i^6 = (i^2)^3 = (-1)^3 = -1
The square, not the square root of -1 is 1. :)
B.t.w., can you prove that -1 times -1 is 1? Hint, try to prove first that for any number X:
-1 times X equals -X
How do you find the square root of -1?
The square of any real number is alwayspositive. This means that if you want to define the square root of minus 1 you have to extend the real number system. Mathematicians have invented imaginary and complex numbers to deal with this.
The square root of minus one is denoted by i. So, by definition:
i^2 = -1
Complex numbers are numbers of the form a + bi, where a and b are real numbers.
So, e.g.
(1 + i)^2 = 1 + 2 i + i^2 = 2 i
Actually, we can't take the square root of a negative number. Mathematicians have a way. To find the square root of a negative number, take the square root, as if it were a positive number, then add the letter "i" (the i stands for imaginary). Therefore, the square root of -1 is 1i.
Suppose / is equal to square root symbol and ^ is the power
Then what would this be
/-1^6
It is 1.You see when you take a negative number and multiply it to a negative number it becomes a positive.
example:
-1*-1=1
when you take a positive number and multiply it to a negative number then it stays negative:
example:1*-1=-1
it's about distributive property
simplify the expression
12xy-4xy
8xy
To solve the expression /-1^6, we need to follow the order of operations, which states that exponentiation should be done before taking the square root.
First, let's evaluate -1^6:
Since the exponent is an even number, (-1)^6 means multiplying -1 by itself six times.
-1^6 = (-1)(-1)(-1)(-1)(-1)(-1) = 1
Now, we can find the square root of 1, which is simply 1.
So, /-1^6 = 1