Proof of x=0 or y=0 for x^2+y^2=0

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Discussion Overview

The discussion revolves around the equation x² + y² = 0 and whether it implies that both x and y must equal zero, particularly within the context of real numbers. Participants explore the implications of the equation, considering both real and imaginary solutions.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant argues that from x² + y² = 0, it follows that x² = -y², leading to the conclusion that x² must equal zero, hence x = 0.
  • Another participant questions the assumption that -y² ≤ 0, expressing confusion about its validity.
  • Some participants note that if x and y are real numbers, then x² + y² = 0 implies both x and y must be zero.
  • There is mention of infinite imaginary solutions if complex numbers are considered, suggesting that the context of the numbers affects the conclusions drawn.
  • Participants discuss the logical implications of the equation, debating whether the correct interpretation is that both x and y must be zero or if one can be zero while the other is not.
  • One participant states that both implications (x = 0 and y = 0) being true is a stronger statement, but does not resolve the disagreement on the interpretations.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the equation, with some asserting that both x and y must be zero, while others suggest that either could be zero. The discussion remains unresolved regarding the interpretation of the implications.

Contextual Notes

Participants emphasize the importance of the assumption that x and y are real numbers, which affects the conclusions drawn. The discussion also highlights the potential for complex solutions, which introduces additional complexity to the problem.

evagelos
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Let x^2+y^2 =0,then x^2=-y^2\Longrightarrow x^2\leq 0 since -y^2\leq 0

but also x^2\geq 0 ,hence x^2=0\Longrightarrow x=0.

Thus x=0 v y=0
 
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evagelos said:
x^2\leq 0 since -y^2\leq 0
I don't see why this is true.

And of course there are an infinite number of imaginary solutions.
 
Perfection said:
I don't see why this is true.

And of course there are an infinite number of imaginary solutions.

We have x^2=-y^2\leq 0

I am sorry ,i should have pointed out that x and y belong to the real Nos
 
Last edited:
evagelos said:
-y^2\leq 0

Why presume this?
 
Perfection said:
Why presume this?

y^2\geq 0\Longrightarrow y^2-y^2\geq 0-y^2\Longrightarrow -y^2\leq 0
 
Looks fine to me
 
gb7nash said:
Looks fine to me

Then x^2+y^2=0\Longrightarrow x=0=y is correct or wrong?
 
evagelos said:
Then x^2+y^2=0\Longrightarrow x=0=y is correct or wrong?

Assuming x and y are real numbers, x^2+y^2=0\Longrightarrow x=0=y. If you're still not sure, just do a proof by contradiction. There's not much to do.
 
gb7nash said:
Assuming x and y are real numbers, x^2+y^2=0\Longrightarrow x=0=y. If you're still not sure, just do a proof by contradiction. There's not much to do.

So which is right x^2+y^2=0\Longrightarrow x=0\wedge y=0

OR

x^2 +y^2=0\Longrightarrow x=0\vee y=0
 
  • #10
evagelos said:
So which is right x^2+y^2=0\Longrightarrow x=0\wedge y=0

OR

x^2 +y^2=0\Longrightarrow x=0\vee y=0


I'm not sure what this means. Assuming that x and y are real numbers, if x^2 +y^2=0, both x and y are equal to 0. If you're allowing complex numbers, then there are infinite solutions. It just depends what you're allowed to work with.

Or are you still trying to figure out your proof? Your proof looked fine.
 
  • #11
evagelos said:
So which is right x^2+y^2=0\Longrightarrow x=0\wedge y=0

OR

x^2 +y^2=0\Longrightarrow x=0\vee y=0
Both implications are true, but the first one is a stronger statement and a more interesting result.
 

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