# Number operator and, annihilation and creation operators

While reading Ryder chapter on quantization of Klein-Gordon field I got stuck at the following:

1. It can be shown that, $$[a(k),a^{\dagger}(k')]=(2\pi)^32\omega_k \delta^3(\mathbf{k}-\mathbf{k'})$$ and that $[a(k),a(k')]=0=[a^{\dagger}(k),a^{\dagger}(k')]$, where $\omega_k=(\mathbf{k}^2+m^2)^{1/2}$.
2. Then consider the operator, $$(2\pi)^32\omega_k \delta^3(0)N(k)=a^{\dagger}(k)a(k),$$ then one can show, $$\hspace{-3cm}[N(k),N(k')]=\{a^{\dagger}(k)[a(k),a^{\dagger}(k')]a(k')$$ $$\hspace{5cm}+ a^{\dagger}(k')[a^{\dagger}(k),a(k')]a(k)\}/((2\pi)^64\omega_k \omega_k'[ \delta^3(0)]^2)$$
3. I can show the above equation but I am stuck at this next step. It goes on to say that, $$\hspace{-3cm}[N(k),N(k')]=\{a^{\dagger}(k)[a(k),a^{\dagger}(k')]a(k')$$ $$\hspace{5cm}+ a^{\dagger}(k')[a^{\dagger}(k),a(k')]a(k)\}/((2\pi)^64\omega_k \omega_k'[ \delta^3(0)]^2)$$ $$\hspace{1.5cm}\sim [a^{\dagger}(k)a(k)-a^{\dagger}(k)a(k)]\delta(\mathbf{k-k'})$$ $$\hspace{-4.5cm}=0$$ the $\sim$ means '"transforms as" in the notation of this text, but I don't know "transforms as" under what? Also not sure what division by $\delta^3$ means.
4. My attempt assumed the following (which may be incorrect), (i) first I assume that $\delta^3(\mathbf{k-k')}=\delta^3(\mathbf{k'-k)} \implies [a(k),a^{\dagger}(k')]=[a(k'),a^{\dagger}(k)]$, (ii) second is assumed that, $[a(k),a^{\dagger}(k')]=-[a^{\dagger}(k),a(k')]$. Then (i) and (ii) $\implies$

$$\hspace{-3cm}[N(k),N(k')]=\{a^{\dagger}(k)[a(k),a^{\dagger}(k')]a(k')$$ $$\hspace{5cm}+ a^{\dagger}(k')[a^{\dagger}(k),a(k')]a(k)\}/((2\pi)^64\omega_k \omega_k'[ \delta^3(0)]^2)$$ $$\hspace{0cm}=\{a^{\dagger}(k)\delta^3(\mathbf{k-k'})a(k')$$ $$\hspace{5cm}-a^{\dagger}(k')\delta^3(\mathbf{k-k'})a(k)\}(2\pi)^32w_k/((2\pi)^64\omega_k \omega_k'[ \delta^3(0)]^2)$$ I don't think this is the correct approach, some hints would be appreciated.