Let x and y be real numbers

And I know that

$x-y < x*y < 0$

Then which one is true?

(I) $x^2-(x*y)<0$

(II) $y(x-1)<0$

(III) $y+ \frac{y}{x}<1$

My Solution is

(III) is wrong. Why?

If $x-y < xy$ then $$x < y+ xy$$ $$ \frac{x}{x} < \frac{y}{x} + \frac{xy}{x}$$ $$1< \frac{y}{x} + y$$

(II) is true.

$$x-2y < x-y < xy -y =(x-1)y$$

I have no idea for (I).

However, the book’s answer key says that (II) and (III) are true. Why? I don’t see this. Please help me to prove that. Thanks a lot.

  • 3
    $\begingroup$ You can only divide an inequality by a positive number. If you divide by a neagtive number the inequality gets reversed. Ex: $ 1<2$ but $\frac 1 {-1} >\frac 2 {-1}$. $\endgroup$ Jul 8 '20 at 10:10
  • $\begingroup$ Thanks. Then (III) is true. Well, please show me the part (I). I have no idea for (I). @KaviRamaMurthy $\endgroup$
    – 1190
    Jul 8 '20 at 10:14
  • $\begingroup$ Since $xy \lt 0$, then one of $x$ and $y$ is negative and the other is positive. With $x - y \lt 0$, this determines which specific value is negative & which is positive. $\endgroup$ Jul 8 '20 at 10:14
  • $\begingroup$ So, $x<y$. Then x is Negative, y is positive. Is it right? @JohnOmielan $\endgroup$
    – 1190
    Jul 8 '20 at 10:16
  • 1
    $\begingroup$ @B11b Yes, $x \lt 0$ and $y \gt 0$. Also, for (I), note $x^2 \gt 0$ and $xy \lt 0 \implies -xy \gt 0$, so what does that say about the sum of those $2$ terms? $\endgroup$ Jul 8 '20 at 10:18

(I) is wrong. $x^2 -xy < 0$ and $xy < 0$ implies that $x^2 < xy<0$ but $x^2 \ge 0$.

(II): If $x>0$, then $y<0$ since $xy<0$. However, this would imply that $x-y >0$ which is a contradiction. Hence we must have $x<0$, $y>0$. Since $x-1 < 0$ and $y>0$, we must have $(x-1)y < 0$.

(III): As argued earlier, $x<0$,

Hence $$1>\frac{y}{x}+y$$

since we reverse the direction when we divide by a negative number.


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