74-hw9s

# 74-hw9s - MATH 74 HOMEWORK 9 DUE MONDAY APRIL 23 1 We...

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MATH 74 HOMEWORK 9 - DUE MONDAY, APRIL 23 1. We have (using the fact that 1 = 1 · 1, and then using (3)) L (1) = L (1 · 1) = L (1) + L (1) . Subtracting L (1) (whatever it is) from both sides we conclude that L (1) = 0. 2. For our fixed y (0 , ) we have y · 1 y = 1, so by Exercise 1 L y · 1 y = L (1) = 0 . On the other hand, by (3) we have L y · 1 y = L ( y ) + L 1 y . We conclude that L ( y ) + L 1 y = 0 and subtracting L ( y ) from both sides we see L ( 1 y ) = - L ( y ) as desired. 3. We prove that L ( x n ) = nL ( x ) for all n N by induction. When n = 1 the statement to be proved is that L ( x ) = L ( x ), which is certainly true. Supposing that L ( x k ) = kL ( x ) for some k N , we have L ( x k +1 ) = L ( x k · x ) = L ( x k ) + L ( x ) by (3) = kL ( x ) + L ( x ) since L ( x k ) = kL ( x ) = ( k + 1) L ( x ) showing that the statement is true for k + 1 as well. By the induction principle we conclude L ( x n ) = nL ( x ) for all n N . Remark. You will see problems (and proofs) very similar to exercises 1, 2, and 3 in Math 113 and other algebra classes, where a very common problem type is “given

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