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?
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.
So I was correct the answer is i right?
To determine the value of I^793, you need to understand the properties of the imaginary unit, denoted by "i."
The imaginary unit, i, is defined as the square root of -1. It has the property that i^2 equals -1. Based on this property, we can determine the possible values of i raised to any power.
To find i^793, we can first see that powers of i repeat with a cycle of four. This means that i^1 = i, i^2 = -1, i^3 = -i, and i^4 = 1. Therefore, any power of i can be simplified by dividing the exponent by 4 and considering the remainder.
In the case of i^793, we can write 793 as 792 + 1. Dividing 792 by 4 gives a remainder of 0, so i^792 equals 1. Therefore, i^793 equals i^792 multiplied by i^1.
Since i^792 = 1 and i^1 = i, we have i^793 = 1 * i = i.
So the value of i^793 is i.