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# let n be an integer greater than 1- In a ring in which x^n-x for all x.docx

let n be an integer greater than 1. In a ring in which x^n=x for all x, show that ab=0 implies ba=0.
Solution
For part a), note that ba=(ba)^n=(ba)(ba)(ba)...(ba) n-times, but since multiplication is associative, this can be re-written as b[(ab)(ab)(ab)...(ab)]a with n-1 times of the term (ab) inside, but our hypothesis is that ab=0, so this whole product is equal to 0 (since 0 times anything is 0) For part b) we can write the product ab in two different ways, first ab=(ab)^2=abab and secondly ab=a^2b^2=aabb. Thus abab=aabb, and now you can cancel the a\'s on the left and the b\'s on the right to obtain ba=ab
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let n be an integer greater than 1. In a ring in which x^n=x for all x, show that ab=0 implies ba=0.
Solution
For part a), note that ba=(ba)^n=(ba)(ba)(ba)...(ba) n-times, but since multiplication is associative, this can be re-written as b[(ab)(ab)(ab)...(ab)]a with n-1 times of the term (ab) inside, but our hypothesis is that ab=0, so this whole product is equal to 0 (since 0 times anything is 0) For part b) we can write the product ab in two different ways, first ab=(ab)^2=abab and secondly ab=a^2b^2=aabb. Thus abab=aabb, and now you can cancel the a\'s on the left and the b\'s on the right to obtain ba=ab
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### let n be an integer greater than 1- In a ring in which x^n-x for all x.docx

1. 1. let n be an integer greater than 1. In a ring in which x^n=x for all x, show that ab=0 implies ba=0. Solution For part a), note that ba=(ba)^n=(ba)(ba)(ba)...(ba) n-times, but since multiplication is associative, this can be re-written as b[(ab)(ab)(ab)...(ab)]a with n-1 times of the term (ab) inside, but our hypothesis is that ab=0, so this whole product is equal to 0 (since 0 times anything is 0) For part b) we can write the product ab in two different ways, first ab=(ab)^2=abab and secondly ab=a^2b^2=aabb. Thus abab=aabb, and now you can cancel the a's on the left and the b's on the right to obtain ba=ab