Tina1785 wrote:
If x^3 > y^3, is it necessary that x > y? The question does not say anything about x, y being integers.
Hello
Tina1785This is a nice query and I think all of us should be clear in our minds about this. I am pasting one handwritten note it might be helpful.
Let us look at two broad cases
\(Case I\): \(x^3<y^3\)
\(Case II\): \(x^2<y^2\)
These cases if you notice are the ones which confuse us. So let's look at it closer.
Also let us assume only non zero real numbers between \(-1\) and \(+1\). For numbers not in this range the solution also follows the same approach.
\(Case I\): \(x^3<y^3\)
In this case no matter what the variable \(x\) or \(y\) is. They will never cross the \(0\) line because as you guessed it right the sign is the reason. Look at the left side of the attached picture. If \(x<y\) then no matter what \(x^3<y^3\) will hold true.
\(Case II\): \(x^2<y^2\)
This is the case which creates confusion in our minds and rightly so. The sign changes when we square the variable and moreover the combined change in absolute value and sign may change the relationship between \(x\) and \(y\) in any way. Look at the right side of the attached diagram. In the first you see a clean inverse relationship, in the second same relationship but in the last two you will see that the absolute value matters.
Conclusion:Don't be worried when the \(variables\) are raised to an \(odd\) \(power\) but be very careful when they are raised to \(even\) \(power\).
p.s. the bottom left diagram has 0.009 it is supposed to be 0.027
Attachments
IMG_20150531_073948549_HDR.jpg [ 1.16 MiB | Viewed 2516 times ]
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Regards
J
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