Asked by Muffy
My question is I^793
I think the answer is just I because it is the only number that can go into it. Is that correct?
I think the answer is just I because it is the only number that can go into it. Is that correct?
Answers
Answered by
bobpursley
are you looking for factors of x^793?
or i^793, where i is the sqrt(-1)?
factors of x^793 are very numerous. x^793 is not a prime number.
i^793=i^792 * 1
Now consider
i^2=-1
i^4=1
i^6=-1
so the conclusion is that i^(4n-2) is -1
otherwise, i to even is 1
WEll, is 792=4n-2
4n=794, or n is not an integer, so
i^792=1
finally, i^792*i= i and it has only one factor, i.
or i^793, where i is the sqrt(-1)?
factors of x^793 are very numerous. x^793 is not a prime number.
i^793=i^792 * 1
Now consider
i^2=-1
i^4=1
i^6=-1
so the conclusion is that i^(4n-2) is -1
otherwise, i to even is 1
WEll, is 792=4n-2
4n=794, or n is not an integer, so
i^792=1
finally, i^792*i= i and it has only one factor, i.
Answered by
Muffy
So I was correct the answer is i right?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.