Is in ( square root x 2 exponent) = x an identity (true for all nonnegative values of x)?

Question

Is in ( square root x 2 exponent) = x an identity (true for all nonnegative values of x)

Answer 1

This somewhat tricky. If you consider sqrt as a function, that is, the function that to each non-negative real number assigns it's positive square root, then the answer is sure yes. Simply because +sqrt(x^2) = x.

But if you think of sqrt as a relation, that is, as the number whose square is x, then you have 2 values for sqrt(x^2): x itself and also -x. Then, in this case the answer is no, not always.

But, by convention, when we just have sqrt(x), x>=0 real, it's understood the positive square root.

Answer 2

the square root of x to the second power is x. because it is the same as saying x to the power of 2/2. the cube root of x to the power of 3, will be x. any number rooted to the same power as the exponent of the base will leave you with just the base, but if the root number and the exponent are different then that's another story all together, for example, if you have the 7 root of x to the power of 4, it's the same as x to the 4/7 power. in that case, x will be taken to the power of 4, then rooted to the power of seven.

Answer 3

so which you prefer to be attentive to if sqrt(x^2) = x is an identity for all nonnegative values of x. the respond is sure. extremely, sqrt(x^2) = |x| (absolute fee of x), yet once you limit this to nonnegative values, then for nonnegative values, |x| = x, so sqrt(x^2) = x An occasion of displaying the way it is fake for destructive numbers is that if we permit x = -a million, then the left hand element is comparable to sqrt([-a million]^2) = sqrt(a million) = a million, yet, the desirable hand element is comparable to -a million. this is sqrt([-a million]^2) = (-a million) sqrt(a million) = -a million a million = -a million this is a pretend assertion.

Answer 4

Yes, for all non-negative values of x, sqrt(x^2) = x