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0

Wikipedia has a list of six different ways to round halves. To see the logic of round to even, consider adding a long column of numbers that are all an exact number of tenths-that all have exactly one decimal place. Now consider the effect of rounding to integers before adding. If the decimals are evenly distributed, the increases in the ones with $.6, ...


3

We first consider the case $n$ odd, since the solution is simpler. Let $n = 2m + 1$. Choose $a_i: 1 \le i \le m$ nonzero distinct rational values such that the sum of squares is itself a square: $$a_1^2 + a_2^2 + \ldots + a_{m}^2 = A^2$$ (One way to generate such a list would be to iteratively use Pythagorean triples, ie something like $a_1 = 3, a_2 = 4, ...


3

As per TonyK's answer, it is only necessary to find the smallest denominator for which a fraction in $[L,U)$ exists. To avoid what he calls "messy" oscillations, we can proceed as follows: Start with $(a,b,c,d)=(0,1,1,1)$. *[Remark: We have $\frac ab<L<U\le \frac cd$, and $ad-bc=-1$, and any fraction between $\frac ab$ and $\frac cd$ has denominator ...


3

It is enough to find the smallest possible $b$, and then the smallest possible $a$ given $b$. This is not completely obvious, so I will try to prove it: Suppose we have positive integers $a,b$ such that (i) $b$ is the smallest denominator of all fractions in $[L,U)$; and (ii) $a/b$ is the smallest multiple of $1/b$ in $[L,U)$. This means ...


1

Why is your algorithm exponential? Here's an option: Don't think about the numbers as fractions, but as pairs. Your goal is to find the pair $(a,b)$ such that $L\leq a/b\leq U$ such that $a\times b$ is minimal. In this case, we do NOT reduce fractions. For a fixed $b$, we can easily find the smallest possible $a$ that satisfies the conditions by ...


0

A recurring decimal expansion is a geometric series with common ratio $10^{-\tau}$, where $\tau$ is the period in number of decimal places. Thus, through the relationship $\frac{a}{1-r} = a\,(1+r+r^2+\cdots);\,r<1$ it is a number of the form $\frac{p}{q};\,p,\,q\in\mathbb{N}$. To infer that rationals have finite or periodic $N-ary$ expansions: if a ...


1

In the comments I proposed following formula for $K\gg 1$ (about $15$ digits precision for $K>30$) $$K-\gamma-\log(K+1/2)-\frac 1{24K^2}+\frac 1{24K^3}-\frac{23}{960K^4}+\frac 1{160K^5}+\frac {11}{8064K^6}+\frac 1{896K^7}$$ and to compute directly $\,K-H_K\;$ for small values of $K$ ($H_K=\displaystyle\sum_{i=1}^K\frac 1i\;$ is an harmonic number). This ...


1

It's easier to first talk about endomorphisms because these have a ring structure. In fact, as has been mentioned a few times already, this group is a countable direct sum $\bigoplus_p \mathbb{Z}$ of a copy of $\mathbb{Z}$ for every prime, and hence in some respects it behaves like a vector space. In particular, its endomorphism ring $\text{End}(\bigoplus_p ...


3

Hint $(\Bbb Q_{> 0}, \cdot)$ is isomorphic to $(\Bbb Z, +) \oplus (\Bbb Z, +) \oplus \cdots$.


1

The standard proof that the real numbers are uncountable does not need to be phrased as a proof by contradiction. To say that $\mathbb{R}$ is uncountable is to say that if $f : \mathbb{N} \to \mathbb{R}$ is a map, then it is not a bijection. The standard proof in fact proves that $f$ is not a surjection by explicitly exhibiting a real number which is not in ...


0

A non-rational root is simply a root that is not a rational number (i.e. it is irrational). The problem has nothing to do with the intermediate value theorem, it is an exercise in pure algebra. First, note that if $r$ is an irrational root, then the other root $r'$ must also be irrational (assume that $r'$ is rational; since $r+r'= - \frac b a$ which is ...


1

$$\sum_{\alpha=2}^{K} \frac{\alpha-1}{\alpha}=K-\digamma^{(0)}(K+1)-\gamma$$ With $\digamma^{(0)}$ is the derivative of the digamma function and $\gamma$ is the eulergamma


1

No there isn't, because finding a compact form for the sum is equivalent to find a compact form for the harmonic sum $$ \sum_{k=1}^n \frac1k $$ which we know that doesn't exists. But you can approximate your sum by using $$ \sum_{k=1}^n \frac1k=\log(n)+\gamma+O(n^{-1}) $$


7

No, homeomorphisms of $\mathbb{Q}$ need not be monotone. For an irrational $c > 0$, let $$h_c(x) = \begin{cases}x &, \lvert x\rvert < c\\ -x &, \lvert x\rvert > c. \end{cases}$$ Then $h_c$ is a non-monotonic homeomorphism of $\mathbb{Q}$.


