Thursday, August 27, 2015

algebra precalculus - On Assumptions In Induction

I'm currently learning mathematical induction from this site: https://www.mathsisfun.com/algebra/mathematical-induction.html




What I'm confused about is how it presents mathematical induction. It says that there are 3 steps to induction:




  • Show it true for $n=1$.

  • Assume it is true for $n=k$

  • Prove it is true for $n=k+1$ (we can use the $n=k$ case as a fact.)



There are many things I am confused about here, about all $3$ steps.




Starting from the first step, why is it necessary to prove it true for $n=1$? I don't get why this step is needed. Second, why choose $1$ of all numbers; can't a number like $2$ be chosen?



Moving on to the second step, why is it legitimate to assume it true for all $n=k$? Is this assumption proved true by the third step, if so, how?



On the final step, first, how can we prove it true for all $n=k+1$ ? Because this prove fundamentally assumes that it is true for $n=k$, but there is no way to verify this. Second, what happens if the set we're doing induction on has only a limited amount of numbers, let's say 100 numbers. So if we go up to the 100th number, then how can $n=k+1$ still be true? Because there is no 101st term for it to be true on; there are only 100 numbers!



Please explain this as simply as possible; I'm still a beginner. I will not be able to understand complicated proof notation.



DETAILS: This question is different from the question above since the question above uses $\lim_{x\to\infty}$ notation and other pieces of calculus knowledge. However, I have not learnt calculus, and nothing about this question suggest prior calculus knowledge. An answer where induction is explained without calculus would benefit me greatly.

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...