Question

# For Problems #5 – #9, you willl either be asked to prove a statement or disprove...

For Problems #5 – #9, you willl either be asked to prove a statement or disprove a statement, or decide if a statement is true or false, then prove or disprove the statement. Prove statements using only the definitions. DO NOT use any set identities or any prior results whatsoever. Disprove false statements by giving counterexample and explaining precisely why your counterexample disproves the claim. ********************************************************************************************************* (5) (12pts) Consider the < relation defined on R as usual, where x < y if and only if x represents a point on the number line that is to the left of the point that y represents. Prove if true, and disprove with a counterexample if false: The < relation is neither reflexive nor symmetric, but it is transitive. (6) (12pts) Prove if true and disprove if false, using NO prior results: For all subsets A, B, C of a common universe U, (A ∪ B) c ∩ C = Ac ∩ Bc ∩ C. (7) (12pts) Let A be any set with at least two elements. Consider the relation R defined on P(A) defined as follows: ∀B, C ∈ P(A), BRC ↔ B − C 6= ∅. For each of the three properties Reflexivity, Symmetry and Transitivity, determine whther or not this relation has the property. For each property either prove that R has the property, or show why it does not have the property with a counterexample. Be thorough. (8) (12pts) Prove if true and disprove if false, using NO prior results: For all sets A, B and C, if B ⊆ A and C ⊆ B, then (C − A) ∪ (B − A) = ∅

5) xRy means x<y

Reflexive property-

for all x€A we know x can't be less than x , thus R is not reflexive.

Symmetry property-

let for x,y€A we have x<y then we can clearly say y can't be less than x, thus R is not symmetric.

Transitive property-

Let x,y,z€A , and xRy,yRz that means we have

x<y and y<z , thus we may write

x<y<z implies x<z i.e xRz

Thus R is transitive.

6).

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