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There are several questions already addressing this topic, but after reading through many of them, I didn't find any addressing my specific question.

According to Wikipedia, the square root symbol refers to the principal (positive) root and is equivalent to x^(1/2).
However, by the rules of exponents, (x^a)^b = x^(a*b) -> (x^2)^(1/2) = x^(2*1/2) = x.
But if we define the square root to be only positive, then ((-4)^2)^(1/2) = 16^(1/2) = 4 <> -4.

None of the other answers I've seen have addressed the contradiction of the rules of exponents inherent in a positive-only root definition. All of them have boiled down to "Because someone said so and we want to pass the vertical line test". My question is:

  1. Which of the following is false:

    a) sqrt(x)>=0

    b) sqrt(x)=x^(1/2)

    c) (x^a)^b=x^(a*b)

    d) 2*(1/2)=1

    e) (x^2)^(1/2)<>x^1 if x<0

  2. If the square root is indeed defined to be only positive, why would we choose to define it this way? Why is that better than only negative and what utility do we get out of having it be a function anyway?

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  • $\begingroup$ Careful: to find ((-4)^(1/2))^2, you need to first define ((-4)^(1/2)), which is never defined to be $2$. $\endgroup$
    – Ben Grossmann
    Commented Apr 8, 2018 at 3:11
  • $\begingroup$ Good point, I'll rearrange the order to make my point without bringing complex numbers into it. $\endgroup$
    – Kalev Maricq
    Commented Apr 8, 2018 at 3:13

1 Answer 1

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Just a reminder, that (-4)^(1/2) is not 2, but is 2i.

Okay, now the answers

  1. e

It's because the former one is not defined in real numbers, but the latter is defined.

  1. The reason square roots are only defined in positive is because there are no number that can be squared to get a value of negative. However, later on, you will learn i, which becomes -1 when squared.

    ex) (-9)^(1/2)=3i

Oh, and as a function, the root function starts from 0, since the coordinate plane is only defined in real numbers. Therefore, we can get the conclusion that in real numbers, rads cannot be negative.

And, I don't understand why the vertical test is mentioned. Vertical test is only used to determine if a graph is a function or not. If a graph's a function, it should have one y value for one x. That's how the vertical test works! If you draw a line on the graph vertically, there shouldn't be more than one point passing that line if it is a function.

hope this helps!

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  • $\begingroup$ I edited my post to avoid complex numbers by switching the order of the exponents. Does that change your answer at all, or is it still e) that is false? $\endgroup$
    – Kalev Maricq
    Commented Apr 8, 2018 at 4:00
  • $\begingroup$ still false, because they are whole different kinds of number $\endgroup$
    – user547075
    Commented Apr 8, 2018 at 4:02
  • $\begingroup$ oh wait (x^2)^(1/2) becomes -x, so they are different. $\endgroup$
    – user547075
    Commented Apr 8, 2018 at 4:04
  • $\begingroup$ I mean if x is negative, which is the requirement in the problem $\endgroup$
    – user547075
    Commented Apr 8, 2018 at 4:04
  • $\begingroup$ and if that's the case, the first example you gave me will have the value of 4, of course. $\endgroup$
    – user547075
    Commented Apr 8, 2018 at 4:08

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