• s3a

(Proof by Induction Problem - Solution included but I still am confused) Question (with solution): http://f.imgtmp.com/OPRWj.jpg What confuses me is that there are three letters b, k and n. Normally, these problems only have an n and a k. The n being the variable and the k being a constant plugged into n and then I attempt to show that it works for k+1 but here the b throws me off completely. If I pretend k is a variable and then make b the constant number, then that makes more sense but why is the definition of T(n) being used? If you aren't sure what I'm asking, tell me and I'll elabor

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  • s3a

(Proof by Induction Problem - Solution included but I still am confused) Question (with solution): http://f.imgtmp.com/OPRWj.jpg What confuses me is that there are three letters b, k and n. Normally, these problems only have an n and a k. The n being the variable and the k being a constant plugged into n and then I attempt to show that it works for k+1 but here the b throws me off completely. If I pretend k is a variable and then make b the constant number, then that makes more sense but why is the definition of T(n) being used? If you aren't sure what I'm asking, tell me and I'll elabor

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  • s3a
elaborate.
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

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  • s3a
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.
  • s3a
than*
  • s3a
ok so k is an arbitrary constant and we choose it to have value b?
  • s3a
that seems redundant
proofs do seem rather redundant yes; but its all for a good cause :)
  • s3a
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
  • s3a
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
  • s3a
(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
  • s3a
sorry my computer lagged
  • s3a
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 :)
  • s3a
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?
  • s3a
Sorry if I am being redundant lol :P
  • s3a
Actually, wait, I think I am starting to get it.

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