- #1
MorallyObtuse
- 45
- 0
Hi,
Are these correct?
a.) Given that x > y, and k < 0 for the real numbers x, yand , show that kx < ky.
b.) Show that if x, y ∈ R, and x < y , then for any real number k < 0,kx > ky
2. The attempt at a solution
a.) 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...[tex]-12 + 8 \not = +ve[/tex]
-12 > 8
therefore kx < ky must be true
b.) kx < ky...1
x < y, x - y is -ve
k < 0 ...3
If kx < ky then kx - ky is -ve
Putting in values using lines 2 and 3
x = 2, y = 3, k = -4
kx < ky...OR kx - ky = -ve
-4(2) < -4(3)...OR...[tex]-8-(-12) \not = -ve[/tex]
-8 < -12
Are these correct?
Homework Statement
a.) Given that x > y, and k < 0 for the real numbers x, yand , show that kx < ky.
b.) Show that if x, y ∈ R, and x < y , then for any real number k < 0,kx > ky
2. The attempt at a solution
a.) 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...[tex]-12 + 8 \not = +ve[/tex]
-12 > 8
therefore kx < ky must be true
b.) kx < ky...1
x < y, x - y is -ve
k < 0 ...3
If kx < ky then kx - ky is -ve
Putting in values using lines 2 and 3
x = 2, y = 3, k = -4
kx < ky...OR kx - ky = -ve
-4(2) < -4(3)...OR...[tex]-8-(-12) \not = -ve[/tex]
-8 < -12