Show that if $b>0; \mathrm{cor}(x,z)=\mathrm{cor}(x,y)$. Did I take a wrong turn somewhere?  I don't know where to go from here...  Can I not do the division in step 6?  Can standard deviation or cor(x,y) ever be zero?


*Let x and y be jointly distributed numeric variables and let z=a+by, where a and b are constants.  Show that cov(x,z)=b*cov(x,y).  <-done already


finished Proof
 Thanks, Thomas!


Your step 6 is fine, because when any of the standard deviations is zero, correlation is undefined.
So you need to prove that $b\cdot\mathrm{sd}(y)=\mathrm{sd}(a+by)$.
Your next line is odd, because you have $x_i$ on both sides, but:
$$\mathrm{sd}(y)=\sqrt{\sum (y_i-\mu(y))^2}\\
\mathrm{sd}(a+by)=\sqrt{\sum (a+by_i - \mu(a+by))^2}$$
Then use that $\mu(a+by)=a+b\mu(y)$.
And since your proof is actually the reverse, you don't need to be worried about multiplying both sides by $\mathrm{cov}(x,y)$. That is, the "real" proof is to start by proving that $\mathrm{sd}(a+by)=b\cdot\mathrm{sd}(y)$, then reversing your argument above. So the only thing you need is that $\mathrm{sd(x)}$ and $\mathrm{sd(y)}$ are non-zero.
