What is wrong with $x=x^{2/2}=\sqrt {x^2}=\lvert x\rvert$

Question

The title says it all, $x=x^{2/2}=\sqrt {x^2}=\lvert x\rvert$ cannot possibly be true, so what am I missing?

Answer 1

I think there is nothing wrong with $x=x^{2/2}$. The order of operatins say you should evaluate $2/2$ first. So this is the same as $x=x^1$.

The problem is with $x^{2/2}=\sqrt{x^2}$. You have changed the order of execution. The right side is the same as $\left(x^2\right)^{1/2}$. So this comes down to whether $x^{(2/2)}$ is the same as $\left(x^2\right)^{1/2}$. You are familiar with the identity $\left(x^a\right)^b=x^{ab}$, which you might think applies here, but that identity only holds when $x$ is non-negative and/or when both $a$ and $b$ are integers.

For example, $(-1)^{2/2}=(-1)^1=-1$, but $\left((-1)^2\right)^{1/2}=\left(1\right)^{1/2}=1$.

Answer 2

$x^{\frac{2}{2}}=\sqrt{x^2}$ only if $x\ge0$,

in fact $x^{\frac{n}{m}}$ (where $n,m\in \mathbb{N}-\left\{0\right\}$) is defined as $\sqrt[m]{x^n}$ only for any $x\ge0$,

so $x=|x|$ is not wrong because $x$ cannot be negative if we use the equality $x^{\frac{2}{2}}=\sqrt{x^2}$.

Answer 3

I have rethought my previous answer. I think a better way to think of this as functions. The function f(x) = $xˆ2$ takes all negative values to positive values. This means that it's inverse function does not exist for all x. We would think was g(x) = $xˆ(1/2)$ But that only undoes the action for positive x. That is it is not the case that g(f(x)) = x if x is negative. So the step you took from $xˆ(2/2)$ to root $(xˆ2)$ is not correct.

Answer 4

The problem is that you assume $\sqrt{} $ and $^2$ commute (with respect to composition) for all values, but they don't. I. e. $$\sqrt{x^2} =\sqrt{x}^2$$ is true only for nonegative values.

Answer 5

An (I think necessary) aside.

It is true that you solve $x^2=4$ by taking the square root of both sides.

If you do that by substituting $\sqrt{x^2} = x$ and $\sqrt 4 = \pm 2$, you will get the right answer but you will have done it wrong.

Because you get the right answer by doing it wrong, you end up wondering why you can prove obviously wrong things like, for example, $x=|x|$ and $x=-x$.

To solve $x^2=4$ correctly you proceed as follows.

• Since $x^2 \ge 0$ and $4 \ge 0$, then you can apply the square root function to them.

• $\sqrt{x^2} = |x|$ and $\sqrt 4 = 2.$

• So you get |x| = 2$

• So $x \in \{2, -2\}$.

Now as to $x=x^{2/2}=\sqrt {x^2}=|x|$

The expression $x^\frac 22$ can be interpreted three ways.

1. $x^{(\frac 22)} = x^1 =x$. (True for all x.)
2. $\left(x^2 \right)^\frac 12 = \sqrt{x^2} = |x|$. (True for all x.)
3. $\left(x^\frac 12 \right)^2 = \left(\sqrt{x}\right)^2 = x$. (True for all $x \ge 0$.)

If all three interpretations must work, the rules for $x^\alpha$ can only be applied when $x \ge 0$.

There are exceptions! For example, $x=x^{3/3}=\sqrt[3] {x^3}=x$ is true for all real numbers, $x$. It is useful to note these exceptions and exploit them when appropriate. It is even more useful to be aware of what to you can and can't do with $x^2$ when $x$ is a negative real number.