Hi,

Someone give me a hand in checking these please.

a.) Given that $\displaystyle x > y$ , and $\displaystyle k < 0$ for the real numbers $\displaystyle x$, $\displaystyle y$ and $\displaystyle k$, show that $\displaystyle kx < ky$.

$\displaystyle kx>y$...1

$\displaystyle x>y$ $\displaystyle x - y$ is +ve...2

$\displaystyle k<0$...3

If $\displaystyle kx>ky$ then $\displaystyle kx - ky$ is +ve

Putting values using lines 2 and 3

$\displaystyle x=6, y=4, k= -2$

$\displaystyle kx>ky$

$\displaystyle -2(6) > -2(4)$OR$\displaystyle kx-ky=+ve$

...................OR$\displaystyle -12 + 8$ $\displaystyle \not =$ +ve

$\displaystyle -12 > 8$

therefore $\displaystyle kx < ky$ must be true

b.) Show that if x, y ∈R, and $\displaystyle x<y$, then for any real number $\displaystyle k<0$, $\displaystyle kx>ky$.

$\displaystyle kx<ky$...1

$\displaystyle x<y, x-y$ is -ve

$\displaystyle k<0$...3

If $\displaystyle kx<ky$ then $\displaystyle kx-ky$ is -ve

Putting in values using lines 2 and 3

$\displaystyle x=2, y=3, k=-4$

$\displaystyle kx<ky$ .......................OR$\displaystyle kx -ky=-ve$

$\displaystyle -4(2)<-4(3)$................OR$\displaystyle -8-(-12)$ $\displaystyle \not =$ $\displaystyle -ve$

$\displaystyle -8<-12$

therefore $\displaystyle kx>ky$ must be true