If a and b are integers and a^b

Question

If a and b are integers and a^b0 II. b

- CJ

RMD007 · Answer

From the question stem, it is quite clear that a and be should be  negative odd  integers.

Statement I and III satisfies.

For II take a= -1 b = -3 so a>b

a^b = -1^-3 = -1 
 b^a = -3^-1 =   \frac{1}{-3}  = -0.66

so  a^b< b^a

Please correct me if I am wrong.

arnabmukherjee10 · Answer

I think E should be the answer. While I and II are pretty evident. Taking two odd negative integers such as a=-5 and b=-3, such that b>a, we can prove that in all such cases b^a<a^b. Hence, it's incorrect a case to consider. All three have to be true.
Please correct me if I am wrong.

Luckisnoexcuse · Answer

+ stands for positive integer - stands for negative integer E stands for even O stands for Odd (a,b) can only be (-O,-O), (-E,-E), (-E,-O), (-O,-E) by putting numbers: Case 1 (-O, -O) = (-3, -5) or (-5, -3) (-3,-5) does not satisfy a^b0, b>a, ab+1 is divisible by 2 Fight between option A & C left Case 2 (-E, -E) = (-6,-2) or (-2,-6) (-2,-6) does not satisfy a^b0, b>a, ab+1 is not divisible by 2 Only option left is A Please correct me if i am wrong (Noob).

ChiranjeevSingh · Answer

Hi,

While you can prove that a statement is not "Must be true" by producing even one counter-example, you cannot prove that a statement is "Must be true" just by taking one or a finite number of examples. For the second statement, if you take a=-5 and b=-3, it'll satisfy the given condition but a<b.

Feel free to ask if you have any questions.

- CJ

ChiranjeevSingh · Answer

Hi,

I see that all the users on this thread have made the same mistake: they have tried to prove that a statement is "Must be true" using numbers.

We need to understand that we cannot prove a statement "Must be true" just by showing that it holds in one case. A statement is "Must be true" only if it holds in every case. Right?

On the other hand, we can easily prove that a statement is not "Must be true" by producing even one case in which it is not true. If it is not true even in one case, then it cannot be "Must be true". Right?

Now, coming to our question, we see that  a^b<b^a<0

Now, a positive number raised to any power (whether positive or negative) will never give you a negative result. Since both  a^b  and  b^a  are less than zero, we can be sure that both a and b are negative.

Also, any number (positive or negative) raised to an even power will give you a positive result and since  a^b  and  b^a  are negative, the powers have to be odd. So, both a and b are odd. Right?

Our analysis above proves statements I and III correct. Now, we are left with Statement II.

We can start trying out different numbers for a and b to come up with a case in which statement II doesn't hold. Now, while thinking of values of a and b, we have to take them to be odd negative integers. Why? Because if we do not take them so, our given inequality (Which puts both  a^b  and  b^a  as negative) will not hold.

Now, since we are trying to disprove statement II, we can start with a=-3 and b = -1. In this case, we have a<b. Now, if this can satisfy the given inequality, we'll have a case against Statement II. However, as it turns out, this combination doesn't satisfy the given inequality  a^b<b^a<0 . Thus, we cannot take this combination.

Next, we try a=-5 and b=-3. This combination satisfies the given inequality. So, now we have a case which satisfies the given inequality but a<b. So, Statement II is not "Must be true".

Therefore, the answer is option C.

ganand · Answer

Hi, If a and b are integers and a^b 0 \Rightarrow a extrm{ and } b have same sign. Hence, this is true . (II) b < a . Try to construct a counter example. Consider a = -7 and b = -3 . a^{b} = (-7)^{(-3)} = \frac{-1}{7^{3}} = \frac{-1}{343} and b^{a} = (-3)^{(-7)} = \frac{-1}{2187} b^{a} > a^{b} but b less a . Hence, this is not true. (III) (ab+1) is divisible by 2. Since, a extrm{ and } b are odd integers => ab is an odd integer => (ab+1) is an even integer. Hence, true. Answer (C) I and III. Thanks.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If a and b are integers and a^b<b^a<0, which of the following

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions