 Solution

Write $x=\Delta E/(k_{\mathrm{B}}T)$ and $r=N_{\uparrow }/N_{\downarrow }=e^{-x}$. Since $N=N_{\uparrow }+N_{\downarrow }$, eliminate the individual populations:The Taylor expansion $e^{-x}=1-x+O(x^2)$ gives the same result, and the odd expansion of the hyperbolic tangent yieldsEquivalently, $N_{\uparrow }\simeq (N/2)(1-x/2)$ and $N_{\downarrow }\simeq (N/2)(1+x/2)$. Their sum is still $N$. The negative sign follows directly from the stated higher-energy up-state population, and does not mean a negative magnitude of the equilibrium magnetization.