Errata on Dummit and Foote, Abstract Algebra

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This is from page 55 of Dummit and Foote's Abstract Algebra, 3rd edition. I am pretty sure the third line of the proof "$x^b-a=x^0=1$" is a typo and it should be "$x^b-a=1$, where $0<b-a<n$, contradicting the fact $|x|=n$".



I am not sure why $x^0$ is there. Could someone please clarify this? I looked up the list of errata but couldn't find anything about this.




abstract-algebra




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  • 1




    I can't comment on the intention of making the remark, but it always holds (by definition) that $x^0 = 1$ and since $x^b-a = 1$, the equality is true (even though it may not be clarifying anything).
    – Guido A.
    Aug 2 at 1:21











  • Thank you for your reply.
    – kmiyazaki
    Aug 2 at 1:23














up vote
1
down vote

favorite












enter image description here



This is from page 55 of Dummit and Foote's Abstract Algebra, 3rd edition. I am pretty sure the third line of the proof "$x^b-a=x^0=1$" is a typo and it should be "$x^b-a=1$, where $0<b-a<n$, contradicting the fact $|x|=n$".



I am not sure why $x^0$ is there. Could someone please clarify this? I looked up the list of errata but couldn't find anything about this.




abstract-algebra




share|cite|improve this question















  • 1




    I can't comment on the intention of making the remark, but it always holds (by definition) that $x^0 = 1$ and since $x^b-a = 1$, the equality is true (even though it may not be clarifying anything).
    – Guido A.
    Aug 2 at 1:21











  • Thank you for your reply.
    – kmiyazaki
    Aug 2 at 1:23












up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



This is from page 55 of Dummit and Foote's Abstract Algebra, 3rd edition. I am pretty sure the third line of the proof "$x^b-a=x^0=1$" is a typo and it should be "$x^b-a=1$, where $0<b-a<n$, contradicting the fact $|x|=n$".



I am not sure why $x^0$ is there. Could someone please clarify this? I looked up the list of errata but couldn't find anything about this.




abstract-algebra




share|cite|improve this question











enter image description here



This is from page 55 of Dummit and Foote's Abstract Algebra, 3rd edition. I am pretty sure the third line of the proof "$x^b-a=x^0=1$" is a typo and it should be "$x^b-a=1$, where $0<b-a<n$, contradicting the fact $|x|=n$".



I am not sure why $x^0$ is there. Could someone please clarify this? I looked up the list of errata but couldn't find anything about this.





abstract-algebra





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 2 at 1:14









kmiyazaki

15811




15811







  • 1




    I can't comment on the intention of making the remark, but it always holds (by definition) that $x^0 = 1$ and since $x^b-a = 1$, the equality is true (even though it may not be clarifying anything).
    – Guido A.
    Aug 2 at 1:21











  • Thank you for your reply.
    – kmiyazaki
    Aug 2 at 1:23












  • 1




    I can't comment on the intention of making the remark, but it always holds (by definition) that $x^0 = 1$ and since $x^b-a = 1$, the equality is true (even though it may not be clarifying anything).
    – Guido A.
    Aug 2 at 1:21











  • Thank you for your reply.
    – kmiyazaki
    Aug 2 at 1:23







1




1




I can't comment on the intention of making the remark, but it always holds (by definition) that $x^0 = 1$ and since $x^b-a = 1$, the equality is true (even though it may not be clarifying anything).
– Guido A.
Aug 2 at 1:21





I can't comment on the intention of making the remark, but it always holds (by definition) that $x^0 = 1$ and since $x^b-a = 1$, the equality is true (even though it may not be clarifying anything).
– Guido A.
Aug 2 at 1:21













Thank you for your reply.
– kmiyazaki
Aug 2 at 1:23




Thank you for your reply.
– kmiyazaki
Aug 2 at 1:23










1 Answer
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Consider how one would get to $x^b-a$ from $x^b$. The most straightforward way is to multiply by $x^-a$. Thus, given the equation $$x^b=x^a$$ we get $x^b-a=x^a-a=x^0$ after multiplying by $x^-a$ on both sides. This is likely to be the origin of the mysterious $x^0$ term.






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  • That makes sense. Thank you so much.
    – kmiyazaki
    Aug 2 at 1:28










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active

oldest

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up vote
2
down vote













Consider how one would get to $x^b-a$ from $x^b$. The most straightforward way is to multiply by $x^-a$. Thus, given the equation $$x^b=x^a$$ we get $x^b-a=x^a-a=x^0$ after multiplying by $x^-a$ on both sides. This is likely to be the origin of the mysterious $x^0$ term.






share|cite|improve this answer





















  • That makes sense. Thank you so much.
    – kmiyazaki
    Aug 2 at 1:28














up vote
2
down vote













Consider how one would get to $x^b-a$ from $x^b$. The most straightforward way is to multiply by $x^-a$. Thus, given the equation $$x^b=x^a$$ we get $x^b-a=x^a-a=x^0$ after multiplying by $x^-a$ on both sides. This is likely to be the origin of the mysterious $x^0$ term.






share|cite|improve this answer





















  • That makes sense. Thank you so much.
    – kmiyazaki
    Aug 2 at 1:28












up vote
2
down vote










up vote
2
down vote









Consider how one would get to $x^b-a$ from $x^b$. The most straightforward way is to multiply by $x^-a$. Thus, given the equation $$x^b=x^a$$ we get $x^b-a=x^a-a=x^0$ after multiplying by $x^-a$ on both sides. This is likely to be the origin of the mysterious $x^0$ term.






share|cite|improve this answer













Consider how one would get to $x^b-a$ from $x^b$. The most straightforward way is to multiply by $x^-a$. Thus, given the equation $$x^b=x^a$$ we get $x^b-a=x^a-a=x^0$ after multiplying by $x^-a$ on both sides. This is likely to be the origin of the mysterious $x^0$ term.







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answered Aug 2 at 1:26









Karl Kronenfeld

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  • That makes sense. Thank you so much.
    – kmiyazaki
    Aug 2 at 1:28
















  • That makes sense. Thank you so much.
    – kmiyazaki
    Aug 2 at 1:28















That makes sense. Thank you so much.
– kmiyazaki
Aug 2 at 1:28




That makes sense. Thank you so much.
– kmiyazaki
Aug 2 at 1:28












 

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