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Unformatted text preview: MATH 413 HW 4: Solution to selected problems. September 26, 2007 Problem 1 Prove that the only map f : R → R satisfying f (1) = 1 and for all x, y ∈ R f ( x + y ) = f ( x ) + f ( y ) , f ( xy ) = f ( x ) f ( y ) is the identity. Is the same true for the complex numbers C ? Solution: (i) It is easy to prove by induction that f ( n ) = n for all n ∈ N . Now since, f (0) = f (0 + 0) = 2 f (0), we conclude that f (0) = 0. Also, 0 = f (0) = f (1 + ( 1)) = f (1) + f ( 1) = 1 + f ( 1), and hence f ( 1) = 1 and by induction f ( n ) = n for all n ∈ Z . Because f preserves the product and it is the identity restricted to the integer numbers, given m, n ∈ Z with n 6 = 0 we have that nf ( m n ) = f ( n m n ) = f ( m ) = m , from which it follows that f ( m n ) = m n and f is the identity map restricted to the rational numbers. (ii) Given x > 0 we have that f ( x ) = ( f ( √ x )) 2 . Now f ( √ x ) 6 = 0, oth erwise f (1) = f ( √ x ) f ( 1 √ x ) = 0, which contradicts our initial hypoth esis, namely f (1) = 1. We conclude then that f ( x ) > 0, i.e f maps positive numbers into positive numbers. This implies that given x < y , f ( y x ) = f ( y ) + f ( x ) > 0, but since f ( x ) = f (( 1) x ) = f ( x ) we infer that f ( x ) < f ( y ). In other words, f preserves order as well. (iii) Finally, given x ∈ R we know that they are sequences of rational numbers n 7→ x n and n 7→ y n converging to x such that x n < x < y n for all n . By (i)(ii) we conclude that x n < f ( x ) < y n for all n , and by taking limit when n goes to infinity we get that f ( x ) = x for x ∈ R , as we wanted to prove. (iv) The function f : C → C defined by f ( a + ib ) = a ib (conjuga tion), maps 1 to 1, and preserves addition and multiplication. However it is not equal to the identity map. Why is this the only function on C with the listed properties different from the identity?the listed properties different from the identity?...
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 Fall '07
 HUBBARD
 Math, Topology, Empty set, Metric space, Closed set, b1, open sets

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