Is every integer the sum of distinct prime numbers? Let $\Bbb{P}$ be the set of prime numbers and $Q=\Bbb{P}\cup\{1\}$. Is it true that every natural numbers ($\neq 0$) is the sum of distinct elements of $Q$? I tried from $1$ to $60$ and it seems true.
 A: By [Bertrand's postulate][1], you can find a prime satisfying $\lfloor n/2\rfloor <p< n$. Proceed by induction.

Update with fuller proof
Prove it for $n\leq 6$ or so.
If true for all $n\leq m$ with $m\geq 6,$ then by Betrand's postulate, there is a prime $$\left\lceil \frac{m+1}2\right\rceil <p<2\left\lceil \frac{m+1}2\right\rceil.$$ But $2\left\lceil \frac{m+1}2\right\rceil\leq m+2,$ so $p\leq m+1.$
If $p=m+1,$ then we are done, since $m+1$ is the sum of one prime.
If $p<m+1,$ then $0<m+1-p<\frac{m+1}{2}<p,$ so $m+1-p$ can be written as a sum of distinct elements of $\mathbb P\cup\{1\},$ and since $m+1-p<p,$ those distinct elements do not include $p,$ so $m+1$ is the sum of distinct elements of $\mathbb P\cup\{1\}.$

You can probably show that for $n$ large, all $n$ can be expressed as a sum of distinct primes, not including $1.$ You'll need something slightly stronger than Bertrand. It might be the case that the only numbers requiring $1$ are $4$ and $6.$
If so, you'll need to find an $N_0$ that for $N\geq N_0$ is a prime between $N$ and $2N-6.$ Then you need to start by showing it is true for $6<n\leq N_0,$ and then you get the same induction. Looks like $N_0=9$ might work, but proving the variant of Bertrand might be tricky.
A: Note that we can not write $2$ in required form.  

Take any integer (say) $x$ with $3\le x.$
  Let $y$ be the largest prime strictly less than $x.$ If $x-y\in Q$ we are already done.
  Suppose $x-y\not\in Q.$ Then, take the largest prime strictly less than $x-y.$
  And continue this..

