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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ² = L (1) = 0 . On the other hand, by (3) we have L ± y · 1 y ² = L ( y ) + L ± 1 y ² . We conclude that L ( y ) + L ± 1 y ² = 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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/02/2008 for the course MATH 74 taught by Professor Courtney during the Fall '07 term at Berkeley.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online