Negating x^2 + y^2 > 0 for All x,y

  • Thread starter Thread starter Caldus
  • Start date Start date
AI Thread Summary
The discussion focuses on how to correctly negate the statement "For all x > 0, x^2 + y^2 > 0 for all y." The proposed negation is refined to "There exists one x > 0 and one y such that x^2 + y^2 ≤ 0." Participants clarify that the original negation attempts were close but needed adjustments, particularly regarding the conditions on x and y. The final formulation emphasizes the existence of specific values that contradict the original statement. Overall, the discussion highlights the nuances in logical negation in mathematical expressions.
Caldus
Messages
106
Reaction score
0
How do I write the negation of:

For all x > 0, x^2 + y^2 > 0 for all y.

I thought it might be this:

There exists x < or = to 0 such that x^2 + y^2 < or = to 0 for one y value.

Thanks.
 
Mathematics news on Phys.org


Originally posted by Caldus
How do I write the negation of:

For all x > 0, x^2 + y^2 > 0 for all y.

I thought it might be this:

There exists x < or = to 0 such that x^2 + y^2 < or = to 0 for one y value.

Thanks.

I think that was close but not exact, it is:

There exist one x > 0 such that x^2+y^2 < or = 0 for one y.

The thing is that there is no statement about x < 0. So that there must be no statement for x <0 in the negation.

******************

Maybe an better formulation (and equivalent) of the problem is:

How do I write the negation of:

For all x>0 and for all y, x^2 + y^2 > 0.

Result:

There exist one x>0 and there one y such that x^2 + y^2 <= 0.

*********************

I hope it did help...
 
Whoops, I did that wrong. The actual statements are (for the problem, not the solution):

For every x >0, x^2 + y^2 > 0 for all y.

Close enough I guess?
 
\forall x&gt;0 \exists y&gt;0 \] s.t. \[ x^2+y^2 \leq 0
 
Originally posted by NateTG
\forall x&gt;0 \exists y&gt;0 \] s.t. \[ x^2+y^2 \leq 0

What is that in English? Thanks.
 
What is that in English? Thanks.
"For all x greater than 0, there exist a y> 0

such that x^2+ y^2\leq0"

(It is, by the way, false.)
 
Last edited by a moderator:
Right, but it is the negation of the statement he made.
 
Back
Top