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,
5. Prove or disprove For all sets A and B, ANCAC B) = A
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) =
8. Let Fn be Fibonacci sequence where Fo =0 F1 = 1 and Fn = Fn-1 + F0-2 .
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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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...