how to prove that binary number is always represent unique decimal number? (uniqueness of binary system) i.e every binary number determines unique decimal number i.e there is one-to-one relationship between binary and decimal.
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Before you state there is a one-to-one correspondence, please be aware of the points like 10.0000000 and 1.11111111. Both of them are 2 in decimal rep. So one should not expect a completely 1-1 correspondence. – Zheng Liu Sep 07 '15 at 05:10
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Not for decimal place value... – LOKENDRA SINGH RATHORE Sep 07 '15 at 05:12
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Yes.The definition of the real number system is usually done without any reference to a number-base.Every real number has a representation in any number-base, base 2 or base ten ,or any other base.It's the same set : the real numbers. – DanielWainfleet Sep 07 '15 at 05:13
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Possible duplicate of Binary expansion Unique – David K Jan 19 '16 at 22:08
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To prove that binary numbers represent unique decimal numbers, simply prove that two unequal sequences of bits will always produce different decimal numbers. This can be done with inequalities.
Define two numbers $b \ne c$ and give their binary representations via:
$b = b_1b_2\cdots b_n$
$c = c_1c_2\cdots c_n$
Define $k$ such that $k$ is the least number such that $b_k \ne c_k$
Let $d = (b_1b_2\cdots b_{k-1}) \times 2^{n-k+2} = (c_1c_2\cdots c_{k-1}) \times 2^{n-k+2}$.
Then define $b' = b - d$, $c' = c - d$
WLOG, say $b_k = 1, c_k = 0$.
The number $c_{k+1}\cdots c_n$ is at most $2^{n-k+1}-1$, while the value $b_k = 2^{n-k+1}$.
This shows that $b' > c' \Rightarrow b' \ne c'$ for any difference in the sequence of bits. They will always yield unique decimal numbers.
Thus, $b \ne c$ since $b' + d \ne c' + d$.
The proof of decimal numbers yielding unique bit sequences uses the same logic.
EDIT: In light of Zheng's comment, this is good for a finite decimal expansion. If the decimal expansion is infinite, it can end up not being a one-to-one.