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Can someone please explain the below text and what it means in simple words? From what I understood it means that if x=m/2^n and if m is odd then x can be represented in binary in two ways? But how is that possible. I tried with 1/2 but how can 0.5 be represented as 0.1 and 0.0111?
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by definition, $x = (.a_1a_2\dots a_n\dots)_2$ iff $$x=\sum_{i=1}^{\infty}a_i2^{-i}$$ if $x = \frac{m}{2^n}$, for $m$ odd, i.e., $m=2k+1$, then $x=\frac{2k+1}{2^n}=\frac{k}{2^{n-1}}+\frac{1}{2^n}=(.a_1a_2\dots a_{n-1}1000\dots)_2$. But, note that $x$ is also equal to $(.a_1a_2\dots a_{n-1}0111\dots)$ by definition, since \begin{align*} \sum_{i=1}^{\infty}a_i2^{-i} & = a_12^{-1}+\dots+a_{n-1}2^{n-1}+\sum_{i=n+1}^\infty2^{-i}\\ & = \frac{k}{2^{n-1}}+\frac{2^{-n-1}}{1-\frac12}\\ & = \frac{k}{2^{n-1}}+\frac{1}{2^n}\\ & = x \end{align*} so the real reason why every real in binary has two decimal expansions is precisely because $\frac12 = 0.1 = 0.0111\dots$, both series converge to the same number. And in fact this occurs on any numerical basis $b$, since $$(0.1)_b=b^{-1}=\sum_{i=2}^\infty (b-1)b^{-i}=\frac{(b-1)b^{-2}}{1-\frac1b}=(0.0(b-1)(b-1)\dots)_b$$ this is why, for example, $1 = 0.999\dots$ in decimal base.
Note that the $1$s go on forever. They don't stop after the first three. This is the binary version of the infamous fact that $$ 0.999\cdots = 1 $$ in decimal. (This equation has its own Wikipedia page.)
I cannot see what equation $(2)$ is in your book, but here's a definition of binary representations which is likely equivalent:
For your example, $x = \frac12$, the $0.1$ representation is short for $0.1000\cdots$. The infinite sum is then $$ \sum_{n = 1}^\infty \frac{a_n}{2^n} = \frac12 + 0 + 0 + 0 +\cdots = \frac12. $$ But for the other representation, $0.0111\cdots$, the sum is $$ \sum_{n = 1}^\infty \frac{a_n}{2^n} = 0 + \frac14 + \frac18 + \frac1{16} + \cdots. $$ This is a geometric series which evaluates to = $1/2$, as desired.