After reading a question made here, I wanted to ask "Why do we do mathematical induction only for positive whole numbers?"
I know we usually start our mathematical induction by proving it works for $0,1$ because it is usually easiest, but why do we only prove it works for $k+1$?
Why not prove it works for $k+\frac12$, assuming it works for $k=0$.
Applying some limits into this, why don't we prove that it works for $\lim_{n\to0}k+n$?
I want to do this because I realized that mathematical induction will only prove it works for $x=0,1,2,3,\dots$. assuming we start at $x=0$, meaning it is unknown if it will work for $0<x<1$ for all $x$.
And why not do $k-1$? This way we can prove it for negative numbers as well, right?
What's so special about our positive whole numbers when doing mathematical induction?
And then this will only work for real numbers because we definitely can't do it for complex numbers, right?
And what about mapping values so that it becomes one of the above? That is, change it so that we have $x\to\frac x2$? Then proving for $x+1$ becomes a proof for all $x$ that is a multiple of $\frac12$!?
Best Answer
$ \def \t {\quad \rightarrow \quad} $ Let's forget about natural numbers for a while and take a look at what mathematical induction is proving. At the base step, it's proven that a property, say $ P $, holds for some special object, say $ a $. At the induction step, it's proven that if $P$ holds for some fixed object, say $ x $, then it holds for some other object related to $ x $ in some way, which can be translated mathematically to: for some function $ S $, $ P $ holds for $ S ( x ) $. Now, since we know that $ P $ holds for $ a $, we conclude that $ P $ holds for $ S ( a ) $, and then for $ S \big( S ( a ) \big) $, and so on: $$ a \t S ( a ) \t S \big( S ( a ) \big) \t S \Big( S \big( S ( a ) \big) \Big) \t \dots $$ this leads to a sequence of objects defined like this: $$ a _ 0 = a $$ $$ a _ { n + 1 } = S ( a _ n ) $$ $$ a _ 0 \t a _ 1 \t a _ 2 \t a _ 3 \t \dots $$ and $ P $ holds for every element of this sequence. If we look at the indices, we find natural numbers. So natural numbers appear naturally in inductive proofs! It may be useful to note that having only the basis and the inductive step, the objects mentioned above are the only objects that we have proven to have the property $ P $.
Now, the standard starting object is $ 0 $ and the standard successor function is $ S ( x ) = x + 1 $. But putting any other objects in a sequence, you can use induction for giving a proof. Note that even in that case, you can define a sequence like we did before, and use the standard induction! So there is no real difference between them. For example, letting $ a = 0 $ and $ S ( x ) = x - 1 $, you can prove a property for every nonpositive integer. In this case we have $ a _ n = - n $. So you can use mathematical induction twice to prove a property for all integers; once with $ a = 0 $ and $ S ( x ) = x + 1 $, and once with $ a = 0 $ and $ S ( x ) = x - 1 $.
Finally, it may be useful to note that not every set of objects can be represented as a sequence. For example, real numbers can't be listed in a sequence and so you can't use this kind of induction for them.