# Homework Help: Set of Ordered pairs question

1. Aug 26, 2010

### NastyAccident

1. The problem statement, all variables and given/known data
(!) Determine the set of ordered pairs (x,y) of nonzero real numbers such that x/y + y/x >= 2.

2. Relevant equations

x/y + y/x >= 2

3. The attempt at a solution
Relatively new to set notation and proving so, I merely am seeking reassurances that what I am doing is correct.

x/y + y/x >= 2
x^2/xy + y^2/xy >=2xy/xy

(x^2+y^2-2xy)/(xy) >= 0
(x-y)^2/(xy)>=0
(x-y)^2 >= 0
x-y >= 0
x >= y

Set would be written therefore as:
$$\left\{\left(x,y\right) \in\Re: x \geq y\right\}$$

But x and y cannot equal zero (not sure how to depict that in set notation).

Sincerely,

NastyAccident

2. Aug 26, 2010

### Staff: Mentor

The inequality above is true for all real x and y.

3. Aug 26, 2010

### ╔(σ_σ)╝

I think the answer is actually all positive real numbers. X need not be greater than Y. To see this set t = x/y then the equation becomes
$$t+ \frac{1}{t}\geq2$$
$$t^{2} -2t+1\geq0$$
$$(t-1)^{2}\geq0$$
Drawing the parabola you can see that x and y simply have to be positive and non zero and that is sufficient. We have to disregard negative t's for obvious reasons.

To show you an example consider x=0. 5 and y=1. This satisfies the inequality but x is less than y.

Last edited: Aug 26, 2010
4. Aug 26, 2010

### Staff: Mentor

Starting from x/y + y/x >= 2, when you multiply both sides by xy, you need two cases: one for which xy > 0, and another for which xy < 0. You get two different inequalities in these two cases.

5. Aug 26, 2010

### ╔(σ_σ)╝

I just realized that x and y can be negative but they have to negative together so that t is not negative. So the actual answer is all real numbers such that x/y is positive.

6. Aug 27, 2010

### NastyAccident

So, cases:

1.) xy <= 0

2.) xy >= 0

x/y + y/x >= 2
x^2/xy + y^2/xy >=2xy/xy
(x^2+y^2-2xy)/(xy) >= 0
(x-y)^2/(xy)>=0
(x-y)^2 >= 0

The first case contains all real numbers in the first and third quadrants so results like (1,1), (-3,-1), et cetera work. The second case contains all real numbers in the second and fourth quadrants so results like (-2,1) and (3,-4) work.

However, wouldn't you have to evaluate the original inequality and determine that (0,0) would not work due to both x and y being found on the denominator?

If this is not the case would the resulting set simply be the union between the two case? Or my original guess: $$\left\{\left(x,y\right) \in\Re: x \geq y\right\}$$

Sincerely,

NastyAccident

7. Aug 27, 2010

### ehild

You have multiplied the first inequality with xy to get the new one in the second line. It is true if xy>0, but fails when xy<0. Multiplying an inequality with a negative number will reverse it. So it is (x-y)^2 <= 0 if one of x and y is negative, the other positive.
(x-y)^2 is always positive so this second inequality cannot be true.

The second case does not work. Yes, you should check in the original inequality x/y + y/x >= 2 if it is true for (-2,1) and (3,-4) ?

ehild

8. Aug 27, 2010

### NastyAccident

Okay, so I think I follow but I want to make sure that my train of thought is correct.
We have one case:
(x-y)^2/(xy)>=0
with two subcases:
xy<0
xy>0

There is no >= in the sub cases since zero no matter what cannot be in the denominator.

When you introduce a negative x or negative y, the first sub case fails since that turns (x-y)^2/(xy) from being greater than or equal to zero to being less than or equal to zero causing the inequality to no longer be true.

In the second subcase, xy>0 holds true when both x & y are positive and when both x&y are negative since a negative times a negative is a positive.

Hence, the set contains values of x & y when they are both positive and the set contains values of x & y when they are both negative.

So, if that is the correct conclusion then the proper set notation when therefore be:
$$\left\{(x,y) \in\Re: -\infty < x < 0 and -\infty < y < 0 or 0 < x < \infty and 0 < y < \infty \right\}$$

Sincerely,

NastyAccident

9. Aug 27, 2010

### ehild

It is correct now.

ehild