# Tag Info

1

Pick any rational $q\in(\sqrt{p},\sqrt{2})$. Then $q^2>p$. Let $r=p/q<q$. So $rq=p$, we've chosen $q$ so that $q\in(0,\sqrt{2})$ and we've chosen $r>0$ with $r<q<\sqrt{2}$. You can find a specific $q$ as follows. If $p=\frac{p_1}{p_2}$ with $p_i$ integers, then $2-p\geq\frac{1}{p_2}$. Find a positive solution to the Pell-like equation ...

0

Assuming $\mathsf{CH}$, one can show that $\Bbb R$ is the union of countably many metrically rigid subsets by viewing it as a vector space over $\Bbb Q$. (A set $D$ in a metric space $\langle X,d\rangle$ is metrically rigid if no two distinct two-point subsets of $D$ are isometric.) This is mentioned in Brian M. Scott and Ralph Jones, Metric rigidity in ...

1

One nice application is proving the following theorem: A rectangle R with side lengths $1$ and $x$ , where $x$ is irrational, cannot be “tiled” by finitely many squares (so that the squares have disjoint interiors and cover all of R ). The proof can be found here: http://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf, as ...

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 $( x - r, x+ r ) \subset \mathbb{domain}(f)$

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 $$22/7 = 3.14285714...$$

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

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

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 !?

1

No. It is not possible. For examples, $$\pi, e, e^\pi, \pi^e$$

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.

0

No. Counterexample: $$\forall u,v \in \mathbb{Q}, \, \sqrt{3} \neq u + v\sqrt{2}.$$

