1. ## Real Numbers

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

2. Originally Posted by Hellbent
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
If x = 3, y = 2, and k = -1, then kx = -3, which is less than y. So your last line above cannot be assumed always to be true.

Try starting like this:

Let x > y, and let k < 0. Assume that kx > ky. Then kx - ky > 0. Also, kx - ky = k(x - y). Since x > y, then x - y > 0.

3. Am I right this time?

a.) Since x > y, so x - y is positive and k is negative.
Product of a negative and positive number is negative, kx - ky
Hence it follows that kx < ky.

b.) Since x < y, so x - y is negative and k is negative.
Product of two negative numbers is equal to a positive number.
Hence it follows that kx > ky.

4. Right...wrong?

5. Originally Posted by Hellbent
Right...wrong?
it seems right for me.

6. $\displaystyle x> y\Rightarrow x-y$ is strictly positive.