Domain of ln(x+y) | x ≠ 0, y ≠ 0 or (x,y) ≠ (0,0)?

[chetzread]

Homework Statement

in my notes , it's stated that the domain of ln(x+y) is x+y> 0 ...
Can i write it as x ≠ 0 , y ≠ 0 ? Or should i write it as (x,y) ≠ (0,0) ?(x,y) ≠ (0,0) represent x , and y can be 0 at the same time ?

Whereas x ≠ 0 , y ≠ 0 only represent x can't be equal to 0 ? for example, when x = 0 , y not = 0 , so , ln(x+y) is defined , when y = 0 , x not = 0 , so , ln(x+y) is defined .

Correct me if i am wrong ?

Homework Equations

The Attempt at a Solution

 

[fresh_42]

Homework Statement

in my notes , it's stated that the domain of ln(x+y) is x+y> 0 ...
Can i write it as x ≠ 0 , y ≠ 0 ? Or should i write it as (x,y) ≠ (0,0) ?(x,y) ≠ (0,0) represent x , and y can be 0 at the same time ?

Whereas x ≠ 0 , y ≠ 0 only represent x can't be equal to 0 ? for example, when x = 0 , y not = 0 , so , ln(x+y) is defined , when y = 0 , x not = 0 , so , ln(x+y) is defined .

Correct me if i am wrong ?

Homework Equations

The Attempt at a Solution

What about the negative numbers?

 

[chetzread]

negative number are ok , as long as x+y >0 , so , Can i write the domain as (x,y) ≠ (0,0) ?

 

[Big Red Car]

Hi Chetzread,
You may be a little confused because typically we see the equation y = f(x).
My advice is to reword the equation to z = f(x + y) = f(u), where u = x + y

So ask yourself what is the domain of f(u)? What values can u take? What values of x and y add to give u?

You are correct, both x and y cannot be equal to 0 at the same time. However, you may be surprised to learn that either x and y can be negative (not both).

Below is a plot of z = ln(x + y) (obtained from Wolfram), where x is the horizontal axis, y is the axis going out of the page, and z is the vertical axis. Note what we have discussed about the cases y<0, x<0, x=0, y=0.

https://www4b.wolframalpha.com/Calculate/MSP/MSP37851g488iahb1eeddb60000639baefd52h542g2?MSPStoreType=image/gif&s=63

EDIT: Sorry I have misunderstood the OP. x, y ≠ (0, 0) seems correct.

 

[fresh_42]

negative number are ok , as long as x+y >0 , so , Can i write the domain as (x,y) ≠ (0,0) ?

$(x,y) \neq (0,0)$ doesn't exclude $(-1,-1)$. You will have to add $x+y>0$ anyway, or an equivalent notation.

 

[chetzread]

Hi Chetzread,
You may be a little confused because typically we see the equation y = f(x).
My advice is to reword the equation to z = f(x + y) = f(u), where u = x + y

So ask yourself what is the domain of f(u)? What values can u take? What values of x and y add to give u?

You are correct, both x and y cannot be equal to 0 at the same time. However, you may be surprised to learn that either x and y can be negative (not both).

Below is a plot of z = ln(x + y) (obtained from Wolfram), where x is the horizontal axis, y is the axis going out of the page, and z is the vertical axis. Note what we have discussed about the cases y<0, x<0, x=0, y=0.

https://www4b.wolframalpha.com/Calculate/MSP/MSP37851g488iahb1eeddb60000639baefd52h542g2?MSPStoreType=image/gif&s=63

Yes, I know what x and y can be negative, as long as x+y> 0 , I'm wondering is it wrong to write the domain as write it as (x,y) ≠ (0,0) ?

 

[Big Red Car]

Look at Fresh_42's comment, he/she has corrected my edit.

EDIT: You can see it in the plot btw, what Fresh_42 is saying :). Good luck.

 

[chetzread]

$(x,y) \neq (0,0)$ doesn't exclude $(-1,-1)$. You will have to add $x+y>0$ anyway, or an equivalent notation.

so , if i write the limit as $(x,y) \neq (0,0)$ and $x+y>0$ at the same time , then , my ans would be correct ?

