# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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.