proof.txt

Assume that there exists a number that is a Two-Bit Number in both decimal and binary. Therefore, we can represent it as:

10^x + 10^y = 2^a + 2^b for some positive integers x<y, a<b.

Establish:

- even * even = even
- odd * even = even
- odd * odd = odd
- even + even = odd + odd = even
- odd + even = odd
- even ^ 0 = odd
- even ^ n = even for n >= 1
- odd ^ n = odd

Firstly, if x < a, then 10^x + 10^y = 10^x(1 + 10^(y-x)) = 10^x * odd = 2^x * 5^x * odd = 2^x * odd, therefore 10^x + 10^y is only divisible by 2 x times. However, this number also equals 2^a + 2^b, so 2^a + 2^b = 2^x(2^(a-x) + 2^(b-x)) = 2^x(even + even) since a>x. However, we just established that this number, divided by 2^x, was odd. Therefore, this is a contradiction, so x >= a.

Now, assume x > a.

10^(x-a) * 5^a + 10^(y-a) * 5^a = 1 + 2^(b-a)

even * odd + even * odd = 1 + even

even + even = odd

even = odd

This is a contradiction, Therefore, x = a.