Solving inequalities in TWO variables?

In summary: Why is the equality(=) true?The equality is true because every element in the first set can be proven to be in the second set.
  • #1
kingwinner
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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!
 
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  • #2
Hi kingwinner! :smile:
kingwinner said:
Suppose -∞ < (u1+u2)/2 < ∞ and -∞ < (u1-u2)/2 < ∞
Then this implies -∞ < u1 < ∞ and -∞ < u2 < ∞

Question 2)
Define u1=y1+y2, u2=y1-y2. If -∞ < y1 < ∞ and -∞ < y2 < ∞, then this implies -∞ < u1 < ∞ and -∞ < u2 < ∞

I don't understand :confused:

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

so these equations don't seem to say anything! :rolleyes:

What is the context?​
 
  • #3
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
kingwinner said:
1) The context is this:
{(u1,u2) : -∞ < (u1+u2)/2 < ∞ and -∞ < (u1-u2)/2 < ∞ } = {(u1,u2) : -∞ < u1 < ∞ and -∞ < u2 < ∞}
Why is the equality(=) true?

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. :smile:
 
  • #5
tiny-tim said:
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. :smile:

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).

Can somebody please help?
 
  • #6
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
kingwinner said:
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).

Can somebody please help?

All real numbers x have the property [tex]-\infty < x < \infty[/tex].

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
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
kingwinner said:
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?

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
Tac-Tics said:
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.

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


R^2 is defined as {(x, y) for all x, y in R}.
 
  • #12
Hi kingwinner! :smile:
Tac-Tics said:
… 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}.

Yes, Tac-Tics :smile: 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". :wink:

(and technically, ∞ and -∞ aren't even in R: you can't really say, for example. "0 < n < ∞", you just say "0 < n")
 
  • #13
Tac-Tics said:
R^2 is defined as {(x, y) for all x, y in R}.
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
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
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
kingwinner said:
… 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?

Because any point (p,q) in R2

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

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

1. How do you graph inequalities in two variables?

To graph inequalities in two variables, first rewrite the inequality in slope-intercept form (y = mx + b). Then, plot the y-intercept at point (0, b) and use the slope to find a second point on the line. Finally, shade the region above or below the line depending on the inequality symbol.

2. What is the solution set for an inequality in two variables?

The solution set for an inequality in two variables is the set of all ordered pairs (x, y) that make the inequality true when substituted into the original inequality. This can be represented graphically as the shaded region on the graph of the inequality.

3. How do you solve systems of inequalities in two variables?

To solve systems of inequalities in two variables, graph each inequality on the same coordinate plane. The solution to the system is the region where all of the shaded regions overlap. This can also be found algebraically by finding the intersection points of the lines or by using substitution or elimination methods.

4. What are the rules for solving inequalities in two variables?

The rules for solving inequalities in two variables are similar to those for solving equations. The only difference is when multiplying or dividing both sides by a negative number, the inequality symbol must be flipped. Another rule to keep in mind is that when graphing, a solid line is used for "greater than or equal to" or "less than or equal to" inequalities, whereas a dashed line is used for "greater than" or "less than" inequalities.

5. How do you determine if a point is a solution to an inequality in two variables?

To determine if a point is a solution to an inequality in two variables, simply substitute the x and y values of the point into the inequality. If the resulting statement is true, then the point is a solution. If the statement is false, then the point is not a solution.

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