# Tag Info

11

It depends if you're talking about a Riemann integral or a Lebesgue integral. If we are talking about a Riemann integral, the answer is that we cannot define the integral because any sub-interval of $[0,1]$ - no matter how small - contains a rational and an irrational. For this reason the upper integral and lower integral will not be the same (here the ...

7

Let $\sqrt3=u+v\sqrt2$. Then by squaring, $3=u^2+2\sqrt2uv+2v^2$ and $\sqrt2=\dfrac{3-u^2-2v^2}{2uv}$ is a rational number !?

6

This is false because if you take $x = \sqrt{2}$ and $r = 0$, $$x \cdot r = \sqrt{2} \cdot 0 = 0,$$ which is rational, not irrational. However, suppose that $r \ne 0$. Then suppose towards contradiction that $x$ is irrational and $r$ is rational, but $rx$ is not irrational, i.e. $rx$ is rational. Then write $rx = s$, where $s$ is rational. Since ...

6

Note that $\sqrt 2=\dfrac 12\left(\sqrt 2+\sqrt 3-\dfrac 1{\sqrt 2+\sqrt 3}\right)\notin\mathbb Q$, hence $\sqrt 2+\sqrt 3\notin \mathbb Q$ OR, Suppose $\sqrt{2} + \sqrt{3}$ is rational, then so is $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6}$. Hence, $\sqrt{6}$ is rational which is of course not true. Hence we are done.

6

There is no such function and one way to show this makes use of the Baire category theorem. Assume that $f:{\mathbb A} \rightarrow {\mathbb Q}$ is continuous. For each $r \in {\mathbb Q},$ let $E_r = f^{-1}(r)$ be the inverse image of the set $\{x\}$ under $f.$ Note that ${\mathbb A} = \cup \{E_r: r \in {\mathbb Q}\}.$ Because $\mathbb A$ cannot be written ...

5

Assume that $\sqrt2 + \sqrt3 =$$p \over q$$p,q \in \mathbb{Z}$. Then $\sqrt3 = $$p \over q -\sqrt2. Squaring gives \sqrt2 =$$p^2-q^2 \over 2pq$ which is a contradiction since $\sqrt2 \notin \mathbb{Q}$

4

$b\ln a=b\int_1^a\frac{1}{x}dx=\int_1^a\frac{b}{x}dx$ Let $u=x^b$ Then $du= bx^{b-1}dx$ and $\frac{du}{u}=\frac{b}{x}dx$ Therefore, $\int_1^a\frac{b}{x}dx = \int_1^{a^b}\frac{du}{u}=\ln {a^b}$

4

Then $yz+xz+xy$ is also an integer. It follows from the Vieta relations that $x$, $y$, and $z$ are the roots of a monic cubic with integer coefficients, so they are integers. Thus we want all triples $(x,y,z)$ of positive integers such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is an integer. Now it is a short search, since the smallest of the integers ...

4

My guess is that you are thinking of the approximation $22/7 \approx \pi$. The two numbers $\pi$ and $22/7$ are not equal, so there's no contradiction in one being rational and the other irrational. Sometimes you use $22/7$ as a number "fairly close" to $\pi$, but: $$\pi = 3.14159265...$$ while $$22/7 = 3.14285714...$$

