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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.

user132181
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    "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
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    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
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    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
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    All digits that appear in it or all digits used to represent a given base? – user132181 Mar 16 '14 at 20:00
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    The second one. – Ben Grossmann Mar 16 '14 at 20:03

2 Answers2

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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.

Ben Grossmann
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    Here is a print reference: http://books.google.com/books?id=mnY8LpyXHM0C&pg=PA76&dq=%22every+nonzero+rational+multiple+of+a+normal+number+is+normal%22&hl=en&sa=X&ei=MQAmU7zLM8GE0AG1yID4Aw&ved=0CCwQ6AEwAA#v=onepage&q=%22every%20nonzero%20rational%20multiple%20of%20a%20normal%20number%20is%20normal%22&f=false – Jason Zimba Mar 16 '14 at 19:50
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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.

Kai Sikorski
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