bsaikrishna wrote:

Hi

I need some confirmation on my thought process regarding some inequalities.

This is primarily to do with an inequality between a and b and how the inequality changes b/w a^n and b^n.

Please confirm the following:

If n is an odd integer:

If a > b then a^n > b^n holds true irrespective of a and b being positive or negative, if n is positive odd integer

If a > b then a^n < b^n holds true irrespective of a and b being positive or negative, if n is negative odd integer

If n is an even integer:

If a > b and if a, b, and n are positive integers then a^n > b^n. If n is negative then it will be the opp. ineqlty.

If a > b and if a, b, are negative integers then a^n < b^n for positive n. If n is negative then it will be the opp. ineqlty.

If a and b are not of the same sign,then for even n we cannot say anything about the relation.

Because 4>-2 and 4>-8 will result in two different inequalities when raised to even powers.

Kindly confirm if I am correct in the above.

Also, please let me know how do we handle the above if 'n' is a real number.

Thanks

Sai

Seems right.

Below is what you need to know for the GMAT about raising inequalities into a power:

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).For example:

2<4 --> we can square both sides and write:

2^2<4^2;

0\leq{x}<{y} --> we can square both sides and write:

x^2<y^2;

But if either of side is negative then raising to even power doesn't always work.

For example:

1>-2 if we square we'll get

1>4 which is not right. So if given that

x>y then we can not square both sides and write

x^2>y^2 if we are not certain that both

x and

y are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example:

-2<-1 --> we can raise both sides to third power and write:

-2^3=-8<-1=-1^3 or

-5<1 -->

-5^2=-125<1=1^3;

x<y --> we can raise both sides to third power and write:

x^3<y^3.

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hope it helps.

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