How if i write the domain as $x+y>0$ only , without $(x,y) \neq (0,0)$ , is it correct ?

 

[Mark44]

so , if i write the limit as $(x,y) \neq (0,0)$ and $x+y>0$ at the same time , then , my ans would be correct ?

How if i write the domain as $x+y>0$ only , without $(x,y) \neq (0,0)$ , is it correct ?

All you need is x + y > 0. That way, x can be 0, but y can't, and y can be 0, but x can't.

Something you might be missing is that the inequality x + y > 0 is an entire half-plane: all the points in the plane that are above the line y = -x. The origin is not in this set, so you don't need to also specify that $(x, y) \ne (0, 0)$.

 

[chetzread]

All you need is x + y > 0. That way, x can be 0, but y can't, and y can be 0, but x can't.

Something you might be missing is that the inequality x + y > 0 is an entire half-plane: all the points in the plane that are above the line y = -x. The origin is not in this set, so you don't need to also specify that $(x, y) \ne (0, 0)$.

so , just leave the final ans as x + y >0 , will do ?

 

[fresh_42]

so , if i write the limit as $(x,y) \neq (0,0)$ and $x+y>0$ at the same time , then , my ans would be correct ?

How if i write the domain as $x+y>0$ only , without $(x,y) \neq (0,0)$ , is it correct ?

$x+y>0$ automatically implies $(x,y) \neq (0,0)$. It doesn't need to be mentioned. However, $x=0$ or $y=0$ is still possible as long as the other one is positive. They even can be negative as you correctly have said (as long as the other one is positive and larger in its absolute value). You might write $x > -y$ if you like.

What is the purpose of writing the domain other than $\{(x,y) \in \mathbb{R}^2\; | \; x+y > 0\}$?

 

[chetzread]

ok , i have one more question here , for the domain of ln (x^2 + y^2 ) , it it given in my notes that the ans is x ≠ 0 , y ≠ 0

IMO , there 's no need to give x ≠ 0 , y ≠ 0 , so just leave the ans (x^2 + y^2 ) > 0 , will do ?

 

[fresh_42]

ok , i have one more question here , for the domain of ln (x^2 + y^2 ) , it it given in my notes that the ans is x ≠ 0 , y ≠ 0

IMO , there 's no need to give x ≠ 0 , y ≠ 0 , so just leave the ans (x^2 + y^2 ) > 0 , will do ?

In this case you don't have to deal with negative numbers, because the squares are always positive: $(-1)^2=1$.
Therefore only $(x,y)=(0,0)$ can violate this condition. So here $x^2+y^2 > 0$ is equivalent to $(x,y) \neq (0,0)$. As soon as one is unequal zero, the point is with in the domain. Drawn as a picture it is the entire plane without the origin.

 

[chetzread]

In this case you don't have to deal with negative numbers, because the squares are always positive: $(-1)^2=1$.
Therefore only $(x,y)=(0,0)$ can violate this condition. So here $x^2+y^2 > 0$ is equivalent to $(x,y) \neq (0,0)$. As soon as one is unequal zero, the point is with in the domain. Drawn as a picture it is the entire plane without the origin.

so , just choose anyone of $x^2+y^2 > 0$ or $(x,y) \neq (0,0)$ , the ans would be correct ?

 

[chetzread]

In this case you don't have to deal with negative numbers, because the squares are always positive: $(-1)^2=1$.
Therefore only $(x,y)=(0,0)$ can violate this condition. So here $x^2+y^2 > 0$ is equivalent to $(x,y) \neq (0,0)$. As soon as one is unequal zero, the point is with in the domain. Drawn as a picture it is the entire plane without the origin.

how about x ≠ 0 , y ≠ 0 is it wrong to write the limit as x ≠ 0 , y ≠ 0 ?

 

[chetzread]

so , just choose anyone of $x^2+y^2 > 0$ or $(x,y) \neq (0,0)$ , the ans would be correct ?

@fresh_42 correct?

