In decimal I can discard zeros after the radix point, e.g.: $$ 0.250_{10} = 0.25_{10} $$ It seems to me that I can do the same with binary: $$ 0.10_2 = 0.1_2 $$ Because
$$ 1\times\frac{1}{2}+0\times\frac{1}{4} = 1\times\frac{1}{2} $$ Am I right?
In decimal I can discard zeros after the radix point, e.g.: $$ 0.250_{10} = 0.25_{10} $$ It seems to me that I can do the same with binary: $$ 0.10_2 = 0.1_2 $$ Because
$$ 1\times\frac{1}{2}+0\times\frac{1}{4} = 1\times\frac{1}{2} $$ Am I right?
On
While Roberts Frost's answer is perfectly correct there is however one small catch!
When engaged in practical computations which seek to obtain an approximation $A$ to the solution $T$ of a complicated equation, then there is a profound difference between the statement $T \approx 1$ and the statement $T \approx 1.0$. In the first case, we implicitly state the error $E = T-A$ satisfies $|E| \leq 5\times10^{-1}$. In the second case, we implicitly make the stronger statement that $|E| \leq 5 \times 10^{-2}$.
By dropping the "extra" 0 we are selling ourselves short because we give the wrong impression of the quality of the approximation.
Yes it's exactly the same system. The only restriction on decimal relative to binary is no digits 2-9. As per your proof, removing a trailing zero is always equivalent to subtracting zero and therefore has no effect.