# Real Numbers

• Nov 26th 2009, 02:30 AM
Hellbent
Real Numbers
Hi,

Someone give me a hand in checking these please.

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

$kx>y$...1

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

$k<0$...3

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

Putting values using lines 2 and 3

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

$kx>ky$

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

...................OR $-12 + 8$ $\not =$ +ve
$-12 > 8$
therefore $kx < ky$ must be true

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

$kx...1

$x is -ve

$k<0$...3

If $kx then $kx-ky$ is -ve

Putting in values using lines 2 and 3

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

$kx .......................OR $kx -ky=-ve$

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

$-8<-12$

therefore $kx>ky$ must be true
• Nov 26th 2009, 03:18 AM
stapel
Quote:

Originally Posted by Hellbent
a.) Given that $x > y$ , and $k < 0$ for the real numbers $x$, $y$ and $k$, show that $kx < ky$.

$kx>y$...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.

• Nov 26th 2009, 09:58 PM
Hellbent
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.
• Nov 27th 2009, 02:54 AM
Hellbent
Right...wrong?
• Nov 27th 2009, 02:58 AM
Raoh
Quote:

Originally Posted by Hellbent
Right...wrong?

it seems right for me.(Nod)
• Nov 27th 2009, 03:34 AM
Raoh
$x> y\Rightarrow x-y$ is strictly positive.(Nod)