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

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

• 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. – Ethan Bolker Apr 10 at 18:09
• 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. – Leonard Blackburn Apr 10 at 18:14
• The reason you can't in general is that addition and multiplication will always yield a rational number, but radicals can be irrational. – David G. Stork Apr 10 at 18:14
• $x^{1/2} = t$ such that $t \cdot t= x$. – Robert Israel Apr 10 at 18:19
• 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. – 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.