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in this setup they are just changing n into k for an arbitrary integer and then assigning k=0 to establish a basis step
then they say since k=0 works then choose some other k such that k=b ... or the k+1 part that we all know and love
Ok, that's different then what I usually do which is just plugin in 0 or whatever the first number is into the n but I get the first post you made.
ok so k is an arbitrary constant and we choose it to have value b?
that seems redundant
proofs do seem rather redundant yes; but its all for a good cause :)
why not just say n (the varaible) is equal to any number denoted as k
pascal was keen to notice that people liked to use words to mean different things and decided that if you could not have a rigoroous way of defining/proving your concepts that your concepts had no value
the way in which people go about making their proofs rigorous can be confusing, but as long as the logic flows from start to finish we consider it valid lol
lol well, does that mean my way is wrong?
wrong? no, just different way to "look" at it
k = 0, b = any int greater than 0 so it satisifies b=k+1
(just plugging in some k for n and showing the k+1 stuff)
not to butt in (ok to butt in) it is not clear why they used "b" in this case. you usually say "for all k < n" then prove it for n by reducting to the case n - 1
sorry my computer lagged
I never did it as k < n but rather n = k. and then assuming that's true and showing n=k+1
best bet, do it your way and see if its proofified :)
lol "proofified". my way is the same mechanically but the assumptions, etc differ. So, basically what this solution is doing is saying "the variable is n, we choose any number k and show that the equation T(2^k) = 3^(k+1) - 2^(k+1) will hold so we then we say that the number k is equal to the number b (which is useless and redundant) and then we prove that n=k+1=b+1 works and then all is good?
Sorry if I am being redundant lol :P
Actually, wait, I think I am starting to get it.