Understanding De Broglie Matter Waves: How to Simplify E/p=c^2/v?

[AStaunton]

Just a very quick question:

in my notes I have the velocity of a DeBroglie wave is given by:

$$\nu\lambda=2\pi\nu\frac{\lambda}{2\pi}=\frac{\omega}{k}=\frac{E}{p}=\frac{c^{2}}{v}$$

I can't figure out how he want from E/p=c^2/v.

I think the assumption is made that E is approximately equal to mc^2+(1/2)mv^2 and p=mv:

$$\implies\frac{E}{p}=\frac{m_{0}c^{2}+\frac{1}{2}m_{0}v^{2}}{m_{0}v}$$

but I still can't see how the above simplifies to c^2/v!

 

[vela]

Try using the relativistic formulas for E and p, rather than the low-speed approximations.