Whats an example of a predicate where if $P(1)$ is False, and $(\forall k\geq 1)(P(k)\implies P(k+1))$ is True for all natural numbers

Question

The textbook solution just says that there is a predicate which satisfies those conditions. Wouldn't this be false since if the base case doesn't hold then by induction principle its false? Is there an example where $P(1)$ is False, and $(\forall k\geq 1)(P(k)\implies P(k+1))$ is True?

Answer 1

Try $$k>42$$ or try $$ k^3-k+1 \text{ is a multiple of }3$$ or try $$ \text{There exists an even $m$ with $2<m<k$ that is not the sum of two primes}.$$

Answer 2

You could try $n\gt n$ or $n=n+1$ which are false for all $n$.

The point is that you need the base case to found the induction. Otherwise you can prove anything (ie all the consequences of a false statement)