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Two suggestions: (a) Show that eventually, there will be a distinction. (b) Once you assume that $W$ is countable, you can go on to say that $W$ can be expressed as $\ldots$ (Also, feel free to use $w$ instead of $\omega$, but this doesn't matter.) On a similar note, giving a bijection $f: W \to \mathbb{N}$ is exactly the same thing as listing the elements ...

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Define $X_n$ to be all the combinations of size n that end in 0 and let $Y_n$ be all the combinations that end in 1. Now define $T_n = X_n + Y_n$ to be the total number of combinations. $T_0 = 1,\quad T_1 = 2$. Let's start with a few identities. $$X_{n+1} = X_n+Y_n\\ Y_{n+1} = X_n\\ X_{n+1} = X_n+X_{n-1}\\ T_{n+1} = X_{n+1} + X_n = X_{n+2}\\ T_0 = 1 = X_1, ... 6 Every number has a binary representation$$ n = \sum_{i=0}^m b_i 2^i, \qquad b_i \in \{0,1\}, $$for an appropriate m. For example,$$ 10 = 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0. $$Here are some hints on how the algorithm works: How can you tell whether b_0 = 0 or b_0 = 1? Suppose that (b_k \cdots b_0)_2 is the binary ... 2$$\log{\binom n{\gamma n}} = \log{\frac{n!}{(n-\gamma n)!(\gamma n)!}}\\ = \left(n+\frac 12\right)\log n - n - \left(n\gamma +\frac 12\right)\log \gamma n + \gamma n \\- \left(n - n\gamma +\frac 12\right)\log (n - \gamma n) + (n - \gamma n) + O(1) \\= n\log n - (n - n\gamma)\log (n - \gamma n) - (n\gamma )\log \gamma n + \frac 12\log n - \frac 12(\log n + ...

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Let there be $n$ colors of hat. Match them to the numbers $0$ through $n-1$. The rear person adds up the numbers corresponding to the hats he sees modulo $n$ and guesses that color. He has $\frac 1n$ chance to survive. Each other person in turn does the same computation and can determine the color of his hat to make the sum match what the person behind ...

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The 1 that is immediately left of the binary point (,) is the unit column and has the value 1. Every other 1 is worth twice as much for each place it is to the left of the unit column, or half as much for each place it is to the right of the unit column. $$\def\one#1{\color{#1}{1}} \one{darkred} \one{red} \one{orange} 0 \one{blue} \one{lightblue} ,0 ... 1 Here's a website which can help you understand how it works. If you just want to know how much it is I suggest using Wolfram Alpha. In your case it is 59.25. 1 As you pointed out, irrationals never have a terminating representation. So let's restrict our attention to rationals. I believe that a rational number p/q (in lowest terms) is representable as a terminating floating point expression in base b if and only if each prime factor of q is a divisor of b. This is because if you write the fractional part ... 0 Just like in decimal subtraction, if the digit you want to borrow from is 0, you borrow 1 from the next digit. If that digit is 0, you go on to the next, and so on. 0 Okay, I just figured it out. You can go all the way to the left and cut the 1 backwards. Thing is, if you have 10 in binary = 2. That means you can cut it down to 1 and then move another 1 to the right. If you keep doing this, you can subtract the equation. 1 Hint: Without using complements, subtraction in binary is just like decimal subtraction: a-b=-(b-a). 1 James Stirling has the wisdom. A Stirling number of the second kind (or Stirling partition number) is the number of ways to partition a set of n objects into k non-empty subsets. In this case, we multiply by k! since we have labeled subsets. So the number is$$4!\ S(9,4)=186480.$$2 Same as any other subtraction with two's complement: a-b is computed by adding a to -b. If b is negative, then its negation -b is positive, but that doesn't really change anything. The steps are the same: first negate b, then add the result of that to a. The only arguable corner case is if b is the most negative representable number. Then ... 0 Multiplication in bit level can be as easy as shifting bits which is O(m). This is the case of a=2^k. (Note here the bit level arithmetic that left-shift is equivalent to multiplying by 2. m here is the number of bits.) Or, plain shifting-and-adding and that's in O(m^2). If you are performing a^n as plain multiplication of a^{i-1} \times a in ... 0 Hints Convert -37.8751 forces the first bit to be set; now convert 37.8751 to binary and express this as 1.??? \times 2^{???} and this is where you will get your mantissa and exponent to store in the arrangement you described. First bit is not set, so the number is positive. Exponent is the next 6, i.e. 100111. The remaining bits are the mantissa ... 0 When you square an n-digit number you get either 2n-1 or 2n digits in the product: this is true for binary, decimal or any other base. Which one you get depends on whether or not there is a "carry" into the 2n column when you make the calculation. Let your number, k, have n binary digits. We want to find how large k must be so that k^2 has ... 0 BCD says you take each decimal digit and represent it with four binary bits, so 6789_{10}=0110,0111,1000,1001 BCD, where I used commas to separate the decimal digits. You add just like decimal, carries and all. You may want to delete the leading 0 0 as using the Binary code decimal, 1 + 1 = 10, 1 + 0 = 1, 0 + 0 = 0 is the rule . so the answer should be: 1101010000101 + 1101111110100 - 101101101 = 11010100001100 0 For i, you are converting 291.5_6 to base 10 (despite the fact that base 6 doesn't have a 9). You were supposed to convert from base 10 to base 6. We have 291.5_{10}=1\cdot 6^3+2\cdot^2+3+3\cdot 6^{-1}=1203.3_6 For ii,iii,iv you have converted correctly to base 10, but the assignment is to convert to base 6. 1 Each bit is the highest order bit of what remains so far, right shifted by four places because the dividend has highest term 2^4. So the first bit is 1 (as always). Because the first subtraction results in a 0 in the next column, the second bit of the quotient is 0. It is just like base 10 division, if you get a zero in the next column over you ... 2 The combinatorial species here seems to be$$\mathcal{T} = \mathcal{Z} + \mathcal{Z}(\mathfrak{M}_1(\mathcal{T}) + \mathfrak{M}_2(\mathcal{T})).$$This is one of several interpretations of the question. The one we chose here is that the trees are not labelled and two trees that differ only in the left and right children being exchanged are considered the ... 2 For a CRC32 algorithm with the polynomial$$x^{32} + x^{31} + 2x^{18} + 2x^{17} + 3x^{16} + x^{15} + x^4 + x^3 + 2x^2 + x + 1$$you first omit the x^{32} term and reduce the remaining coefficients mod 2. This gives$$x^{31} + x^{16} + x^{15} + x^4 + x^3 + x + 1 Now simply substitute $x=2\;$ and evaluate, this gives the 32 bit number for the CRC ...

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This boils down to a sequence of shift and xor operations, see for example here for an explanation, including an example calculation.

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\begin{align} t_1,t_2 &\in \{0,1\}\\ |y_1-y_2|&=2t_1 + 4t_2\\ t_1 + t_2 &= 1 \end{align} The first constraint states that the variables are binary. Second one allows the LHS to be either 0,2,4 or 6 (depending on $t_1$ and $t_2$). The third one prevents the sum required term from being $0$ or $6$.

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To expand upon Gerry Myerson's excellent comment, If you're working in base 10, and want to get 1/10, you do two steps, as shown in the diagram below Write the numerator, 1, as 01. (add in the decimal point and a leading zero) Move the decimal point one space to the left, to get 0.1 So if you're working in base 2, and want to get 1/2, you need to do ...

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