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Unformatted text preview: Solutions to Assignment Section 2.1 #4 Solution 1 Consider the real number a & ( a + ( 1)) : We have a & ( a + ( 1)) = ( a & a ) + ( a & ( 1)) ; by distributive law = a + ( a ) ; by our assumption a & a = a = Hence by Theorem 2.1.3 (b), we have a = 0 or a + ( 1) = 0 : In other words, a = 0 or a = 1 : #4 Solution 2 Suppose a 6 = 0 : Then 9 1 a 2 R such that a & 1 a = 1 : Hence 1 = a & 1 a = ( a & a ) & 1 a = a & ( a & 1 a ) = a & 1 = a: In other words, a = 1 : (Technically we don&t need to verify that and 1 are solutions to the equation a & a = a , although this is very easy to do.) #8 (a) We can write x and y in terms of fractions x = a b ; and y = c d ; such that a;b; c; d 2 Z ; and bd 6 = 0 : then using the properties of real numbers and the fact that Z is closed under + and & ; we see that x + y = ad + bc bd 2 Q xy = ac bd 2 Q as well. Hence Q is closed under + and & : #8 (b) Suppose y is irrational. Note by the same argument as above, Q is closed under taking additive inverse, and Q nf g is closed under taking multiplicative inverse. So if x 2 Q and ( x + y ) 2 Q ; then ( x ) 2 Q and y = ( x + y ) + ( x ) 2 Q ; which is a contradiction. If x 6 = 0 ; x 2 Q and xy 2 Q ; then 1 x 2 Q ; and y = ( xy ) & ( 1 x ) 2 Q ; which is a contradiction. #20 (a) By Theorem 2.1.7 (c), the condition < c < 1 ; if the inequalities are multiplied by c; means < c 2 < c: Hence < c 2 < c < 1 by Theorem 2.1.7 (a). #20 (b) Again, 1 < c leads to c < c 2 ; and hence 1 < c < c 2 : Section 2.2 #4 Here it is implicit that " > ; otherwise two statements are trivially true and hence they imply each other. By Theorem 2.2.2 (c), j x a j < " if and only if " < x a < ": By Theorem 2.1.7 (b), this is equivalent to a " < x < a + "; by adding a or adding ( a ) to go back and forth between the two statements....
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