By Definition a positive integer $N$ is d-deficient if $\sigma(N)=2N-d$.
Am I correct if I say that the inequality $N>d$ always hold for this definition?
Here is my attempt to show that it is true. First note that for any $N$, $\sigma(N)>N$.
If $N=d$ we have $\sigma(N)=N$ a contradiction. If $d>N$ we have $\sigma(N)=2N-(N+k)$ for some positive integer k, and thus $\sigma(N)=-k$ a contradiction.
Am I correct Sir/Mam?
thank you in advance.
If by $\sigma(N)$ you mean the sum of the divisors of $N$, then yes, if $N$ is $d$-deficient with
$$\sigma(N) = 2N - d,$$
then since $\sigma(N) \geq N$ (where equality holds if and only if $N = 1$), then you have
$$N \leq \sigma(N) = 2N - d \Longrightarrow d \leq N.$$
Again, equality holds if and only if $N = 1$.