Does $X^{2/2} = X^{1/1}$?

Question

I'm having a bit of a hard time wrapping my head around how the following that I have just learned:

$\sqrt{X^2} = |X|$, and I totally understand why.

But, when expressed as an exponent, doesn't this really just mean the following:

$X^{2/2} = |X|$, if this is the case, and I simplify the rational exponent, I would get:

$X^{1/1}$ or $X^1$, which does not equal $|X|$.

Also, if I apply the following rule of a radical function:

$\sqrt[n]{P^Q} = (\sqrt[n]P)^Q$ where n is the index of the root and $Q$ is the power of the radicand, then this should mean that:

$\sqrt{X^2} = (\sqrt{X})^2$, but the $(\sqrt{X})^2$ does not equal $|X|$ and has a domain where $X > 0$, while the $\sqrt{X^2}$ has a domain equal to all real values for $X$.

Does this mean that when $X$ is raised to an even-numbered power and is the radicand in a radical expression, that one should not simplify the rational exponent or one should not rewrite the radical expression such that the power of the radicand $X$ now lies outside of the root function?

Any replies will be greatly appreciated.

Answer 1

This is due to the fact that there is a slight difference between $\sqrt{x}$ and $x^{1/2}$
I recommend looking up the term "Principal Root", with a basic introduction here.
In essence, for positive numbers there are always two answers to $x^{1/2}$, namely $\pm \sqrt{x}$.
From that, note that the $\sqrt{x}$ function always gives the positive root for a positive argument, and is thus a true function... it outputs one root for each argument $x$. However, the function $x^{1/2}$ outputs two values for each argument $x$, and is thus NOT a true function. Nuances like these are what is messing with your argument!

Answer 2

The exponent rules you are referring ( $(a^{b})^c=a^{bc}$ ) work well when the base $a$ is positive, that is, $a>0$. In general, handling exponents of negative numbers should be done with care, precisely because of domain issues of the even-numbered roots.

Answer 3

Any exponent function are defined from $\mathbb{R}^+$ to $\mathbb{R}^+$, because the definition of $f(x)=x^{1/p}$ is the inverse (in algebraic sense) of $x^p$ so if $p$ is even we know that $x^p$ is not injective in $\mathbb{R}$ but is bijective in $\mathbb{R}^+$, for general $p$ ($p$ odd) we can define easily $x^p$ in $\mathbb{R}$ as one to one function but the calculus can't be done easily in fact : $$ (-1)=(-1)^1=(-1)^{2/2} $$ but to say that $(x)^{ab}=(x^a)^b=(x^b)^a$ we need to see (as function ) if this is well defined, that mean if $x^a$ is well defined and $x^b$ is too well defined!! so in this case we can't say that $$ (-1)^{2/2}=((-1)^{1/2})^2 $$ because the second term have no sense.

Answer 4

More slowly, $\sqrt{x^2} = (x^2)^{1/2}$. But the standard identity $$(x^2)^{1/2} = x^{2 \times 1/2}$$ holds for $x\ge 0$.

So Case 1: if $x\ge 0$, then $(x^2)^{1/2} = x^{2 \times 1/2} = x = |x|$.

But if Case 2: $x < 0$, then setting $y := - x$, so that $y > 0$, then $(x^2)^{1/2} = ((-y) ^2)^{1/2} = (y^2)^{1/2}$. Since $y > 0$, we can then use the displayed identity with $y$ instead of $x$ to get $(y^2)^{1/2} = y^{2 \times 1/2} = y = |x|$. Therefore $(x^2)^{1/2} = |x|$ in this case also.

So one point to make is that a general form of the displayed identity, namely, for $n, m > 0$, then $(x^n) ^m = x^{nm}$ holds for $x \ge 0$.

Answer 5

If we are only talking about real numbers, $a^{m/n}=\sqrt[n]{a^m}$ is only for those $a\ge0$. In this case, $|a|=a$. We don't apply this to $a<0$.

If complex numbers are allowed, which introduce the imaginary unit $\mathrm i ^2=-1$, for $a<0$, we rewrite $a^{m/n}$ to $(a^{1/n})^m$ (but not $(a^m)^{1/n}$). The result of $a^{1/n}$ is a complex number. Note that you cannot simply swap the positions of the powers in a context with complex numbers. In this definition, $(-1)^{1/2}$ evaluates $\mathrm i$. Then, $(-1)^{2/2}=((-1)^{1/2})^2=\mathrm i ^2=-1$.

The fractional power of complex numbers is a little complicated. You are likely to learn the details in the future study (considering the tag precalculus).