0

Hint.-I would do it this way: You can start with a lesson about countable and uncountable sets. Rinse thoroughly that $\mathbb Q$ is countable and $\mathbb R$ is uncountable. Then tell them the numbers with finite decimal expansion form a "very small" subset of $\mathbb Q$. Play with that "very small" (it is infinite and equals $\mathbb Z[\frac ... 0 Well, depends what the meaning of "disaster" is. In the rationals,$ \mathbb Q$, every non-zero quantity is invertible. Any fraction can be expressed as a finite or infinite repeating decimal. Thus$1.5$has as inverse$0.(6)$. In a more theoretical POV, any linear transformation with coefficients in$ \mathbb Q$acting on elements of$ \mathbb Q$will ... 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. 0$\newcommand{\Q}{\mathbb{Q}}$Let me try and describe all the equivalence classes. Write$\alpha \in \Q$as $$\alpha = \frac{a}{b},$$ with$a, b$coprime integers. If$a$is even, then$\alpha \in A$, so it is in the class of$0$. All$\alpha$such that both$a, b$are odd are in the class of$1$, as $$\alpha - 1 = \frac{a}{b} - 1 = \frac{a - 1}{b} \in ... 1 This isn't a complete answer, as I can't classify all equivalence classes. Your professor isn't right. The equivalence relation which you have provided in fact has infinitely many equivalence classes. Take for example \frac{1}{2^n},n\in\Bbb N. I claim these numbers are pairwise inequivalent under this relation. Indeed, suppose n<m. Then we have ... 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 ... 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 ... 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 ... 1 Argue by contradiction: Suppose there is some irrational r such that f(r) \neq 0; let d := |f(r)|. If f is continuous on \Bbb{R}, then there is some \delta > 0 such that |x-r| < \delta only if |f(x) - f(r)| < d/2. But there is some rational x such that |x-r| < \delta, implying that for that rational x we have |f(x) - f(r)| ... 0 Since f is continuous, given \epsilon > 0 there is some \delta > 0 so that |f(x) -f(y)| < \epsilon whenever |x-y|<\delta. Let x \in \mathbb{R}. Then since \mathbb{Q} is dense in \mathbb{R} there is some q \in \mathbb{Q} with |x-q| < \delta. Since f(q) = 0 and since f is continuous, we have |f(x) - f(q)| = |f(x)| < ... 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)$$This allows you to see that \frac{1}n\star 1=\frac{1}n. Essentially, all you can say about the ... 2 No reason to partition. First, show 0\star q=0. Then show that (-p)\star q=-(p\star q). Then show that if n is natural, then n\star q=nq, by induction. Then show that if r=\frac{m}{n} with m an integer and n natural, then mq=n(r\star q). 0 Assuming knowledge of (a^m)^n=a^{mn}\tag{1} Now let a^m=b and a^n = c Then from (1) we have b^n=(a^m)^n=a^{mn} so \log_a(b^n)=mn therefore n\log_a b=\log_a(b^n) Hence \color{blue}{\fbox{\log_a (b^n) = n\log_a b}} I did this for general logarithms, as I think it's good practice to generalize proofs, but the above proof works ... 0 \ln(x)=\int_1^x \frac{1}{t} dt \\ \text{ so } \ln(x^r)=\int_1^{x^r} \frac{1}{t} dt \\ \text{ and then differentiating both sides gives} \\ [\ln(x^r)]'=(x^r)' \cdot \frac{1}{x^r}=\frac{r}{x} \\ \text{ now integrating both sides ... see if you can finish from here } \ln(x^r)=... \\ 4 b\ln a=b\int_1^a\frac{1}{x}dx=\int_1^a\frac{b}{x}dx Let u=x^b Then du= bx^{b-1}dx and \frac{du}{u}=\frac{b}{x}dx Therefore, \int_1^a\frac{b}{x}dx = \int_1^{a^b}\frac{du}{u}=\ln {a^b} 0 I guess you have already proved the main property$$ \ln(xy)=\ln x+\ln y $$(for positive reals x and y). By an easy induction you get that$$ \ln(a^m)=m\ln a $$for a positive integer m. If m<0, you have$$ \ln(a^m)=\ln\frac{1}{a^{-m}}=-\ln(a^{-m})=-(-m\ln a)=m\ln a $$Now suppose b=m/n, where, without loss of generality, n>0. Then$$ ... 0 $$x = \log_b(a^n)$$ $$b^x=a^n$$ $$(b^x)^{1/n}=(a^n)^{1/n}$$ $$b^{x/n}=a$$ $$x/n = \log_b(a)$$ $$x = n \log_b(a)$$ $$\log_b(a^n) = n\log_b(a)$$ 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 Hint: The vector space idea will work. Choose a basis$H$that contains$1$, map$1$to itself, interchange two other basis elements, and map the others to themselves. Extend by$\mathbb{Q}$-linearity. 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$\mathbb{Q}$, ... 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$2$. 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. 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 ... 0 Assume to get contradiction$\sqrt{3} + \sqrt{2}=\frac{p}{q}$where$p,q\in \mathbb{N}$.$\sqrt{3} + \sqrt{2}=\frac{p}{q} \Rightarrow (\sqrt{3} + \sqrt{2})^2=(\frac{p}{q})^2 \Rightarrow 3+2\sqrt{6}+2=\frac{p^2}{q^2} \Rightarrow \sqrt{6}=\frac{1}{2}(\frac{p^2}{q^2}-5)$. So now we have$\sqrt{3}+\sqrt{2}\not \in \mathbb{Q} \iff \sqrt{6} \not \in \mathbb{Q}$. ... 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}$6 Note that$\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. 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 ...

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

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.

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 $6$.

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 ...

1

If $f$ is not continuous, the Riemann integral may not exist. In this case, by splitting the range into intervals and picking numbers in the range, it is possible to pick only rational numbers, forcing the value of the integral to $0$. It is also perfectly valid to pick only irrational numbers, forcing the value of the integral to be $1$. Since no such value ...

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

$f$ 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.

0

r is rational so r = a/b for some integers a and b and assume r not equal zero. (If r is zero xr is zero and so rational) Assume $xr$ is rational and so $xr = m/n$ for some integers m and n lets divide $xr$ by $r$ ie $xr/r = (m/n)/(a/b)$ so $x = mb/na$ and $x$ would be rational which is a contradiction. So the statement "rational r by irrational x is ...

0

$\pi$ is irrational and $\dfrac{10}\pi$ is irrational, but their product is rational. In other words, you are right that it's not easy to prove that the product of two irrational numbers is rational. Proving $\pi$ is irrational is not so easy: https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational However, if $\dfrac{10}\pi$ is rational, then ...

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 ...

0

Following Jyrki Lahtonen's suggestion, let's write $f(x)=xg(x)+t$ and try to prove that $f(x)=0$ (mod m) for all $x \in \{0,1,\ldots,m-1\}$. If $gcd(x,m)=1$, then we can find a prime $p=g_q$ (mod q) for all prime powers $q|m$ and $g_q$ being a primitive root mod q. Hence we can write $x=p^n$ (mod m) for some $n$ and we are done. If $gcd(x,m)>1$, we first ...

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