# Homework Help: Proof of real numbers

1. Nov 26, 2009

### MorallyObtuse

Hi,

Are these correct?

1. The problem statement, all variables and given/known data

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

-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...$$-8-(-12) \not = -ve$$

-8 < -12

2. Nov 26, 2009

### LCKurtz

You don't solve problems like these by plugging in numbers. Also using baby talk notation like +ve for "positive" isn't likely to be received well.

I would suggest you start with a clear statement of what you are given and what you are to prove. For example for the first problem:

Given x > y means there is a positive number h such that x = y + h
Given k < 0 (k is negative)

Prove: kx < ky which means there is a positive m such that ky = kx + m

Work with the equations instead of the inequalities and use the fact that k is negative.

3. Nov 27, 2009

### MorallyObtuse

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. Nov 27, 2009

### HallsofIvy

the basic problem here is that we don't know what you have to work with. What properties of the real numbers do you know that you can use? I suspect you are using the fact that the real numbers are an ordered field: that there is a ">" relation defined such that
1) If x> y then x+z> y+z.
2) If x> y and z> 0 then zx> zy.
3) For any two real numbers, x and , one and only one of these is true:
a) x> 0
b)-x> 0
c) x= 0.
and now you want to prove that if x> y and k< 0, then kx< ky.

Of course, "a< b" means "b> a". "k< 0" means "0> k" and so, by (2), adding -k, -k> 0. Then from (1), -kx> -ky. Adding ky to both sides, (2) gives ky-kx> 0 and adding -kx to both sides ky> kx which means kx< ky.

I honestly don't see any difference between your problems (a) and (b) except that the names of "x" and "y" have been swapped. You might want to simply prove that "if a> b and k< 0 then ka< kb" first. Then prove (a) by letting x= a, y= b, and prove (b) by letting x= b, y= a.

5. Nov 27, 2009

### MorallyObtuse

The questions are close. So, not much difference in the answers.
You might want to simply prove that "if a> b and k< 0 then ka< kb" first. Then prove (a) by letting x= a, y= b, and prove (b) by letting x= b, y= a. This part I'm not understanding. I'd have to input values and the teacher says that proves nothing. Yeah, I did it in the original post...couldn't solve it any other way.

6. Nov 27, 2009

### MorallyObtuse

The teacher uses 'baby talk notation' like +ve :surprised

7. Nov 27, 2009

### Staff: Mentor

It seems silly to me to use nonstandard notation like "x - y is +ive" when you can say the same thing more economically with "x - y > 0."

8. Nov 27, 2009

### HallsofIvy

Proving "if a> b and k< 0 then ka< kb", then setting a= x, b= y so that you have proved "if x> y and k< 0 then kx< ky" and setting a=y, b= x so that you have proved "if y> x and k< 0 then ky< kx" is NOT the same as setting "x=6, y=4, k= -2"!

9. Nov 28, 2009

### MorallyObtuse

Yeah, I agree with you, Mark :)

10. Nov 28, 2009

### MorallyObtuse

Forget it! I barely understand whenever you help me.

11. Nov 29, 2009

### HallsofIvy

Well, "barely understanding" is still better than "not understanding"!

12. Nov 29, 2009

### MorallyObtuse

That's true, maybe I'm a little too ungrateful.
Put it this way, I'm not the fastest learner.