0
$\begingroup$

How can you express radicals as multiplication/addition?

Most mathematical operations clearly reduce to multiplication/addition (same thing), but how do you do that for exponentials/radicals? Thank you.

Example: $x^{1/2}$ = ?

$\endgroup$
  • $\begingroup$ You can't in general. If you have a particular expression you want help with please edit the question to include it, and tell us where it comes from and what you have tried. $\endgroup$ – Ethan Bolker Apr 10 at 18:09
  • 1
    $\begingroup$ This might be tangentially related: Suppose you want to relate 16^(3/4) to multiplication. Write 16 = (2)(2)(2)(2) and then 16^(3/4) equals the product of three-fourths of those factors: 16^(3/4) = (2)(2)(2) = 8. $\endgroup$ – Leonard Blackburn Apr 10 at 18:14
  • 1
    $\begingroup$ The reason you can't in general is that addition and multiplication will always yield a rational number, but radicals can be irrational. $\endgroup$ – David G. Stork Apr 10 at 18:14
  • $\begingroup$ $x^{1/2} = t$ such that $t \cdot t= x$. $\endgroup$ – Robert Israel Apr 10 at 18:19
  • $\begingroup$ In high school and college in the late '60s (before calculators) we used to have to find square roots with paper and pencil. Here is one link to a square root algoritm and there are many more such as this one for cube root etc. $\endgroup$ – poetasis Apr 10 at 18:22
1
$\begingroup$

$x^k; k\in \mathbb N$ is $\underbrace{x\cdot x\cdot .... \cdot x}_{k\text{ times}}$

With $x^0 = 1$ and $x^{-k} = \frac 1{x^k}$.

$x^{\frac 1k}$ the real number $y$ (if any... it is assume $x > 0$ and $y > 0$) so that $\underbrace{y\cdot y \cdot...\cdot y}_{k \text{ times}}= y^k = x$.

And $x^{\frac mk} = (x^{\frac 1k})^m$ or in other words if $y$ is the $y$ so that $\underbrace{y\cdot y \cdot...\cdot y}_{k \text{ times}}= x$ then $x^{\frac mk} = \underbrace{y\cdot y\cdot ...\cdot y}_{m \text{ times}}$.

Need to keep in mind: 1) The doesn't define $x^{v}$ where $v$ is not rational. 2) if $x < 0$ then ... this doesn't always work. 3) There are might be more than one $y$ so that $y^k =x$. $x^{\frac 1k}$ refers specifically to the positive one, and 4) This makes a lot of assumptions that will need to be proven such as i) that such a $y$ so that $y\cdot ..... y=x$ actually exists; that there is exactly $1$ such thing and not many; ii) that if $\frac mk = \frac jl$ that $(x^{\frac 1k})^m = (x^{\frac 1l})^j$ etc.

$\endgroup$

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.