 

[fresh_42]

how about x ≠ 0 , y ≠ 0 is it wrong to write the limit as x ≠ 0 , y ≠ 0 ?

I'm not sure I understand what you mean by limit. The difficulty with $x \neq 0\, , \,y \neq 0$ is, that it is not clear whether it has to hold for both or just for one of them. Therefore $(x,y) \neq (0,0)$ is clear: not both zero at the same time.

 

[fresh_42]

@fresh_42 correct?

Yes.

 

[chetzread]

I'm not sure I understand what you mean by limit. The difficulty with $x \neq 0\, , \,y \neq 0$ is, that it is not clear whether it has to hold for both or just for one of them. Therefore $(x,y) \neq (0,0)$ is clear: not both zero at the same time.

sorry , i mean domain as x ≠ 0 and y ≠ 0 ... Here's the original ans that the author gave ...

 

[fresh_42]

sorry , i mean domain as x ≠ 0 and y ≠ 0 ... Here's the original ans that the author gave ...

Even this is ambiguous: For $(x,y)=(1,0)$ there is $[\;(x \neq 0) \text{ and } (y \neq 0) \;] = \text{ false }$, but the point is allowed!
Langauge often isn't very clear, especially when it comes to negations and even more if two of them are combined by and or or.
It definitely should be avoided. In general it is always better to express something positively, if possible, rather than using "not". E.g. it is usually harder to prove, that something cannot occur, than it is to prove that something is true. Think of how long it took to prove Fermat's last theorem: more than 300 years!

So $\{(x,y)\in \mathbb{R}^2 \,\vert \, x^2 +y^2 > 0\}$ is better than something with $x \neq 0 </whatever> y \neq 0$.
The pairing $(x,y) \neq (0,0)$ avoids these troubles.

 

[chetzread]

Even this is ambiguous: For $(x,y)=(1,0)$ there is $[\;(x \neq 0) \text{ and } (y \neq 0) \;] = \text{ false }$, but the point is allowed!
Langauge often isn't very clear, especially when it comes to negations and even more if two of them are combined by and or or.
It definitely should be avoided. In general it is always better to express something positively, if possible, rather than using "not". E.g. it is usually harder to prove, that something cannot occur, than it is to prove that something is true. Think of how long it took to prove Fermat's last theorem: more than 300 years!

So $\{(x,y)\in \mathbb{R}^2 \,\vert \, x^2 +y^2 > 0\}$ is better than something with $x \neq 0 </whatever> y \neq 0$.
The pairing $(x,y) \neq (0,0)$ avoids these troubles.

So , for this type of question , always stick to (x^2 +y^2 ) > 0 will be enough ?

 

[fresh_42]

So , for this type of question , always stick to (x^2 +y^2 ) > 0 will be enough ?

Yes. It is pretty clear. But so is $(x,y) \neq (0,0)$. In this case it isn't much of a difference.
However in $x+y>0$ dealing with the cases makes it more complicated.

 

[chetzread]

However in x+y>0x+y>0x+y>0 dealing with the cases makes it more complicated.

what do you mean ?

 

[fresh_42]

what do you mean ?

I mean, if you have $D=\{(x,y) \in \mathbb{R}^2 \,\vert \, x+y > 0\}$ and try to describe it like

$$D=\{(x,y) \in \mathbb{R}^2 \,\vert \, \begin{cases} x \neq 0 & \text{ if } y =0 \\ y \neq 0 & \text{ if } x =0
\\ x > -y & \text{ if } y \leq 0 \text{ and } x>0 \\ x > 0 & \text{ if } y \geq 0 \\ y > -x & \text{ if } x \leq 0 \text{ and } y > 0 \\ y > 0 & \text{ if } x \geq 0 \end{cases} \quad \} $$
it is obvious what is clearer. And it took me quite a while to write this. And I'm still not sure, whether there is a mistake in it.

 

[SammyS]

So , for this type of question , always stick to (x^2 +y^2 ) > 0 will be enough ?

It's much clearer to write (x,y) ≠ (0,0) . Especially, anybody grading your solution will be more certain the you understand what is the domain.