I am doing the exercise from my textbook teaching the induction and stuck for a long time about the induction hypothesis part.
The question is the following:
Let n∈N∖{0}, use some form of induction to prove that for all such n there exists an odd natural m and a natural k such that n=2km.
From my lecture my prof told us that we have to use P(n) to prove P(k+1) and can't use P(k+1) as a precedent, I understand this part, and the following is my approach.
Define predicate P(n)= there exists an odd natural m and a natural k such that n=2km for all such n.
Check P(1), if I choose k=0 and m=1 this holds.
Assume P(h) is True, namely, I can find such m and k so that h=2km.
Prove p(h+1), I know I have to show two cases since P(h+1), so I assume P(h)+1 is odd first, then I can write h+1=2km+1, and get h=2km, and by I.H. this is true. (Although I don't think I get this correctly it just looks weird). Then I assume h+1 is even, this implies that h must be odd, so induction hypothesis must be changed to h=2km+1 in this case, so h+1=2km+2, then get h+2km+1, by I.H this is true.
I don't know what goes wrong in my proof, but when I just stare at it I just feel it is not correct, I need help to explain this question, thanks.
No comments:
Post a Comment