sole3 - ANALYSIS EXERCISE 3–SOLUTIONS 1. [in F.T.A.] (a)...

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Unformatted text preview: ANALYSIS EXERCISE 3–SOLUTIONS 1. [in F.T.A.] (a) Reflexive (f (x) = f (x)), symmetric (f (x) = f (y )) ⇒ (f (y ) = f (x)), transitive ([(f (x) = f (y )] ∧ [f (y ) = f (z )] ⇒ [f (x) = f (z )]). (b) Define f : R2 → R by f (x, y ) = x2 + y 2 . Then [(a, b) ∼ (c, d)] ⇔ [f (a, b) = f (c, d)]. So by (a) ∼ is an equivalence relation. Equivalence classes: [(0, 0)] = {(0, 0)}. If (a, b) = (0, 0) then [(a, b)] is the circle through (a, b) with the center (0, 0). 2. (i) 0; (ii) a3 − 1; (iii) a3 + 3a2 + 3a; (iv) a3 − 3a2 + 3a − 2; (v) 16a3 − 2. 3. (a) Ran(f ) = Z; (b) Ran(f ) = N+ ; (c) Ran(f ) = (−∞, −2] ∪ [2, ∞). 1 1 4. Substituting x instead of x in the equality we find that f ( x ) + 1 1 1 2f (x) = x , or 2f ( x ) + 4f (x) = 2 x . Subtracting the given equal2 ity from the last one we obtain f (x) = 1 x − x . 3 5. (a) {(0, 0)}; (b) Z × {0}; (c) {(m, n) | m ∈ N, n ∈ Z} = N × Z; (d) {(m, n) | m ∈ Z, n ∈ Z} = Z × Z. 6. (a) Negation: (∃x1 ∈ X )(∃x2 ∈ X )[(f (x1 ) = f (x2 )) ∧ (x1 = x2 )]. (b) Negation: (∃y ∈ Y )(∀x ∈ X )[f (x) = y ]. 7. (a) Injective. Proof: (2n1 + 1 = 2n2 + 1) ⇒ (n1 = n2 ). Therefore (∀n1 ∈ N)(∀n2 ∈ N){[f (n1 ) = f (n2 )] ⇒ (n1 = n2 )}. Not surjective. (2 ∈ Ran(f )). Not bijective. 1 (b) Not injective. (g (− 8 ) = g ( 1 ) = − 1 ). Surjective. Not bijective. 2 8 x (c) Injective ([( 1−1 1 = x jective. 1−x2 x2 ) ⇒ (x1 = x2 )]). Surjective, hence bi- 1 ...
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