Question

I want to use proper set notation to describe a domain and my domain happens to be y^2-x^2

so in proper set notation which one is more accurate D{(x y)|y^2-x^2>0}
or
D{(x y)|y=-x, y=x}

Thank you.

Answer 1

@inkyvoyd

Answer 2

i think your question is incomplete,
define x and y which type of numbers

Answer 3

\[\cos[1/\sqrt{X^2+Y^2}]\]

Answer 4

SORRY ITS Y^2-X^2

Answer 5

I don't know what i was thinking

Answer 6

\[\cos[1/(\sqrt{y^2-x^2})] find the domain \in proper set notation.\]

Answer 7

@Astrophysics

Answer 8

\[y^2-x^2>o\]

Answer 9

\[y^2=x^2\]

Answer 10

Wait what?

Answer 11

Your domain is just all real numbers then

Answer 12

y^2-x^2

Answer 13

\[y=x and y=-x\]

Answer 14

Oh is it cos(all that stuff)

Answer 15

\[\cos \left( \frac{ 1 }{ \sqrt{x^2-y^2} } \right)\]

Answer 16

exactly

Answer 17

y=x and y=-x ? for the domain

Answer 18

How can i write that in proper set notation?

Answer 19

http://www.wolframalpha.com/input/?i=cos%28%281%2Fsqrt%28x%5E2-y%5E2%29%29+domain

Answer 20

you wanted it like this right

Answer 21

how did wolfram get x^2>=y^2

Answer 22

I would've thought it was > 0 as well but looking at the alternate forms it does not seem to be the case

Answer 23

Yay jim

Answer 24

Set the denominator equal to 0 and solve for y
\[\Large \sqrt{y^2 - x^2} = 0\]
\[\Large (\sqrt{y^2 - x^2})^2 = 0^2\]
\[\Large y^2 - x^2 = 0\]
\[\Large y^2 - x^2+x^2 = 0+x^2\]
\[\Large y^2 = x^2\]
\[\Large \sqrt{y^2} = \sqrt{x^2}\]
\[\Large |y| = |x|\]
\[\Large y = |x|\]
\[\Large y = \pm x\]

So if y = x or y = -x, then the denominator is equal to 0. To avoid division by zero errors, we say that y cannot equal x and y cannot equal -x.


The domain is drawn from the set of real numbers R. In terms of ordered pairs, we're drawing from the set R x R = R^2.


So if you use set notation, the domain D would look like this

\[\Large D = \left\{(x,y) | (x,y) \in \mathbb{R}^2 \ \ , \ \ y \ne \pm x\right\}\]

D is the set of ordered pairs (x,y) allowed to be plugged into the function.

Answer 25

Sorry I'm not thinking. The radicand must also be positive, so

y^2 - x^2 > 0
y^2 > x^2
|y| > |x|
y > |x| or y < -|x|


So the better domain is

\[\Large D = \left\{(x,y) | (x,y) \in \mathbb{R}^2 \ \ , \ \ y > |x| \text{ or } y < -|x|\right\}\]

Answer 26

Haha was going to say

Answer 27

Perfect..And I believe the graph looks something like this