What are the oxidation numbers of ‘N’ in ${\mathrm{NH}}_4{\mathrm{NO}}_3$?

The compound ${\mathrm{NH}}_4{\mathrm{NO}}_3$ exists as $\mathrm{N}{{\mathrm{H}}_4}^{+}$ and $\mathrm{N}{{\mathrm{O}}_3}^{-}$

Let us assume $X$ be the oxidation state of nitrogen in $\mathrm{N}{{\mathrm{H}}_4}^{+}$. Hydrogen oxidation state is $+1$

So, $\mathrm{X}+4\mathrm{x}(+1)=+1$

 $\mathrm{X}+4=+1$

  $\mathrm{X}=+1-4$ 

$\mathrm{X}=-3$

Let us assume $X$ be the oxidation state of nitrogen in $\mathrm{N}{{\mathrm{O}}_3}^{-}$. Oxygen have oxidation state $-2$

So, $\mathrm{x}+3\mathrm{x}(-2)=-1$

  $X-6=-1$ 

  $\mathrm{X}=-1+6$ 

$\mathrm{X}=+5$