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# solutions - Solutions to Assignment Section 2.1#4 Solution...

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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 = 0 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 0 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 0 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 0 < c < 1 ; if the inequalities are multiplied by c; means 0 < c 2 < c: Hence 0 < 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 " > 0 ; 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. #15 Let " = j a ° b j 2 : If x 2 V " ( a ) \ V " ( b ) ; then j x ± a j < " and j x ± b j < ": Hence j a ± b j = j a ± x + x ± b j ² j a ± x j + j b ± x j < j a ± b j 2 + j a ± b j 2 = j a ± b j ; which is a contradiction. Hence V " ( a ) \ V " ( b ) = ; : QED. (Warning: assuming U \ V 6 = ; for all neighborhoods of a and b does not mean a single element x belongs to all of them!) 1

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Section 2.3 #4 Note 1 = 2 ² 1 ± ( ° 1) n n ² 2
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solutions - Solutions to Assignment Section 2.1#4 Solution...

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