# Real Numbers

• Nov 26th 2009, 02:30 AM
Hellbent
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
• Nov 26th 2009, 03:18 AM
stapel
Quote:

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.

• 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
\$\displaystyle x> y\Rightarrow x-y\$ is strictly positive.(Nod)