Explain why any even root of a negative number is not a real number, but an odd root of a negative number is a real number.
let i=sqrt(-1)
then i^2= real number
i^3= realnumber*sqrt(-1)
and so on i^2n always real
and i^3n always has sqrt(-1) as a factor
a negative number, multiplied an odd number of times, is still negative
no real number, multiplied an even number of times, can be still negative
pos times pos is pos
neg times neg is pos
To understand why any even root of a negative number is not a real number, we need to consider the concept of complex numbers and the nature of even roots.
An even root refers to square root (√), fourth root (∜), sixth root (∛∛), and so on. When we take an even root of a number, we are trying to find a number that, when multiplied by itself a certain number of times, gives us the original number.
However, square roots involve a key principle: every positive real number has two square roots, one positive and one negative. For example, the square root of 4 is 2 and -2, because both 2 × 2 = 4 and (-2) × (-2) = 4.
Now, when dealing with even roots of negative numbers such as -1 or -4, we encounter a problem. There is no real number that can be multiplied by itself an even number of times to give a negative number as the result. For example, there is no real number x such that x × x = -4.
This is because squaring any real number will always yield a positive result. Therefore, any even root of a negative number is not a real number since there is no real solution. However, we can find a solution using complex numbers.
On the other hand, when it comes to odd roots of negative numbers, such as ∛(-1) or √(-8), there is a real number that can be multiplied by itself an odd number of times to give a negative result. For instance, ∛(-1) = -1, because (-1) × (-1) × (-1) = -1.
In summary, any even root of a negative number is not a real number because there is no real solution. However, an odd root of a negative number is a real number since there exists a real solution.