Graph of xy=0 and absolute value y =1

[Nazariy]

Is a graph of xy=0 just a coordinate plane?

The way I think about it is in the following way:

In order to get 0 as a result, either x or y or both have to be zero.

If x =0, then y can take any value, i.e. is infinite range of y values
If y=0, then x can take any value, i.e. is infinite range of x values
If both x and y =0, then that is just the origin of the coordinate plane.

So it must be that the graph of the above is a coordinate plane, right?

For another one, is it just simply the graph of y=1, since we cannot have any other y's (i.e. negative ones)?

 

[Nazariy]

is that a silly question?

 

[pka]

Is a graph of xy=0 just a coordinate plane?

No it is not the coordinate plane. But it is the union of the x-axis with the y-axis.

 

[Deleted member 4993]

Which means - ONLY the set of points on the x-axis and the y-axis (x*y = 0) → union of points on the x-y axes.

A point on the x-y plane which do not fall on the axes - is not included in the above set, For example point (2,2,0) is on the x-y plane but cannot be included in the set above (because in this case x*y = 4 and $\displaystyle \ne \ 0$

|y| = 1 → Collection of points on two horizontal lines which are y = 1 and y = -1 (union) ............ added

 

[Nazariy]

No it is not the coordinate plane. But it is the union of the x-axis with the y-axis.

Yeap, that's what I meant

 

[Nazariy]

|y| = 1 → Collection of points on two horizontal lines which are y = 1 and y = -1 (union) ............ added

Hmmm, added?

 

[pka]

|y|=1

$\displaystyle \{(x,1): x\in\mathbb{R}\}\cup\{(x,-1): x\in\mathbb{R}\}~.$

 

[Nazariy]

$\displaystyle \{(x,1): x\in\mathbb{R}\}\cup\{(x,-1): x\in\mathbb{R}\}~.$

Got it, thanks pka.