6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Please TeX it up. $\endgroup$ – user305860 Jul 1 '16 at 17:39
  • $\begingroup$ For a lazy person like me it is hard to follow, please tex it up! $\endgroup$ – Babak Miraftab Jul 1 '16 at 17:45
3
$\begingroup$

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 "Principle 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!

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

any exponent function are defined from $\mathbb{R}^+$ to $\mathbb{R}^+$, because as definition $f(x)=x^{1/p}$ 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 $(x)^{ab}=x^a x^b$ 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 $$ (-1)^{2/2}=(-1)^2(-1)^{1/2} $$ because the second term have no sense

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.