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s3a
 4 years ago
(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
s3a
 4 years ago
(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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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0in this setup they are just changing n into k for an arbitrary integer and then assigning k=0 to establish a basis step

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0then 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

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok so k is an arbitrary constant and we choose it to have value b?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0proofs do seem rather redundant yes; but its all for a good cause :)

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0why not just say n (the varaible) is equal to any number denoted as k

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0pascal 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0the 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
 4 years ago
Best ResponseYou've already chosen the best response.0lol well, does that mean my way is wrong?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0wrong? no, just different way to "look" at it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0k = 0, b = any int greater than 0 so it satisifies b=k+1

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0(just plugging in some k for n and showing the k+1 stuff)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not 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
 4 years ago
Best ResponseYou've already chosen the best response.0I never did it as k < n but rather n = k. and then assuming that's true and showing n=k+1

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0best bet, do it your way and see if its proofified :)

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0lol "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
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry if I am being redundant lol :P

s3a
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, wait, I think I am starting to get it.
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