Sketch the following set of points in the xy plane

  • #1
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Homework Statement


[itex] {(x,y) ∈ ℝ^2 : (y-x)(y+x) = 0 } [/itex]

Homework Equations




The Attempt at a Solution


[/B]
Is the reason the answer is the union of y = -x and y = x because if you devide (y-x)(y+x) = 0 by y-x you get y = -x and if you devide (y-x)(y+x) = 0 by y+x you get y = x?

and if you multiply (y-x)(y+x) = 0 by any constant, after you devide by the factors you will get the same solutions anyways?
 

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  • #2
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That is right. I'm not sure it you have to go into that much detail. When numbers are multiplied together to get zero, one of the factors must be 0. Your book or class notes may state that as a theorem or fact that you can reference without an explanation. Your explanation goes into detail about it, what is good to know in any case, but I don't know if it is required.

PS. A good thing to learn is to use the theorems in the book rather than trying to figure everything out from scratch. Use the tools.
 
  • #3
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Homework Statement


[itex] {(x,y) ∈ ℝ^2 : (y-x)(y+x) = 0 } [/itex]

Homework Equations




The Attempt at a Solution


[/B]
Is the reason the answer is the union of y = -x and y = x because if you devide (y-x)(y+x) = 0 by y-x you get y = -x and if you devide (y-x)(y+x) = 0 by y+x you get y = x?

and if you multiply (y-x)(y+x) = 0 by any constant, after you devide by the factors you will get the same solutions anyways?
You can not divide the equation by an expression which can be zero. So explain the solution {x=y} ∪ {y=-x } in the following way: When a product is zero one of the factors must be zero. So either x-y = 0 or x+y =0 (or both) .
You can multiply an equation by any non-zero constant. But why do you want to multiply it?
 
  • #4
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So

( (y-x)(y+x) ) / (y+x) = 0 / (y+x)

would not be a valid expression? I thought in the case when the numerator is already 0 you can divide by an expression thats = 0


>You can multiply an equation by any non-zero constant. But why do you want to multiply it?

Ah it was just an observation I noticed I guess it doesn't really have any relevance.
 
  • #5
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You can not divide the equation by an expression which can be zero.
CORRECTION: I had to cross out some errors and insert corrections.
It's ok where the divisor is not zero. So that is only a problem at (0,0) when ##y=\pm x##. The other points are ok and the (0,0) point is easy to handle separately.

PS. This logic is only needed to show that there can not be other points in the graph. To prove that ##(y-x) \cdot 0 = 0## only requires the definition of 0.
 
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  • #6
ehild
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So

( (y-x)(y+x) ) / (y+x) = 0 / (y+x)

would not be a valid expression? I thought in the case when the numerator is already 0 you can divide by an expression thats = 0
No. it is not valid. You must no divide with zero.
If y+x=0, y=-x. If you perform the division, you get y-x=0, that is, y=x. And both equations can be valid only when x=y=0. But y=-x means that y-x= -2x, with any y, so y-x is not zero.
 
  • #7
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It's ok where the divisor is not zero. So that is only a problem at (0,0). The other points are ok and the (0,0) point is easy to handle separately.

PS. This logic is only needed to show that there can not be other points in the graph. To prove that ##(y-x) \cdot 0 = 0## only requires the definition of 0.
well, what is sin(x)/x at x=0?
You can multiply anything with zero, and the result will be zero. But you must not divide by zero.
 
  • #8
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well, what is sin(x)/x at x=0?
You can multiply anything with zero, and the result will be zero. But you must not divide by zero.
It's undefined at x=0, but defined everywhere else. Just because x=0 is a special case does not prevent division everywhere else. And the special case of x=0 is easy to handle separately, if needed.
 
  • #9
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So I would be fine to do it as long as I'm not looking at the point that gives me x=0?
 
  • #10
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So I would be fine to do it as long as I'm not looking at the point that gives me x=0?
Sorry, I was wrong in post #5 and will correct it.
In post #8, when I said x=0, I was referring to the sin(x)/x example of post #7. You would have to worry about any step where you divide by ##0=(y-x)## or by ##0=(y+x)##. So that is why I stated in post #5 that your logic would work better to show that any other point (where ##y \neq \pm x##) was not in the graph. At those points, there is no division by 0.
 
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  • #11
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So I would be fine to do it as long as I'm not looking at the point that gives me x=0?
Dividing by x+y is allowed only if you state that x≠-y, and dividing by x-y is allowed if you say that x≠y.
Note what I said in Post #3. Do not divide by anything, just refer to the law that multiplication with zero yields zero. The reverse is also true: when a product is zero at least one of the factors must be zero.
 
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