

(1) b^a is negative
b^a < 0 ,then b must be negative and a must be odd
.: when a = 1 ,b = -1
a^b = (1)^(-1) = 1 = Integer? Yep!
when a= 3 ,b =-2
a^b= (3)^(-2) = 1/(3^2)= 0.111=Integer? Nope!
(Not sufficient)

(2) a^b is negative
a^b < 0, then a must be negative and b must be odd.
But a^b can be negative Integer such as -1 or negative non-Integer such as -1/2
(Not sufficient)

(1+2) We know a= odd and negative, b= odd and negative
Still not sufficient to determine wether a^b =Integer/Non-Integer
Eg -1^(-1) vs -3^(-1)
(Not sufficient)

Hit that E

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(1) b^a is negative sufic

b^a<0 then b<0, and a=anything≠0.
a,b=1,-1: -1^1=-1 and 1^-1=integer
a,b=3,-2: -2^3=-1/8 and 3^-2=1/9≠integer

(2) a^b is negative insufic

a^b<0 then a<0, and b=anything≠0.
a,b=3,2: -3^2=9=integer
a,b=3,-2: -3^-2=1/9≠integer

(1&2) insufic

a<0 and b<0;
a,b=-1,-1: -1^-1=-1=integer
a,b=-2,-3: -2^-3=-1/8≠integer

Ans (E)

If a and b are nonzero integers, is \(a^b\) an integer?

(1) \(b^a\) is negative
(2) \(a^b\) is negative

(1) \(b^a\) is negative
=> b<0; a is odd
if a = -1, b = -1 => \(a^b = -1\) is an integer
if a = -3, b = -3 => \(a^b = -1/9 \) is not an integer
=> Not Suff
(2) \(a^b\) is negative
=> a<0; b is odd
if a = -1, b = -1 = > \(a^b = -8\) is an integer
if a = -3, b = -3 => \(a^b = -1/9 \) is not an integer
=> Not suff
Combine (1) and (2)
=> a<0, b<0, a and b are odd
Same example as only (1) or only (2)
=> Not suff

=> Choice E

If a and b are nonzero integers, is \(a^b\) an integer?

(1) \(b^a\) is negative
\(b^a\) < 0 only when b < 0 where a > 0 OR a < 0 (a is odd integer)
Example:
a = -1 b = -2
\(b^a = (-2)^{-1} = -\frac{1}{2}\); \(a^b = (-1)^{-2} = 1\) YES

a = -3, b = -10
\(b^a = (-10)^{-3} = -\frac{1}{1000}\); \(a^b = (-3)^{-10} = -\frac{1}{3^{10}}\) NO

INSUFFICIENT.

(2) \(a^b\) is negative
\(a^b\) < 0 only when a < 0 where b > 0 OR b < 0 (b is odd integer)

Example:
a = -1 b = -3
\(a^b = (-1)^{-3} = -1\) YES

a = -3, b = -1
\(a^b = (-3)^{-1} = -\frac{1}{3}\) NO

INSUFFICIENT.

Together 1 and 2.
Both a < 0 and b < 0 and are odd.
So,

Example:
a = -1 b = -3
\(b^a = (-3)^{-1} = -\frac{1}{3}\); \(a^b = (-1)^{-3} = -1\) YES

a = -3, b = -1
\(b^a = (-1)^{-3} = -1\); \(a^b = (-3)^{-1} = -\frac{1}{3}\) NO

a = -3, b = -5
\(b^a = (-5)^{-3} = -\frac{1}{125}\); \(a^b = (-3)^{-5} = -\frac{1}{3^5}\) NO

INSUFFICIENT.

Answer E.

Given that a and b are non-zero integers, we are to determine if a^b is an integer.

Statement 1: b^a is negative.
For b^a to be negative, then a must be odd and b must be a negative number. Let a=3 and b=-2
then b^a = -2^3 = -8.
a^b = 3^-2 = 1/9 No.
However when a=-1, and b=-3, then b^a = -3^-1 = -1/3
a^b = -1^-3 = -1. Yes.
Statement 1 is insufficient.

