 # 10 Problems with Functions, Sets, Recurrence Relations, Modular Operations, and Equivalence

## Transcribed Text

1. 1. Solve the following recurrence relation : as = 7ax-1 -10a1-2 Vk> 2 with ao - a, - 2 2. Solve the following recurrence relation : as = -4a1-1 -4a1-2 with a =0a1 =-1 3. a) Define F:Z->: Z by the rule F(n) = 2 -3n, for all integers n, i) Is Fone-to-one? (Prove or give a counterexample.) ii) Is Fonto? (Prove or give a counterexample.) b) Define G:R> R by the rule F(x) = 2 -3x, for all real number x, (i) Is R one-to-one? (Prove or give a counterexample.) ii) Is R onto? (Prove or give a counterexample.) 4. Supply a reason for each step in the derivation For all sets A, B and C, (AUB)nc=(ANC)U(BNC) = Proof: by (a) =(CNA)J(C)B) by (b) = (A7CJ(BNC) by (c) 5. Prove or disprove For all sets A and B, ANCAC B) = A 6. Determine which of the properties are true for all functions / from a set x and which are false for some functionf Justify your answers. a) For all subsets A and B of X, if Ag B, then f(A) n f(B). b) For all subsets A and B of X, f(AJB)=f(A)Uf(B) = 7. Compute 2014 2015 mod 11 8. Let Fn be Fibonacci sequence where Fo =0 F1 = 1 and Fn = Fn-1 + F0-2 . Prove that 9. Let A be set of non-zero integers and let R be the relation on AxA defined by (a,b) R (c,d) whenever ad = bc. Prove that R is an equivalence relation. 10. Let R and S be relations on a set A. Assume A has at least three elements, prove or disprove each of the following statements : If R and S are symmetric, then R ns is symmetric. If R and S are symmetric, then R US is symmetric. If R and S are reflexive, then R ns is reflexive. If R and S are reflexive, then R US is reflexive. If R and S are transitive, then R ns is transitive. If R and S are transitive, then RUS is transitive.

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9)
We need to prove the relation is reflexive, symmetric and transitive.
For reflexivity, this means (a,b) R (a,b) This is true since a*b=b*a => R is reflexive.
For symmetry, this means if (a,b) R (c,d), then ad=bc.
On the other hand, cb=da (we reverted the order)=> (c,d) R (a,b)=> R is a symmetric relation.
For transitivity, we assume that (a,b) R (c,d) and also (c,d) R (e,f).
In this case it means that ad=bc and also cf=de
We multiply the two equalities side by side => adcf=bcde => adcf-bcde=0=> (af-be)cd=0...
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