Just to be clear: you think that there are many irrational numbers that exist independently of which number system you use, but that there are infinitely many that do depend on the number system you use? Is that right?
Irrational numbers, by definition, include decimal numbers that have infinitely many digits after the decimal point, and there are no rules about what those digits have to be. This is powerful enough to represent any irrational number regardless of the number base. However, if you're talking about number systems, not all number systems have equal ability to represent irrational numbers. Whatever the system, to represent all the irrationals, it would have to be capable of going on forever. I think the part of my point that you're asking about is my statement that they're a side effect of decimal numbers. Irrational numbers like sqrt(2) and e have concise representations as the limits of Taylor series. Only when converted into decimals do they appear to have infinite amounts of information. So if we used Taylor series as the number system instead of decimals, simple irrationals would look simple, and we might be less inclined to treat them the same as numbers that have no finite descriptions at all.
I was under the impression that irrational numbers are numbers that simply can't be represented as a ratio. This is independent of the number system used. I'm not familiar with the Taylor series: can you use it as a number system? Isn't it simply a representation of functions (in which case, you might as well just write "sqrt(2)" as invoke anything else)?
It's tough to Google for "Taylor series number system" as you just get pages about the guitars. ;-)
I don't know if you're still reading this, but... Using Taylor series as a number system is just a hypothetical. Any systematic way of describing numbers could be a number system. Irrationals can't be represented as ratios, but beyond that, some have other finite representations, and others have none.