I am currently self-studying number theory in high school, and came across this question:
Assuming that for every integer $n>1$, there exists a prime number between $n$ and $2n$, prove that every positive integer can be written as a sum of distinct primes or $1$ more than a sum of distinct primes.
My attempt:
We use induction on $\Bbb N$.
Clearly, $1$ can be written simply as $1$, thus the conjecture is true for $1$. We assume that the conjecture is true for all numbers up to some $k\in \Bbb N$.
Now, we take $k+1$ and find out $\Bigl \lfloor \frac{k+1}{2} \Bigl \rfloor.$ By Bertrand's Postulate, there must exist primes between $\Bigl \lfloor \frac{k+1}{2} \Bigl \rfloor$ and $2\Bigl \lfloor \frac{k+1}{2} \Bigl \rfloor$, where $2\Bigl \lfloor \frac{k+1}{2} \Bigl \rfloor \leq k+1.$ By the well-ordering principle, the set of primes has a least element, say $p$. We find out $(k+1)-p$.
Now by our induction hypothesis, the number $(k+1)-p$ can be written as the sum of distinct primes, none of them $p$ since $p>(k+1)-p$. Let them be $p_1, p_2... p_n$.
Thus we can write
$$k+1=p+p_1+p_2+...+p_n$$
which concludes our proof.
Is the proof correct? Is there a better way to do it? I have a distinct feeling that it is possible to do this through a proof by construction, in which we find the primes through a similar argument.
Edit: My question is not a dupicate because the other question's top answer is not clear enough and only provides a hint. The other answers are not satisfactory.
Further, this is a proof verification question, and my final statement 'can this be proved by a proof by construction' is not addressed in the other question.