Unformatted text preview: 94 Solutions to Exercises (b) This is true. From the identity r a  sa = (r  s)(r a l + r a 2 s + r a 3 s 2 + .. . rs a 2 + saI), we see that n I (r a  sa). 16. Following the hint, we note that 10 == 1 (mod 11) and so 10 k == (_l)k (mod 11) for any natural number k. Since (anIan2 .. . a3a2alaoho = lOnI an_ 1 + lOn2an_2 + ... + 103a3 + 102a2 + lOal + ao. we have (anIan2 ... a3a2al aoho == (_l)nI an_1 + (It 2 an_2 + ...  a3 + a2  al + ao (mod 11) and this is 0 (mod 11) if and only if ao + a2 + ... == al + a3 + ... (mod 11). 17. (a) [BB] Suppose V2 = ~ for some integers a and b. If a and b have any factors in common, these can be canceled leaving us with an equation of the form V2 = ~ where a and b have no common factors (except ±1). So we now make this assumption. Then a 2 = 2b 2 , so a 2 + b 2 = 3b 2 == 0 (mod 3). By Problem 23, a and b have 3 as a common factor, which is not true....
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 Summer '10
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 Graph Theory, Trigraph, Divisor, Norrbottens regemente

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