# Solving inequalities in TWO variables?

1. Nov 27, 2008

### kingwinner

Question 1)
{(u1,u2) : -∞ < (u1+u2)/2 < ∞ and -∞ < (u1-u2)/2 < ∞ } = {(u1,y2) : -∞ < u1 < ∞ and -∞ < u2 < ∞}
Why is the equality(=) true? How can I see that the two sets describe the same region?

Question 2)

2) Define u1=y1+y2, u2=y1-y2, so the mapping (or function) is (u1,u2)=f(y1,y2)=(y1+y2,y1-y2). If -∞ < y1 < ∞ and -∞ < y2 < ∞ are the DOMAIN of this mapping, then this implies the RANGE is -∞ < u1 < ∞ and -∞ < u2 < ∞. WHY?

Can someone explain, please? Any help would be appreciated!

Last edited: Nov 28, 2008
2. Nov 28, 2008

### tiny-tim

Hi kingwinner!
I don't understand

if a number is well-defined, then it must be between -∞ and ∞ …

so these equations don't seem to say anything!

What is the context?​

3. Nov 28, 2008

### kingwinner

1) The context is this:
{(u1,u2) : -∞ < (u1+u2)/2 < ∞ and -∞ < (u1-u2)/2 < ∞ } = {(u1,y2) : -∞ < u1 < ∞ and -∞ < u2 < ∞}
Why is the equality(=) true?

2) Define u1=y1+y2, u2=y1-y2. If -∞ < y1 < ∞ and -∞ < y2 < ∞ are the DOMAIN of this mapping, then this implies the RANGE is -∞ < u1 < ∞ and -∞ < u2 < ∞

Why?

4. Nov 28, 2008

### tiny-tim

oh i see … they're sets!

this is an equality of sets.

Why is it true? because every element (u1,u2) in the first set can be proved to be in the second set,

and every element (u1,u2) in the second set can be proved to be in the first set …

and that's what you have to prove.

5. Nov 28, 2008

### kingwinner

Yes, for sure I know this.

This is from an example from a textbook (just a very small fraction of a very long example) and it just state the above equality with no explanation or proof and I got confused.
So my trouble is I have no idea how to prove this (or at least to "believe" or convince myself this. I don't want to use the word proof here, since this part is just one-millionth of a very long example, and not supposed to be a main stream topic).

6. Nov 28, 2008

### Office_Shredder

Staff Emeritus
Are these real numbers? This just looks like a pretty trivial tautology (are there any other kinds? Yes) seeing how every pair (u,v) would satisfy all the inequalities

7. Nov 28, 2008

### Tac-Tics

All real numbers x have the property $$-\infty < x < \infty$$.

Your first question is worded strangely. It *sounds* like it's asking you to show that "if x and y are real numbers, then the average of x and y are real numbers".... but it's stated so poorly, I would go talk to your professor or whomever gave you this problem.

The second question is equally poor. Domains and range apply to functions, but it's not clear what function you're working with. And still, it's talking about being between negative and positive infinity, which is always true of real numbers. Saying a real number is less than infinity is the same as saying nothing at all about that number.

8. Nov 28, 2008

### kingwinner

Sorry for the bad wording! I have reworded my questions in my first post, please check it. So the thing is that in this context -∞ < u2 < ∞ does not just mean it is a single real number, it means u2 can take on EVERY real number. (similar to the idea: the "range" of a function is the set of ALL "output" values produced by that function)

9. Nov 28, 2008

### Tac-Tics

This still doesn't make sense. If u1 and u2 are real numbers, then (u1 + u2) / 2 is also a real number. So is (u1-u2)/2. Real numbers are always between negative and positive infinity, and so to say "-∞ < (u1+u2)/2 < ∞ and -∞ < (u1-u2)/2 < ∞" is silly, because it's always true, no matter what numbers u1 and u2 you use. You might as well be saying {(u1, u2) for all u1, u2 in R}.

Both sets are in fact equivalent to R^2, the set of ordered pairs of reals.

10. Nov 29, 2008

### kingwinner

1) But the thing I don't understand is WHY is {(u1,u2) : -∞ < (u1+u2)/2 < ∞ and -∞ < (u1-u2)/2 < ∞ } equal to R^2?

11. Nov 29, 2008

### Tac-Tics

R^2 is defined as {(x, y) for all x, y in R}.

12. Nov 29, 2008

### tiny-tim

Hi kingwinner!
Yes, Tac-Tics is right …

it's a tautology … it's automatically true …

saying "-∞ < (u1+u2)/2 < ∞" is the same as saying "(u1+u2)/2 is a number which is a number".

(and technically, ∞ and -∞ aren't even in R: you can't really say, for example. "0 < n < ∞", you just say "0 < n")

13. Nov 29, 2008

### kingwinner

I can certainly see why {(u1,y2) : -∞ < u1 < ∞ and -∞ < u2 < ∞} is equal to R^2 (just trival), but why is {(u1,u2) : -∞ < (u1+u2)/2 < ∞ and -∞ < (u1-u2)/2 < ∞ } equal to R^2? How can you be so sure that every point in R^2 is in the set?

14. Nov 29, 2008

### Office_Shredder

Staff Emeritus
How could a point fail to be in the set? Given two real numbers, when you add them you get a real number, and when you divide by two you get a real number

15. Nov 29, 2008

### Fredrik

Staff Emeritus
That second set is just the set of all ordered pairs (x,y) of real numbers that have the property that x+y and x-y are real numbers. And we know that no ordered pair (x,y) of real numbers have the property that x+y isn't a real number, since the sum of any two real numbers is a real number.

16. Nov 29, 2008

### tiny-tim

Because any point (p,q) in R2

(btw, do use the X2 and X2 tags … they're just above the reply field )

can be written ((p+q)/2 + (p-q)/2 , (p+q)/2 - (p-q)/2).