Let a e Z. Then the translation function by a is defined to be the function A sub a : Z ---> Z defined by the rule x l--> x + a. Prove: If a,b e Z, then A sub a o A sub b = A sub (a+b) o = composed with
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Originally Posted by jzellt Let a e Z. Then the translation function by a is defined to be the function A sub a : Z ---> Z defined by the rule x l--> x + a. Prove: If a,b e Z, then A sub a o A sub b = A sub (a+b) o = composed with Let me start you off. note that $\displaystyle (A_a \circ A_b) (x) = A_a (A_b(x)) = A_a (x + b) = \dots $ can you finish?
Let me start you off. note that can you finish? So, Aa(x + b) = (x + b) + a = x + (b + a) = A sub (a+b) (x) .... Is this correct??
Originally Posted by jzellt Let me start you off. note that can you finish? So, Aa(x + b) = (x + b) + a = x + (b + a) = x + (a + b) = A sub (a+b) (x) .... Is this correct?? yup, that's it
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