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2

You can use strong induction. $1=2^03^0$ is the base case. Now assume all numbers up through $m$ can be represented. If $m+1$ is even, take the representation of $\frac 12(m+1)$ and increase all the $x$'s by $1$. If $m+1$ is an odd multiple of $3$, take the representation of $\frac 13(m+1)$ and increase all the $y$'s by $1$. If $m+1$ is coprime to $6$, ...


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As described in the comments: the answer, let's call it $f(n)$, is one less than the greatest power of $2$ which divides $5^n-1$. Thus $$f(n)=v_2(5^n-1)-1$$ where, as usual, $v_2(k)$ denotes the greatest power of $2$ dividing $k$. To compute $v_2(5^n-1)$... first note that this is $2$ if $n$ is odd: indeed $5^n-1=(5-1)(5^{n-1}+\dots+1)$ regardless ...


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You can use the same technique as with repeating decimals: $$\begin{array}{rcrl}x &=& 0.\overline{0011}\\ 10000_2 x &=& 11.\overline{0011}\\ 1111_2x &=& 11.0000\\ x&=&\dfrac{11_2}{1111_2}&=\dfrac{3}{15}=\dfrac15\end{array}$$


2

Imagine you write the $n$ zeros and the $n$ ones along a line. The possible configurations can be created by permuting them in all possible ways. There are $(2n)!$ permutations of $2n$ digits. However, permutations that only exchange ones among themselves do not produce new configurations. You must discard them. There are $n!$ permutations among the $n$ ...


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Taken from comments: You have correctly proved that $\sum_{i=0}^{n-1} 2^i = 2^n - 1.$ To complete the proof of the initial claim, you should elaborate on why $\sum_{i=0}^{n-1} 2^i $ is the greatest number you can make with $n$ bits.


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$\color\green1\color\red0\color\red0\color\red0\color\red0\color\red0\color\red0\color\red0\color\green1\color\green1\color\red0.\color\red0\color\green1\color\green1=$ $\color\green1\cdot2^{10}+$ $\color\red 0\cdot2^{ 9}+$ $\color\red 0\cdot2^{ 8}+$ $\color\red 0\cdot2^{ 7}+$ $\color\red 0\cdot2^{ 6}+$ $\color\red 0\cdot2^{ 5}+$ $\color\red ...


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I always manage to confuse myself with this process since it is not done manually too often, so refer to this handy algorithmic like approach. In both cases, the number we are subtracting is larger in magnitude. $1's$ Complement: Determine the $1's$ complement of the larger number: $00101$ Add the $1's$ complement to the smaller number: $01001 + 00101 = ...


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I'm not sure what you mean. In 'why 10 in any base number system write as 10' what is the supposed base of each '10'...? For example in base 2, 'two' is written as 10, but it is not decimal 'ten'; it's two. We use subscript to denote a base (with a common convention that the base in subscript is in decimal), so: two in binary is $2 = 10_2$, ten in decimal ...


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You can find many explanations of the IEEE-754 format on the Web. In short, each 32-bit single-precision floating-point number consists of three parts (we assume that the bits numbering begins from zero): sign, bit 31 exponent, bits 30-23 (eight bits in total) significand (or "mantissa"), bits 22-0 (twenty three bits in total) However, in your case the ...


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No. If we use $n$ bits to represent unsigned integers in the range $0,1,\ldots, 2^n-1$ then each bit pattern corresponds to exactly one of these integers and vice versa. Assume that some integer $k\ge 0$ allowed two distinct patterns of $n$ bits, where $2^n>k$. Among those integers we let $k$ be the minimal one. And among all matching numbers of bits, ...


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No, each nonnegative integer has one and only one binary representation. (That is, if we don't care about representations that only differ in how many leading zeroes they have). Suppose you have two different bit strings and we want to prove that they represent different integers. Let's look at the leftmost position where the two bit strings differ; there ...


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Partial answer: Any nonzero rational in $[0,1)$ with a terminating representation has two binary expansions. Suppose $x\in (0,1)$ is rational, $x\ne 0$, and $$ x = 0.d_1 \dotsm d_n $$ for binary digits $d_i$. As $x\ne 0$, we have $d_n = 1$: $$ x = 0.d_1 \dotsm d_{n-1} 1 $$ But now it's easy to see and show that $$ x = 0.d_1 \dotsm d_{n-1} 0 \overline{1} ...


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As BrianO explains in his answer, any rational number that can be written with a terminating binary expansion has two distinct binary representations: one in which the binary representation terminates in a $1$, and one in which the binary representation terminates in $0\overline{1}$. So the question in the OP reduces to: which rational numbers in $[0,1)$ ...



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