Nice try epix, but numbers of the form .###... (where # is a placeholder for some digit in some base > 1) are not members of R or Q sets.
I believe that there is a reason for you to claim so. First, let's consider a finite case and select number 0.372 as a guinea pig for the purpose of showing how things can get treacherous when someone wanders off the path right into the wooded area of Organic Mathematics (OM). Standard Math (SM) says that 0.372 is a rational number. The adjective is derived from the word ratio, which describes this symbol: p/q. SM establishes this equivalency:
A) 0.372 = p/q
where p and q are non-zero integers. So, what are the p and q?
There is no way of finding out, because (A) is
one equation with
two different variables. You can employ all the rules that govern over equations solving, but to no avail.
See, Doron? One considers just a finite case and runs straight into a legitimate obstacle. What is the problem here apart from that one described above?
The problem is that you are attempting to
solve an equation instead of making a progress toward
converting 0.372 into p/q . The process of conversion is similar to equation solving, but it includes steps that are not a part of equation solving. Let me demonstrate...
First, you dismiss this part of (A): "= p/q".
That move leaves you with 0.372. Since you know that p and q are supposed to be integers and 0.372 is not, you start to scheme and ask this question: Can I convert 0.372 into an integer?
Of course you can. There are actually many ways to do so. You can add 0.628 to 0.372 and voila! 0.372 + 0.628 = 1 with 1 being a nice integer. But if you go other way than multiplication, you hit the dead end. Trust me. See, if you use multiplication as a tool and multiply 0.372 by 10
3, where the exponent is equal to the number of digits in the fractional part of 0.372, then you obtain a wonderful integer.
0.372 * 10
3 = 372
Since 372 is an integer and p is supposed to be integer as well, you can set 372 = p. But you need q as well on the right side of the identity. You get it by catching two flies in one swipe. It would be more than advisable to get rid of that "*10
3" helper after it did its job. But how? Well, you divide both sides of the identity by 10
3.
Left: 0.372 * 10
3/10
3 = 0.372
Right: 372/10
3 = 93/250
Written in one line, the result looks like this.
372 = 93/250
where 93 and 250 are integers organized in ratio p/q.
So remember that conversion of numbers displayed in different formats is not identical to equation solving, even though SM says 372 = p/q, which is an equation.
Now when we learn the finite case, we apply our knowledge in the infinite case.
Get this guy: 0.666...
So we repeat the steps taken in the finite case. We convert 0.666... into an integer by multiplying the number by 10
n where n is the number of repeating sixes in the fractional part of 0.666...
And here is the problem that OM cannot solve: n→∞, and since infinity is not a point on R - it is not a number - there is no way to multiply and convert 0.666... into an integer. It follows that 0.666... cannot be converted to its equivalent p/q and that means that 0.666... and all 0.###... are not rational numbers, exactly as OM claims. That claim rests on a firm logical ground:
Finite is to Can as Infinite is to Cannot.
But there is something that may undermine all the rigor

that OM paddles on. You may come accidentally across a particular long division case involving 2 in the numerator p, and 3 in the denominator q:
[p/q = 2/3] => [LD(2, 3) = 0.666...]
Is it possible that if a conversion from 2/3 into 0.666... exists via an algorithm, then there is an algorithm capable of converting 0.666... into 2/3 ?
Since OM claims that all 0.###... are not in Q, you have not found such an algorithm. Maybe you didn't look with powerful enough spatial_visual binoculars.
I know the argument by heart:
If a/b = c, then a = b*c
and since there is no last digit in 0.666... the check by multiplication cannot be done. But that doesn't prove that 2/3 = 0.666... is wrong - infinity merely prevents to perform the check.
Your claim is a proof-free statement. It lacks the necessary intro: Suppose that 0.###... is in Q. If it is so ... and a contradiction would follow.