Jtdylktuy

I am looking for an example where you have $P(n)$ implying $P(n+1)$. However there is no base case. For which there is therefore no solution at all for the induction problem even though the inductive step itself works.

Suppose we want to prove $n=n+1$ for all (positive) integers $n$. We omit the base case. The induction hypothesis is $k=k+1$ for some $kin mathbb N$. Adding $1$ to both sides gives $k+1=k+1+1$, or $(k+1)=(k+1)+1$, which is the statement to be proven for $n=k+1$. Thus, we have completed the induction step, but there is no base for which this is true, so the statement won't have to be true too.

$$1+2+3+dots+n=frac{n(n+1)}2+pi$$
$$1+2+4+dots+2^{n-1}=2^n$$

(False) Theorem. Let $P$ be literally any property of elements of $mathbb N$. Then every element of $mathbb N$ has property $P$.
Proof. This is true if all nonempty finite subsets of $mathbb N$ consist of elements with property $P$. Suppose $nge 1$ and all elements of subsets of $mathbb N$ with $n$ elements satisfy $P$. Let $Ssubsetmathbb N$ have $n+1$ elements, and pick any $xin S$. Since $S'=Ssetminus{x}$ has $n$ elements, all members of $S'$ satisfy $P$. Moreover, since $n=|S'|>0$ there exists $yin S'$. Then $(S'setminus{y})cup{x}$ has $n$ elements, so $x$ has property $P$ as well. q.e.d.

Let $P(n)$ be the statement "$1=2$". Assume $P(k)$ is true, hence $1=2$. By assumption, $1=2$, so $P(k+1)$ is true.

Then what?

Simply let $P(n)$ be the statement that $n=n+1$. Suppose this is true for some positive integer $k$. Then $k=k+1$, so $k+1=k+1+1$, and we also have $P(k+1)$.

Let $P(n)$ be a false statement, for each $n$. Then $P(k) implies P(k+1)$ vacuously, because a false statement implies any other, but certainly $P(0)$ is false.

1. Suppose we want to show $n=n+1$ where $nin mathbb{N}$. Now suppose the statement holds for some $n$
   $$n=n+1$$
   then we want to show $$n+1=n+1+1$$
   which is easily true since $(n+1)+1=n+1$.




2. Suppose we want to show for all $nin mathbb{N}^+$, $n+1<n$. Now suppose the statement holds for some $n$
   $$n+1<n$$
   then $$n+1+1=(n+1)+1<n+1$$