4

Here's an explicit construction of a number that is in $D$ but not in $V$: Start by setting $a_0=0, b_0=1$, and repeat the following steps for each $k\ge 1$: Let $n$ be the smallest $n$ such that $a_{k-1}<v_n<b_{k-1}$ and $v_n\ne v_k$. Let $\varepsilon = \min(2^{-(k+n_k)}, |v_n-v_k|, v_n-a_{k-1}, b_{k-1}-v_n)$. Let $a_k = v_n-\varepsilon/2$ and $b_k ... 4 A space is called ultraparacompact if every open cover can be refined to a cover by disjoint open sets. Note that clearly any fiber bundle on an ultraparacompact space is trivial (just apply the definition to an open cover on which it is trivial). So it suffices to show that$\mathbb{Q}P^n$is ultraparacompact (with either the standard topology or the ... 4 Suppose there was a smallest rational number greater than$2$. Call it$k =p/q$. Then consider$ k' = \frac{2+k}{2}$. This is a number bigger than$2$and less than$k$. Also$k'$is rational. Therefore there is no smallest rational number greater than$2$. 3 $$x\in \mathbb{Q} \Rightarrow x^2-2\in \mathbb{Q} \Rightarrow \frac{x}{x^2-2}\in \mathbb{Q}.$$ So actually$f$is$\mathbb{Q}\rightarrow\mathbb{Q}$, hence it is not on to. $$f(x)=f(y) \Longleftrightarrow x(y^2-2)=y(x^2-2) \Longleftrightarrow (y-x)(xy+2)=0.$$ So if$xy=-2$and$x\neq y$, we also have$f(x)=f(y)$, hence it is not one-one. 3 No. The polynomial$p(x)=\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)}{4}$takes integer values with$p(x) \equiv 0 \bmod 8$for all$x \not\equiv 6 \bmod 8$, but$p(6) \equiv 4 \bmod 8$. If$x$is a prime power, then$x \not\equiv 6 \bmod 8$. It follows that$\frac{p(x)}{8}$takes integer values at all prime powers, but not at$6$. 3$f$isn't Riemann-integrable but Lebesgue-integrable and indeed its integral is$1$, because$f=1$almost everywhere on$[0,1]$, since$\mathbb{Q}$is countable. 3 It depends on your definition of integral. The Riemann integral, the first one taught in calculus classes, does not have a value because the lower sum is always zero and the upper sum is always one. The Lebesgue integral of this function exists and is$1$as your intuition suggests. 3 Another reason that not all irrationals can be written as$u+v\sqrt 2$with$u,v\in \Bbb Q$: there are only countably many reals of that form, but the irrationals are uncountable. 2 (1)Look up a proof of the Baire Category Theorem for completely metrizable spaces. It's quite simple.(2) A corollary is that if$F=\{f_n :n\in N\}$is a non-empty countable family of dense open subsets of$[0,1]$, then$\cap F$is uncountable, because if$S=\{r_n : n\in N\}$is any countable set then$G=\{f_n\backslash \{r_n\} \}$is also a non-empty ... 2 Sometimes when trying to prove/disprove an "if...then" statement, it's useful to quickly check if the contrapositive is any easier. In this case, the contrapositive is: If$x$is rational, then$x^{1/3}$is rational. Is that true? 2 Given a rational number$m/n$one can go to the localization$R$(which is a PID) of$\mathbf{Z}$at a suitable prime$p$not a factor of$n$, and so$m/n$will be an element of$R$. Being a PID one can define divisibility there. 2 You can look the notion of discrete valuation, if p is a prime, write$x=p^ia/b$gcd(a,b)=1,$v_p(x)=i$. Here you can say that$x$is divisible by p if$v_p(x)>0$, if$x=m/n, gcd(m,n)=1$write$m=p^ia$, gcd(a,p)=1$x$is divisible by p in your sense if and only if$v_p(x)>0$2 It depends on what the subset is, of course. For example, no non-identity automorphism of$\mathbb{R}$fixes all of$\mathbb{R}$! But if$A\subseteq \mathbb{R}$with$|A| < |\mathbb{R}|$, then we can find a non-identity group automorphism of$\mathbb{R}$which fixes$A$. Your idea is exactly right: View$\mathbb{R}$as a vector space over$\mathbb{Q}$, ... 2 No reason to partition. First, show$0\star q=0$. Then show that$(-p)\star q=-(p\star q)$. Then show that if$n$is natural, then$n\star q=nq$, by induction. Then show that if$r=\frac{m}{n}$with$m$an integer and$n$natural, then$mq=n(r\star q)$. 2 You're wrong, the left side is not included in the right side. Since$2^{-(n+k)}$is decreasing in$k$, $$\bigcup_{k\ge 1} I_{n,k} = I_{n,1} = X \cap (v_n - 2^{-n-1},v_n + 2^{-n-1})$$ Take some$v_n$that is near$0$and$v_m$that is near$1$and you'll have$I_{n,1} \cap I_{m,1} = \emptyset$. So the right side is empty. Or did you mean$\bigcap_{k \ge ...

1

You do to complicate: $$(3x-2)(ax+b)=3ax^2+(3b-2a)x-2b$$ and thus, $$(3x-2)(ax+b)=k(x^2-2)+1\iff(3a-k)x^2+(3b-2a)x-2(k-b)-1=0\iff...$$

1

It's easier if you write $[3x-2]=3r-2$, where $r^2=2$. Then $$\frac{1}{3r-2}=\frac{3r+2}{(3r-2)(3r+2)}=\frac{3r+2}{18-4}= \frac{3}{14}r+\frac{1}{7}$$ More generally, any element of your field can be written in a unique way as $a+br$, with rational $a$ and $b$, not both zero. Then \begin{align} (a+br)^{-1} &=((a+br)(a-br))^{-1}(a+br)\\[6px] ...

1

Go through the following PDFs: http://www.ams.org/bookstore/pspdf/mbk-48-prev.pdf & https://www.dropbox.com/s/7ahnld1wtvqiurz/Continued%20fraction%20expansion.pdf?dl=0 This will clarify your issue.

1

The $7$ is irrelevant to the question. $$\sqrt{x}+\frac{1}{5-\sqrt{x}}=\frac{(25-x)\sqrt{x}}{25-x}+\frac{5+\sqrt{x}}{25-x}=\frac{(26-x)\sqrt{x}+5}{25-x}$$ Do you see a value for $x$ that makes something special happen?

1

Consider expressions like: $$1=1\star 1=\left(\underbrace{\frac{1}n+\frac{1}n+\ldots+\frac{1}n+\frac{1}n}_{n\text{ times}}\right)\star 1=\underbrace{\frac{1}n\star 1+\frac{1}n\star 1+\ldots+\frac{1}n\star 1}_{n\text{ times}}=n\left(\frac{1}n\star 1\right)$$ This allows you to see that $\frac{1}n\star 1=\frac{1}n$. Essentially, all you can say about the ...

1

Use the definition of continuity that for every sequence $\{ x_n \} _ { n \ge 1}$ converging to a given $x \in \mathbb{R}$, the image sequence $f(x_n) \to f(x)$ Note: You can use the $\epsilon - \delta$ definition only when there exists some $r > 0$ such that $( x - r, x+ r ) \subset \mathbb{domain}(f)$

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