(10^100) + 1 − (10^100) is 1, not 0.

A "computer algebra system" would have accomplished a similar goal, but been much slower and much more complicated

$ maxima -q

(%i1) (10^100)+1-(10^100);

(%o1)                                  1
(%i2) 

There's no perceptible delay on my laptop here, and I use maxima on my phone and my computers. And a CAS gives you a lot more power to do other things.