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

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.

• "If all digits of $\pi$ were either $0,1,2,3,$ or $4$, the proof is self evident". How do you figure? If all the digits of $\pi$ were $0,1,2,3,4$, then $\pi$ wouldn't be normal. I have no idea what you're going for; does $\pi$ being normal mean something besides the normal definition? – Ben Grossmann Mar 16 '14 at 19:55
• Can't a number be normal if only some of the digits appear in it in some given base? – user132181 Mar 16 '14 at 19:58
• No. A number is normal if (and only if) all digits (and all finite sequences of digits) appear with equal frequency. – Ben Grossmann Mar 16 '14 at 19:59
• All digits that appear in it or all digits used to represent a given base? – user132181 Mar 16 '14 at 20:00
• The second one. – Ben Grossmann Mar 16 '14 at 20:03

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.