No Solution for 3x+2y=7 in Real Numbers

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SUMMARY

The discussion centers on the logical statement (∃x ∈ R)(∀y ∈ R)(3x + 2y = 7) and its negation. The correct negation is (∀x ∈ R)(∃y ∈ R)(3x + 2y ≠ 7), indicating that for every real number x, there exists a real number y such that the equation does not hold. Participants confirm the interpretation of the notation and validate the negation process, emphasizing the importance of understanding logical quantifiers in mathematical statements.

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Statement:

([tex]\exists_{x} \in[/tex] R) ([tex]\forall_{y} \in[/tex] R) (3x + 2y = 7)

Trying to find negation statement. This is what I think it is:

([tex]\forall_{x} \in[/tex] R)([tex]\exists_{y} \in[/tex] R) (3x + 2y ≠ 7)


Is this close?
 
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I'm having trouble reading this notation, but I'm assuming the first statement says "There exists an x in R such that for all y in R, 3x + 2y = 7."

To make this backwards, we would need to say, "For each x in R, there exists a y in R such that 3x + 2y does not equal 7," so if I have interpreted your notation correctly, I think your answer is correct.
 
Thanks, phreak. Yes, that is how it should read.
 

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