If $\pi $ is a normal number, is $\tau $ one?

Question

If $\pi$ is a normal number, would that imply that $\tau =2\pi $ is also a normal number? If so, why? Something tells me that it should be, but I have no idea how to prove it. If all digits of $\pi$ were either $0$, $1$, $2$, $3$ or $4$, the proof is self-evident. Obviously, every decimal digit appears in $\pi$, so that's where that self-evident proof will fall apart.

Answer 1

Yes, a rational multiple of any normal number (with respect to a base $b$, such as $b = 10$) is also normal (with respect to the same base).

The number $\tau = 2 \pi$ is a rational multiple of $\pi$, and visa versa. So, $\tau$ is normal if and only if $\pi$ is normal.

Answer 2

Well according to wikipedia this statement:

"If x is normal in base b and q ≠ 0 is a rational number, then x⋅q is normal in base b."

has been proven by Wall.

Wall, D. D. (1949), Normal Numbers, Ph.D. thesis, Berkeley, California: University of California.