An inquiry on Rational numbers in most basic kind

A function $f(x)$ is specified $\forall x \in \mathbb{Q}$ and also $ x \in (0,1)$ as $f(x) = q $ where $x = (p/q)$ in most basic kind. Locate $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)$ where $ a \in \mathbb{Q}$ and also $ a \in (0,1)$ and also $ \tau \in (\alpha,\beta)$ where $ 0< \alpha < \beta <2 $ and also $ \alpha,\beta \in \mathbb{Q} $. $\sup$ and also $\min$ are taken control of just those $x$ where $f(x)$ is specified. Is the supremum infinity?

EDIT

The supremum might or might not rely on $\alpha$ and also $\beta$ ...this has actually not been dismissed.


EDIT 2

I will certainly attempt something!

Take into consideration $\tau = l/m$ where $l,m \in \mathbb{N}$ in most basic kind and also $\tau \in \mathbb{Q}\cap(0,2)$.

what is minimum feasible value of $q$ where $\tau-a = p/q $ created in most basic kind such that $a \in \mathbb{Q} \cap (0,1)$ and also $(\tau-a) \in (0,1)$.

As Yuval Filmus recommended in his solution ...

one means to pick $a$ is that it has very same (in most basic kind) as that of $\tau$ and also numerator $n$ such that $l-n$ is reasonably prime to $m$, (If such a selection is feasible under offered problems), after that $f(a)+f(\tau-a)$ comes to be $2m$.

The inquiry is whether such a selection bring about $f(a)+f(\tau-a) = 2m$ is the minimum feasible value? Otherwise I request you to offer a counter instance.

If it is, after that the supremum (over $\tau$ differing in any kind of approximate open period had in $(0,2)$) is $\infty$ as we can increase $m$ randomly by differing $\tau$ in any kind of approximate open period $(\alpha,\beta)$.

MODIFY 3

tiny adjustment concerned ...as opposed to $0<\alpha<\beta<2$ we have $-1<\alpha<\beta<1$

this adjustment is not made with any kind of purpose of highlighting a particular solution.Also its impact on the remedy is wanted to be really small.Inconvenience is been sorry for.

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2019-12-02 02:53:14
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Answers: 2

EDIT : This is a combined remedy which formerly showed up partly.

Lemma 1. For each and every all-natural $M$ and also actual $\alpha < \beta$ there is a sensible number in $(\alpha,\beta)$ whose lower kind has a bigger than $M$.

Evidence. Pick some power of 2 $2^n > \max\left(M, \frac{2}{\beta-\alpha}\right)$. Given that $2^{-n} < \frac{\beta-\alpha}{2}$, there go to the very least 2 portions of the kind $x/2^n$ in $(\alpha,\beta)$. Among them has weird $x$ therefore remains in lower kind.

Lemma 2. If $p_1/q_1 + p_2/q_2 = P/Q$, all rationals in lower kind, after that $(q_1+q_2)^2 \geq 4Q$.

Evidence. Taking a common measure, $\frac{p_1 q_2 + p_2 q_1}{q_1q_2} = \frac{P}{Q}$ therefore $Q|q_1q_2$ (given that $P/Q$ is lowered). Specifically $Q \leq q_1q_2 \leq \left(\frac{q_1+q_2}{2}\right)^2$, where the last inequality is the math - geometric mean inequality.

Theory. For every single actual $M$ and also reals $\alpha < \beta$ there is a sensible $r \in (\alpha,\beta)$ such that $\min_{a \in \mathbb{Q}} f(a) + f(r-a) > M$.

Evidence. Making use of lemma 1, take a sensible $r \in (\alpha,\beta)$ with $Q > M^2/4$. Making use of lemma 2, we end that for every single sensible $a$, $f(a) + f(r-a) \geq 2\sqrt{Q} = M$.

Effect. $$\sup_r \min_a f(a) + f(r-a) = \infty.$$

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2019-12-03 04:18:20
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For an offered $\tau=\frac{1}{q}, q$ prime it appears that $a$ needs to be (probably to name a few selections) $\frac{\tau}{2}$ with $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)=2q$. Can you confirm this? If so, after that the sup over $\tau$ is infinity as you can locate $\tau$ with randomly huge .

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2019-12-03 02:31:45
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