Question

If x and y are real numbers such that x > 1 and y < −1,
then which of the following inequalities must be true?
A. (x/y)> 1
B. abs(x^2)>abs(y)
C. (x/3)− 5 >(y/3) − 5
D. x2 + 1 > y2 + 1
E. x^−2 > y^-2
ACT Practice says C, but I don't see why it can't be E

Answer 1

If x = 10 and y = 3, then

\[\Large x^{-2} > y^{-2}\]
\[\Large 10^{-2} > 3^{-2}\]
\[\Large \frac{1}{10^{2}} > \frac{1}{3^{2}}\]
\[\Large \frac{1}{100} > \frac{1}{9}\]
\[\Large 0.01 > 0.1111\]

The last inequality is false, so the first inequality is false when x = 10 and y = 3.

This counter example shows us that the original inequality is NOT true for all x and y such that x > 1 and y < -1.

Answer 2

But y must be negative

Answer 3

do what jim_thompson5910 did, but replace \(y=3\) with \(y=-3\)

Answer 4

Then it would be true that 1/100>-1/9

Answer 5

oh my bad, yeah y is negative. Zarkon has the proper fix

Answer 6

\[\Large (-3)^{-2} = \frac{1}{(-3)^2} = \frac{1}{9}\]