Why is the domain of x^(2/4) different to the domain of x^(1/2) ?

If we have f(x) = x^(1/2), the domain of f(x) is x >= 0.

However if we rewrite f(x) as f(x) = x^(1/2) = x^(2/4) = [x^(2)]^(1/4), we can input a negative number x, square that, and comfortably calculate the fourth root of the resulting positive number. Hence the domain of f(x) becomes all real x.

Yet a third written form of f(x):
f(x) = x^(2/4) = [x^(2/2)]^(1/2) = |x|^(1/2)
This again shows the domain becoming all real x. It also shows that despite the domain change, the range is identical to the original form x^(1/2), x >= 0 (since |x| >= 0).

From this speculation, the graph of y = x^(2/4) should be different to the graph of y = x^(1/2) (specifically, it would be the graph of x^(1/2) on both sides of the x-axis since the domain now includes negative x). I have searched online for a graph of y = x^(2/4), but every graphing programme immediately simplifies this to x^(1/2).

In a calculator, negative numbers can indeed be input into x^(2/4) = [x^(2)]^(1/4) BUT not into the equivalent form [x^(1/4)]^2.

Can anyone explain these observations? Why does the domain depend on the way we write the exponent and the order in which we take to a power / take a root?

Reply 1

16Characters....

13

Original post by Ecasx

Edit: Apologies I did not write my answer well.

x^(1/2) is defined for negative real x, but only if you allow your function to return complex numbers. If you do not allow it to return complex numbers then it is just not defined for negative real x, however you rewrite it.

Rewriting the function will not make it return a real where it would have previously returned a complex number. Let x = -4 and hence the true value of x^(1/2) is 2i. This is the principal root (-2i also satisfies the equation y^2 = -4 but by convention x^(1/2) will return the "positive" root when x is a negative real number)

Using your method of rewriting, x^(1/2) = [x^2]^(1/4) = 16^(1/4). The principal fourth root of 16 is 2. But 2 clearly does not satisfy y^2 = -4 as 2^2 = +4. The method failed because the fourth root function returned the "wrong root" which does not actually satisfy our original equation.There are 4 numbers which when raised to the power 4 equal 16: 2, -2. 2i, -2i. The value we wanted was 2i but by convention the function returned 2 as that is the principal root.

Also, my calculator will return a Non-real error if I input (-4)^(2/4) so it might be different for each calculator.

(edited 8 years ago)

Reply 2

Ecasx

OP

3

Original post by 16Characters....

Edit: Apologies I did not write my answer well.

x^(1/2) is defined for negative real x, but only if you allow your function to return complex numbers. If you do not allow it to return complex numbers then it is just not defined for negative real x, however you rewrite it.

Rewriting the function will not make it return a real where it would have previously returned a complex number. Let x = -4 and hence the true value of x^(1/2) is 2i. This is the principal root (-2i also satisfies the equation y^2 = -4 but by convention x^(1/2) will return the "positive" root when x is a negative real number)

Using your method of rewriting, x^(1/2) = [x^2]^(1/4) = 16^(1/4). The principal fourth root of 16 is 2. But 2 clearly does not satisfy y^2 = -4 as 2^2 = +4. The method failed because the fourth root function returned the "wrong root" which does not actually satisfy our original equation.There are 4 numbers which when raised to the power 4 equal 16: 2, -2. 2i, -2i. The value we wanted was 2i but by convention the function returned 2 as that is the principal root.

Also, my calculator will return a Non-real error if I input (-4)^(2/4) so it might be different for each calculator.

So in other words, by extending the codomain to complex numbers, x^(1/2) and x^(2/4) are clearly identical expressions.

I wasn't considering complex numbers outputs. I was only looking at real numbers, and I had this thought:

Assume x^(1/2) = x^(2/4) = [x^(2/2)]^(1/2) = |x|^(1/2)

Squaring both sides, x = |x|. So the assumption that x^(1/2) = x^(2/4) leads to x = |x|. Since this is only true for x >= 0, x^(1/2) = x^(2/4) only for x >= 0. Hence the domain of x^(2/4) is not extended.

I still don't understand, however, why x^(2/4) cannot deal with negative inputs and return real number outputs.

I can only clearly see x^(1/2) = x^(2/4) for a complex number codomain.

(edited 8 years ago)

Why is the domain of x^(2/4) different to the domain of x^(1/2) ?

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