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# Tag Info

16

The function $f$ is indeed not Riemann-integrable, but it is Lebesgue-integrable. And its integral is $0$. Therefore, the answer is $0$. That is natural, since $\mathbb Q\cap[0,1]$ is countable, whereas $[0,1]$ is uncountable. It follows that there is no bijection between $\mathbb Q\cap[0,1]$ and $[0,1]\setminus\mathbb Q$.

8

Noah Schweber's answer, using Cantor's theorem, was also the first answer that occurred to me, but here's an alternative approach that doesn't need Cantor's theorem. I'll use Noah's convenient list of four types of intervals of rationals. Type 4, where both endpoints are in the interval (and are therefore rational), is the easiest. The linear, increasing ...

7

The answer is almost yes: we need to go a little bit deeper than just closed/open, but not much deeper. It turns out to be a bit clarifying to talk about linear orders in general, rather than specifically sets of real numbers. (Note that "order-preserving bijection" now is just "isomorphism.") The key result is an old theorem of Cantor: Any two ...

7

The idea is the same. Instead of choosing $\sqrt{2}$, choose an irrational number in $(0,1)$ and split it to two open intervals.

6

This answer might miss the real goal, but is inspired by the tag "probability". In your question it is not explicitly mentioned how probability $P$ is defined. So actually the question: "what is the probability that $x$ is rational?" makes no sense in this context. Your try indicates that you are thinking of uniform distribution on interval $[0,1]$ where: ...

5

Pick $U=A\cap (-\infty,\frac{\sqrt2}{2})$, $V=A\cap (\frac{\sqrt2}{2},\infty)$.

2

The standard way consists of: To define integers in a way that doesn't use rational numbers. To define a rational number as an equivalence class $\bigl[(a,b)\bigr]$ of elements of $\mathbb Z\times(\mathbb Z\setminus\{0\})$, where the equivalence relation is$$(a,b)\sim(c,d)\text{ iff }ad=bc.$$ This approach also avoids another problem in what you wrote: it ...

2

Hint: There are no solutions mod $5$. Indeed, if $r=x/y$ then $2x^4 \equiv -6 y^4$ or $x^4 \equiv -3 y^4$. Now use Fermat's theorem.

1

No, it is reversed. It should read if x $\le$ y, then for all z where z $\le$ 0, (y * z) $\le$ (x * z). Alternately, you can reverse the $\le$ to $\ge$ and keep $x$ and $y$ where they are.

1

No, it is not. Either there is a typo in the book, or there is a typo in your transcription. For example, $-2 \leq 3$, and $-5 \leq 0$, but $$-5 \times -2 = 10 \not\leq -15 = -5 \times 3.$$

1

What do we get with continued fractions? $\dfrac{5}{9}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{4}}}$ $\dfrac{4}{7}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}$ The fractions are identical until we get to the last layer where one has a $4$ and the other has a $3$. Were there an integer between $3$ and $4$ we could replace the last layer with the smallest such ...

1

Let $z=y-x$. Since $x<y$, we know that $0<z$. Therefore, there is some positive integer $n$ such that $\frac1n<z$. (For instance, some $n$ such that $n>1$ and $n>\frac1z$.) Therefore, multiplying both sides of the inequality by $n$ (which is positive), we get $$1<nz=n(y-x)=ny-nx$$ Since the gap between $nx$ and $ny$ is greater than 1,...

1

I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets You can't. The set of rationals is countable. The set of irrationals is uncountable (see Proof that the irrational numbers are uncountable). Cantor proved by diagonalization that such that mapping cannot exist.

1

Take $P=[0,1/\pi)\cap\mathbb Q$ and $Q=(1/\pi,1]\cap\mathbb Q$ such that $P\cup Q=A$. Now $P\cap\overline Q=\phi$ and $\overline P\cap Q=\phi$ suffices to prove the result.

1

This is rather simple really. If a submarine is going downwards, it's travelling a distance, regardless of direction. The question says 'How far did it descend?'. This is merely a distance and therefore requires a length. i.e a positive number. Because you can't have a negative length. If they asked 'Where is the submarine?' then your son's answer would be ...

1

Тhat is a beautiful solvable nonic! Here is its smallest real root: With[{α = Sin[ArcSin[(1 + 3 × 3^(2/3))/8] / 3], β = 3 + 9 × 3^(1/3) + 7 × 3^(2/3), γ = 18 + 9 × 3^(1/3) + 2 × 3^(2/3)}, With[{ ρ = -51 - 33 × 3^(1/3) + 121 × 3^(2/3) - 8 α β + 4 α^2 (β - 60 × 3^(2/3)), σ = 12 - 9 × 3^(1/3) - 7 × 3^(2/3) + α γ + 2 α^2 β, τ = 53 + 24 × ...

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