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

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

$x^k; k\in \mathbb N$ 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.


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