document.write( "Question 1110263: Consider the sets A = {x ∈ Z | x = 6s + 1 for some s ∈ Z} and B = {x ∈ Z | x = 3t + 1 for some t ∈ Z}. Prove the following:\r
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document.write( "a: A ⊆ B\r
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document.write( "b: ¬(B ⊆ A)\r
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document.write( "So sorry I haven't tried anything, but I have no idea where to even begin. \n" );
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Algebra.Com's Answer #725303 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Do you understand the definitions of the two sets? \n" ); document.write( "Set A contains all integers that are 1 more than a multiple (positive or negative) of 6: A = {..., -11, -5, 1, 7, 13, ...} \n" ); document.write( "Set B contains all integers that are 1 more than a multiple (positive or negative0 of 3: B = {..., -11, -8, -5, -2, 1, 4, 7, 10, 13, ...} \n" ); document.write( "It is easy to see that every element in A is also in B, but not the other way around; that proves that A is a subset of B but B is not a subset of A. \n" ); document.write( "I'm not going to try to do a formal proof.... You can try it if that's what you need. \n" ); document.write( " \n" ); document.write( " |