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

[evagelos]

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

 

[Perfection]

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.

 

[evagelos]

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

 

[Perfection]

-y^2\leq 0

Why presume this?

 

[evagelos]

Why presume this?

y^2\geq 0\Longrightarrow y^2-y^2\geq 0-y^2\Longrightarrow -y^2\leq 0

 

[gb7nash]

Looks fine to me

 

[evagelos]

Looks fine to me

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

 

[gb7nash]

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.

 

[evagelos]

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

 

[gb7nash]

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.

 

[Fredrik]

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.