Here are a couple of examples, the first with almost full detail, the second with less.
First I’ll convert $156_{16}$ to base ten using repeated division in base sixteen. I’ll use $A,B,C,D,E$, and $F$ for the base sixteen digits corresponding to base ten $10,11,12,13,14$, and $15$. I’ll also use a subscript $s=16$ to indicate that a number is to be interpreted in base sixteen.
Divide $156_s$ by $A_s$. Do this just as you would in base ten: $A_s$ won’t go into $1_s$, but it will go into $15_s$. In fact $15_s=1\cdot 16+5=21$, and $A_s=10$, so it goes twice. The first digit of your quotient is $2_s$, so you need to subtract $2_s\cdot A_s$ from $15_s$.
$2_s\cdot A_s=2\cdot 10=20=1\cdot16+4=14_s$, and $15_s-14_s=1_s$, so after you bring down the $6_s$, you’re left dividing $A_s$ into $16_s$.
Similarly, $16_s=1\cdot 16+6=22$, so $A_s$ goes in twice. After you repeat the previous step (with suitable minor modifications) you have your full quotient $22_s$ and overall remainder $2_s$, as shown below.
$$\begin{array}{}
&&&2&2\\
&&\text{_}&\text{_}&\text{_}\\
A&)&1&5&6\\
&&1&4\\
&&-&-&-\\
&&&1&6\\
&&&1&4\\
&&&-&-\\
&&&&\color{red}2
\end{array}$$
Now divide $22_s$ by $A_s$. $22_s=2\cdot 16+2=34$, so the integer part of the quotient is $3_s$:
$$\begin{array}{}
&&&3\\
&&\text{_}&\text{_}\\
A&)&2&2\\
&&1&E\\
&&-&-\\
&&&\color{red}4\\
\end{array}$$
Finally, divide this last quotient, $3_s$, by $A_s$:
$$\begin{array}{}
&&0\\
&&\text{_}\\
A&)&3\\
&&0\\
&&-\\
&&\color{red}3\\
\end{array}$$
Read off the red remainders in reverse order: $156_s=342$.
Here’s one a little more complicated, the conversion of $2BA_s$ to base three.
$$\begin{array}{ccccc|cccc|cccc|cccc|ccc}
&&&E&8&&&4&D&&&1&9&&&&8&&&\color{red}2\\
&&\text{_}&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}\\
3&)&2&B&A&3&)&E&8&3&)&4&D&3&)&1&9&3&)&8\\
&&2&A&&&&C&&&&3&&&&1&8&&&6\\
&&-&-&-&&&-&-&&&-&-&&&-&-&&&-\\
&&&1&A&&&2&8&&&1&D&&&&\color{red}1&&&\color{red}2\\
&&&1&8&&&2&7&&&1&B\\
&&&&-&&&-&-&&&-&-\\
&&&&\color{red}2&&&&\color{red}1&&&&\color{red}2
\end{array}$$
That last quotient of $2$ is less than the divisor, so the next division will have a $0$ quotient and remainder of $\color{red}2$, so I’ve skipped the step and colored the quotient instead. Reading the remainders in reverse order, we have $2BA_s=221212_t$ (where the subscript $t$ indicates base three).
Check: $$2BA_s=2\cdot 256+11\cdot16+10=698\;,$$ and $$221212_t=2\cdot 243+2\cdot81+1\cdot27+2\cdot9+1\cdot3+2=698\;.$$
Best Answer
I don't think you need to generate much data to realize the problem with fractional number bases. Consider base $2.5$, illustrated here. Just by looking at the top diagram we can see that '$0.22$'$ > 1$. The number line is useful to us because it allows us to measure quantities - we divide it into units and then repeatedly subdivide each of them. Whereas the first division of unity in base $2.5$ yields $0.4$, the second yields $0.16$. Two of these latter subdivisions are greater than the 'resolution' of $2.5$ (i.e. half in base $2.5$, or $0.2$ in base $10$) so it doesn't necessarily help us attain finer resolution. Why not just use the resolution of $2.5$ as a subunit? This gives us base $5$, and as can be seen from the lower diagram multiples of second subdivisions will not 'spill over' to confuse our understanding of what 'number one' actually means (because by definition there can't be enough of them). See here for the formal proofs of these concepts.
EDIT1 This answer follows on from this reddit discussion. User Brightlinger claims that the above problem leads to non-unique representations for numbers in fractional bases and gives the example of $4$ in base $π$ (this is 'fractional' as in 'not an integer').
EDIT2 The following Python3 script allows a number to be represented in a chosen base. Both the number and the base can have decimal parts:
Example session:
chrx@chrx:[tilde]/Desktop$ python3 fracbase.py 1001 10.4
The [tilde] means the tilde character.