0

If you mean to include all the k that are less than $\frac{f_s}{2f}$, you can hedge and use the floor function: $\sum_{k=1}^{\left\lfloor\frac{f_s}{2f}\right\rfloor} \sin(2 \cdot \pi \cdot k) $ This guarantees your bounds are both integers, but contain all the k you want.


1

My intuition says that since distinct integers differ by at least one and rationals are arbitrarily close, the density of the integers should be zero and the density of the rationals should be positive. If you throw in the reals, I'll throw in the towel.


9

It does repeat, but with a very long period: Factoring the denominator into primes gives $$998 = 2 \cdot \color{#3f3fff}{499};$$ since $2$ is a factor of $10$ and $10$ is a primitive root modulo $\color{#3f3fff}{499}$, the period of repetition is $\color{#3f3fff}{499} - 1 = 498$ digits long. Indeed, consulting WolframAlpha gives: $$\color{#bf0000}{ ...


7

Emphasis on "apparently". The same thing happens with $1/7$: $$\begin{array}{crrrrrrrrrl} & 0 & . & 14 \\ + & & & & 28 & & & & & & =2\times 14 \\ + & & & & & 56 & & & & & = 4\times 14 \\ + & & & & & 1 & 12 & & & & =8\times 14 ...


2

Suppose you run a numerical analysis program, and it outputs $$0.5323$$ What does that even mean? Does it mean: $$x = 0.5323$$ $$0.5323 \le x < 0.5324$$ $$0.53225 \le x < 0.53235$$ $$\text{Probability}\left(\frac{|x - 0.5323|}{0.5323} < 0.01\right) > 95 \%$$ The meaning of a numerical program still must be expressed and proven ...


2

It is possible for an irrational number to have a predictable pattern; consider $0.1101001000100001...$. It is also possible to have an irrational number that is another irrational number away from a rational; i.e. $x-y = r $, where $x,y$ irrational and $r $ rational; in fact the equivalence classes of such numbers are dense in the reals. So you can subtract ...


0

In order for you to be able to convert the denominator to a number of the form $2^n 5^m$, it needs to a number that is divisible by $2$ or $5$ and no other primes, and that is not the case for $264600$. But, as it turns out, you can make the denominator that way by "simplifying" the fraction: $$\frac{1323}{264600} = \frac{1}{200}.$$ Now observe that $200 = ...


0

No, it's not the same thing. Note that $\frac{1}{2}$, a rational number, cannot be added to any integer to get 271. The first one is false. The second is true. For any integer, there is some integer I can add to it to yield 271, and all integers are also rational numbers.


4

For all $q \in \Bbb Q$ there exists $n \in \Bbb Z$ such that $q + n = 271$. False. $\Bbb Q$ is the set of rationals, i.e. fractions: $\Bbb Q = \{\ldots, -\frac 1 3,- \frac 2 2,-\frac 1 2, 0, \frac 1 2, \frac 2 2, \frac 1 3, \ldots\}$. $\Bbb Z$ is the set of integers: $\{\ldots, -3, -2, -1, 0, 1,2,3,\ldots\}$. Take $q = \frac 1 2$ for an easy ...


0

Hint: $$264600 = (100)\cdot(2646) = 2^35^2\cdot1323$$


6

Others have focused on the specific value of the integral itself. However, it is clear from your description that you wish to evaluate something along the lines of $$ \frac{\int_0^1 f(x)r(x)dx}{\int_0^1 r(x)dx} $$ where $r(x)$ is the rational indicator function. While the component integrals cannot be evaluated, the overall concept is still meaningful. It ...


4

The integral exists in the Lebesgue sence and it is zero. That is because the rationals have measure zero, being a countable set. $r$ is the characteristic function of the rationals, and for such function the integral is given by $$ \int_0^1 r(x) \, dx = \lambda([0,1] \cap \mathbf Q) $$ where $\lambda$ is the Lebesgue measure. We have for measurable sets $A ...


17

It depends on whether you're considering the Riemann integral or Lebesgue integral. Note that every interval of any partition of $[0,1]$ contains a rational point, so any upper sum for $r(x)$ is $1$. Similarly, any lower sum for $r(x)$ is 0. Thus, $r(x)$ is not Riemann integrable. However, $r(x)$ is the indicator function for $\mathbb{Q}$, which is a ...



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