Voltage for a diode in forward bias
Ideality factor is more like \$2\$ on a real diode so your 2nd number is high by \$4\$ or \$5\$ orders of magnitude. 

With \$V_{T} = 0.025\$ and \$V_{F}= 0.1\$ or \$0.5V\$ the current will generally be pretty low (nA or tens of uA respectively).

The diode is forward biased when \$V_{F} &gt; 0\$. It starts to conduct substantially (for many purposes anyway) when the current exceeds roughly \$0.5V\$. 

They are saying that the diode equation works better when \$V_{F} &gt; 0.1V\$, or at least you can start to ignore that pesky \$-1\$ in the equation.  

Edit: Shockley diode Equation: 
\$I = I_\text{S}( e^\frac{V_\text{D}}{n V_\text{T}}- 1)\$

"\$n\$ is the ideality factor, also known as the quality factor or sometimes emission coefficient" to quote Wikipedia.