Negating a Statement: Understanding the Rules and Implications

  • Thread starter Thread starter cris(c)
  • Start date Start date
Click For Summary
The discussion centers on the correct negation of a mathematical statement involving continuous functions defined on the interval [0,1]. The original statement asserts that for all elements in a countable set J, one function is less than or equal to another for all x in the interval. The user correctly negates this to state that there exists at least one element in J for which the first function exceeds the second at some point in the interval. Additionally, the user inquires whether this negation allows the assumption that all other elements in J still satisfy the original inequality, which is a crucial point for constructing a proof. Clarification on these logical implications is sought to ensure accurate understanding and application in mathematical reasoning.
cris(c)
Messages
21
Reaction score
0
negating a statement...need help urgently!

Hi everyone:

I am not sure about the following thing I did. Let J be a countable finite set, and f_{jk}^{0} and f_{jk}^{1} be two continuous functions defined on [0,1]. Consider the following statement:

\forall lj\in J,\forall x\in[0,1],\: \, f_{jk}^{0}(x)\leq f_{jk}^{1}(x)

Negating the above statement gives me:

\exists lj\in J,\exists \hat{x} \in[0,1],\: \, f_{jk}^{0}(\hat{x})> f_{jk} ^{1}(\hat{x})

Question 1: Am I correct in the way I negate the original statement?

Question 2 (and perhaps the most important): The fact that the negation involves only one member gives the freedom to assume that every other element satisfies the properties in the original statement? i.e., can I assume, while constructing a proof, that \forall hz\in J other than l and j, f_{lz}^{0}(x)\leq f_{lz}^{1}(x)\: \forall x\in[0,1]?

Thanks a lot! I truly appreciate any help you can give me!
 
Physics news on Phys.org


any one out there willing to help?
 

Similar threads

  • · Replies 71 ·
3
Replies
71
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
932
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 11 ·
Replies
11
Views
1K