Statement 2: a^b is negative.
Statement 1 implies a is negative and b is odd.
when a =-1 and b=-3, a^b = -1 yes.
However when a=-2 and b=-3, then a^b = -2^-3 = -1/8, No.
Statement 2 is insufficient.

1+2
To satisfy both 1 and 2, we must have both a and b as odd negative values.
a=-1 and b=-3, a^b = -1^-3 = -1 and b^a = -3^-1 = -1/3. The answer is yes.
when a=b=-3, then a^b = b^a = -1/27. The answer is No.

Both statements even taken together are not sufficient.

The answer is E.

If a and b are nonzero integers, is \(a^{b} \) an integer?

(Statement1): \(b^{a}\) is negative.
--> b <0, a could be any odd integer.
if b=-1 and a = 1, then \(a^{b}= 1^{-1}= 1\) (yes)
if b=-2 and a=3, then \(a^{b}= 3^{-2}= \frac{1}{9}\) (NO)
Insufficient

(Statement2): \(a^{b}\) is negative.
--> a <0, b could be any odd integer.
if a=-1 and b= 1, then \(a^{b}= (-1)^{1}= -1\) (Yes)
if a= -2 and b=-3, then \(a^{b}= (-2)^{-3}= -\frac{1}{8}\) (NO)
Insufficient

Taken together 1&2,
a,b --any odd negative integers (a<0, b<0)
if a=-1 and b=-1, then \(a^{b}= (-1)^{-1}= -1 \)(Yes)
if a=-3 and b=-3, then \(a^{b}= (-3)^{-3}= -\frac{1}{27}\) (No)
Insufficient

The answer is E.

a and b are nonzero integers, is a^b an integer?

(1) b^a is negative
If a=-1 and b=-1, then b^a is negative and a^b is an integer
If a=-3 and b=-3, then b^a is negative but a^b is not an integer
NOT SUFFICIENT

(2) a^b is negative
If a=-1 and b=-1, then a^b is negative and a^b is an integer
If a=-3 and b=-3, then a^b is negative but a^b is not an integer
NOT SUFFICIENT

1)+2)
Using the same scenarios, it is still uncertain whether a^b is an integer.
NOT SUFFICIENT

FINAL ANSWER IS (E)

Ans: E

a)b^a= -ve
(-1)^(-1)=-1(b=a=-1), so, a^b=-1 int yes
(-1)^(3)=-1 (b=-1,a=3), so, a^b=1/3 not int no
not sufficient

b)a^b=-ve
(-1)^(-1)=-1(a=b=-1) yes
(-3)^(-3)=-1/27(a=b=-3) No
not sufficient

taking both a & b
(-1)^(-1)=-1(a=b=-1) yes
(-3)^(-3)=-1/27(a=b=-3) No

Not possible

If a and b are nonzero integers, is a^b an integer?

(1) \(b^a\) is negative
(2)\( a^b\) is negative

For \(a^b\) to be an integer, b has to be positive. but if we can get to know whether a =1, the sign of b will not be a matter.

1) from this statement, we can deduce that b is negative and a is an odd integer. when a = 1, then \( a^b \) can be an integer, but for any other value of a, \(a^b\) is not an integer. insufficient.

2) a is negative and b is odd. no information about the sign of b. insufficient.

Together, a and b are both negative and odd. For any value of a except 1, \(a^b\) won't be an integer. but we don't know about the value of a. insufficient.

E is the answer.

possible when a=1 and b = -3
and a = and b =-3


#1b^a is negative
yes /no insufficient
#2
a^b is negative
yes/no insufficient
nothing in common
IMO E



If a and b are nonzero integers, is a^b an integer?

(1) b^a is negative
(2) a^b is negative

This isn't very tough nut to crack if you commute all the different combination that validate let me show you:-
A b^a negative could be -1^1 accoedingly 1^-1 integer however {-1/2)^3 accordingly 3^-1/2 not an integer insuff
B similarly we get it also as insuff even combined doesn't make the matter any better
Therefore IMO E

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