Forgive me if I have set this up wrong, I haven't done proofs in a long time. I was thinking at lunch today what it would be like if we could write binary behind a decimal point. Imagine
$$110.11 = 6.75$$
This made me think, are there numbers in base-10 decimal that cannot be represented rationally in base-2 decimal? Maybe, but since 2 evenly divises 10, maybe not. How about 3? $n$?
This is my conjecture:
$$\forall ( x_1,l_1,n \in Z | x_1 < 10^{l_1} ) \exists i_2,l_2 (\frac{x_1}{10^{l_1}} = \frac{x_2}{n^{l_2}} ) $$
How would I go about proving it?
Best Answer
I think that you’re confusing rational with having a terminating decimal expansion. The two are not synonymous. Which numbers with terminating decimal expansions depends on the base; which numbers are rational does not.
Added: The rational numbers are those that can be represented as quotients of integers; the actual representation of the integers is irrelevant. In any base these numbers will have expansions that either terminate or eventually repeat. (Actually, termination can be viewed as eventually repeating the digit $0$.)
In order for the expansion of a fraction $m/n$ in lowest terms to terminate in base $b$, it’s necessary and sufficient that all of the prime factors of $n$ also divide $b$.