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This is the kind of thing you learn in the first lecture of a discrete mathematics class that's required for every single CS program in the world. It's very ob
by gnulinux 3y ago
This is the kind of thing you learn in the first lecture of a discrete mathematics class that's required for every single CS program in the world.
It's very obvious if you think about it. If the encoding on the memory is n that doesn't mean it represents "n" data conceptually. It can be n+x for any value of x. For example, you can declare 0 represents $0, 1 represents $0.01, 2 represents $0.02 and so n represents n cents. This is pretty common in tax calculation applications. Then, you can also say n represents n+99^99 so that what GP said holds. They just provided a formal argument for it which proves it by induction, as we call it.
- dj_mc_merlin 3y ago> They just provided a formal argument for it which proves it by induction, as we call it. This is contradiction, I think. Assume n (exists and) is the largest number representable in 64 bits is statement 1. The numbers in the n+x system are larger than n is statement 2. That causes a contradiction and since statement 2 is true, statement 1 must be false. The addition in n+x is not induction, just a way to define a system.