Is there an algebraic method to concat two numbers? I'm searching an algebraic way to concat numbers in base $10$.
Concatening two numbers is to put side by side their notations.
Let $c$ a concatenating function.
$c(2,2) = 22$
$c(8,9) = 89$
$c(11,0) = 110$
$c(14856,183) = 14856183$
I know we can easily do that using programmation, but it would be so elegant to do it with Maths. :) 
I think that if $n(x)$ return the number of digits in a number $x$,
$c(a,b) = a * 10 ^{n(b)} + b $ 
But I don't know how to do it. Perhaps you can help ?
 A: Thanks to @egreg , @peterwhy and @xaviermos2 I can now say that 
$c$ is concatening function for $a$ and $b$ natural numbers ! 
$c(a,b) = a*10^{⌈log10(b+1)⌉}+b$
A: Assuming that all of the numbers are non-negative integers in base $10$, then the problem peterwhy describes can be fixed. We just need a special function to detect if $b$ is $0$.
A general trick to detect strict inequalities like $a>b$ versus $b>a$ is to involve the absolute value of the difference. In our case, $b>1/2$ if $b\ne0$, and $b<1/2$ if $b=0$. Therefore, we should consider the quantity $\left|b-\frac12\right|$. It equals $b-\frac12$ unless $b=0$. To make it easier to detect what's going on and avoid fractions, consider $\left|b-\frac12\right|-\left(b-\frac12\right)$. That's $0$ unless $b=0$ in which case it's $\left|-\frac12\right|-\left(-\frac12\right)=1$. This is perfect for us, because we need to add $1$ to the power of $10$ exactly when $b=0$.
Therefore, our final formula is $$c(a,b)=b+a*10^{\left\lfloor\log_{10}\left(b+1\right)\right\rfloor+\left|b+1/2\right|-\left(b-1/2\right)}$$
(Note: If you want something that looks weirder, you could also use $\left\lceil \left|b-\frac12\right|\right\rceil-b$ in place of the $\left|b-\frac12\right|-\left(b-\frac12\right)$ part.)
