Anyone know how to prove the Strong induction?

Question

Anyone know how to prove the Strong induction?? Because I will need it and I don't know how to do it... please.. thanks ! Can someone please help me.... ~!

Answer 1

I know that you can prove the strong induction is equivalent to "weak" induction (if that is what you meant). The proofs are in a lot of different math books (try number theory, algebra, discrete math, etc).

Sometimes mathematical induction is taken as an axiom of the real numbers but it can be proved (see link below under: Proof or reformulation of mathematical induction).

Answer 2

What we elect to tutor is that for any n>=18, there exist constructive integers a and b such that 4a + 7b = n. initiate the induction with the case even as n=18: 18 = 4*a million + 7*2 consequently (a,b) = (a million,2) fulfill. even as n=ok, assume the actuality is authentic that's 4a + 7b = ok for some (a,b). even as n=ok+a million we elect to exhibit that there exist (a',b') such that 4a' + 7b' = ok+a million. ok + a million = 4a + 7b + a million utilising the induction hypothesis. = 4a + 7*(b-a million) + 8 = 4*(a + 2) + 7*(b - a million) consequently (a+2,b-a million) is a conceivable mix. that's sufficient iff (b-a million) is constructive. contained in the case the position b=0, that's 4a = ok we elect to construct a diverse mix: ok + a million = 4a + a million = 4*(a - 5) + 7*3 consequently for this reason, (a',b') = (a-5,3) is valid. when you consider that (a-5) is constructive for all n>=18 with b=0, the authentic result follows through induction.

Answer 3

I think you are referring a type of proof in which you do the following:

(1) Verify the statement is correct for the first case.

(2) Assume it is true for the nth case, and use this to prove it is true for the (n+1) case.

Example: to prove the sum of the first n integers is n(n+1)/2:

(1) It works in the first case, i.e. n=1: 1(1+1)/2 = 1.

(2) Assuming it works for the nth case, then 1+2+...+n=n(n+1)/2.
Then, for the (n+1) case, 1+2+....+n+(n+1) = n(n+1)/2 + (n+1) = (n^2 +3n +2)/2 = (n+1)(n+2)/2. That proves the formula works.