# Tag Info

73

In base $n$ the numerator is $$p = n^{n-1} - \frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = \frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$ Note that $p = (n-2)q + n-1$ and for the quotient we get $$\frac{p}{q} = n-2 + \frac{(n-1)^3}{n^n} \frac{1}{1 - \frac{n^2-n+1}{n^n}} = n-2 + \frac{(n-1)^3}{n^n} \sum_{k=0}^{\infty} \left(\frac{n^2-n+1}{n^n}\right)^k.$$ ...

34

A nice generalization of the fundamental theorem of arithmetic is that every rational number is uniquely represented as a product of primes raised to integer powers. For example: $$\frac{4}{9} = 2^{2}*3^{-2}$$ This is the natural generalization of factoring integers to rational numbers. Positive powers are part of the numerator, negative powers part of the ...

32

There are infinitely many primitive Pythagorean triples, that is, triples $(a,b,c)$ of positive integers such that $a$, $b$, and $c$ are positive integers $\gt 1$ such that $a^2+b^2=c^2$. Any such triple determines a right triangle. The sines and cosines of the two non-right angles are the rationals $\frac{a}{c}$ and $\frac{b}{c}$. So there are infinitely ...

30

Let $$S_n(a)=1 +2a+\ldots +na^{n-1}=\frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$ $$T_n(a)=a^{n-1}+2a^{n-2}+\ldots +n=a^{n-1}S_n(a^{-1}).$$ Then $$\frac{S_n(a)}{T_n(a)}=\frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$ For $a=10,n=9$ we have $$\frac{S_n(a)}{T_n(a)}\approx\frac{8\cdot 10^{10}+1}{10^{10}}.$$

29

Here's one example of where the difference between rational numbers and irrational numbers matters. Consider a circle of circumference $1$ (in any units you choose), and suppose we have an ant (of infinitesimal size, of course) on the circle that moves forward by $f$ instantaneously once per second. Then the ant will return to its starting point if and only ...

21

Just to add to the excellent answers above, some examples: ${987654321\,/\,123456789}\approx 8.00000007290000066339$ ${{87654321}_9\,/\,{12345678}_9}\approx {7.000000628000056238}_9$ ${{7654321}_8\,/\,{1234567}_8}\approx {6.0000052700046137}_8$ ${{654321}_7\,/\,{123456}_7}\approx {5.00004260036036}_7$ ...

20

The number $\sqrt{-1}$ is not real. Since the rationals are just a particular type of real number, it cannot be rational, either. Another way to look at it: Were there integers $a$ and $b$ such that $\sqrt{-1} = \frac{a}{b}$ then $$-1 = \frac{a^2}{b^2},$$ and so $$a^2 = -b^2.$$ Since $b^2$ is certainly positive, that means $a^2$ is certainly negative, ...

19

In the ancient Greek sense, if we have two quantities $x$ and $y$, then the quantity $z$ is a common measure of $x$ and $y$ if each of $x$ and $y$ is an integer multiple of $z$. The quantities involved were not thought of as numbers, but of course they were what we think of as positive. So for example the diagonal of a square, and the diagonal of a square ...

19

If $x^3 - 5x = y^3 - 5y$ with $x\neq y$ then $x^3-y^3 = 5(x-y)$ or $x^2+xy+y^2=5$. Solving for $x$ in terms of $y$: $$x = \frac{-y\pm \sqrt{y^2-4(y^2-5)}}2$$ So $20-3y^2$ has to be a square of a rational number. This is equivalent to finding integer solutions to $20q^2 - 3p^2= n^2$ with $(p,q)=1$. Show that this isn't possible $\pmod 5$.

18

Let's assume $x = p/q$. $p$ and $q$ integers without a common factor. Then, $$p^{3} + p^{2}q = q^{3}$$ It's is only satisfied whenever $p$ and $q$ are simultaneously even. It contradicts the initial hypothesis that we can set $x = p/q$ where $p$ and $q$ has not common factors. $$\mbox{Then,}\quad x \not\in {\mathbb Q}$$

18

It follows rather easily from this fact: for every $0\neq n\in\mathbb N$ the number $e^{i/n}$ is transcendental. If we suppose it: if $\tan y=2\tan x$ then $(e^{2iy}-1)(e^{2ix}+1)=2(e^{2ix}-1)(e^{2iy}+1)$. If $x\neq0$ and $y$ are rational and $n$ is the least common denominator of $x$ and $y$, this becomes a polynomial equation for $e^{i/n}$, hence we get a ...

15

Hint: Suppose to the contrary that $\dfrac{a}{b}=n$, where $a$, $b$, and $n$ are integers. Suppose also that $a$ is odd, while $b$ is even (and of course non-zero). Then $a=bn$. See whether you can show this is impossible. Here you will be using the fact that $a$ is odd and $b$ is even.

15

Try the function $f:\mathbb Q\to\mathbb Q_{\gt0}$ defined by $f(x)=x+1$ if $x\geqslant0$ and $f(x)=1/(1-x)$ if $x\lt0$. Thus, $f$ sends $0$ to $1$ and sends bijectively $\mathbb Q_{\gt0}$ to $\mathbb Q_{\gt1}$ and $\mathbb Q_{\lt0}$ to $\mathbb Q\cap(0,1)$. Edit: For a bijection $g:\mathbb Q_{\gt0}\to\mathbb Q_{\geqslant0}$, consider $g(x)=x-1$ if $x$ is an ...

14

A supplement to the answer by Chris above: Let $r$ be a positive rational number and $i$ a positive irrational number. If $r^{i}$ is rational, then $r^i=\frac{a}{b}$ for $a,b\in \mathbb{Z}$ such that $b\neq 0$. In particular, $i=\log_{r}\left(\frac{a}{b}\right)$. Therefore, Chris Eagle's answer is, in fact, prototypical. (Note also, that if $r=1$, then we ...

14

No. There isn't even a non-constant continuous function $\mathbb R \rightarrow \mathbb Q$. Recall a continuous function maps connected sets to connected sets. We know that $\mathbb R$ is connected, but $\mathbb Q$ is totally disconnected: each point is its own connected component. So any continuous map $\mathbb R \rightarrow \mathbb Q$ must be constant.

13

Among all fields the rationals can be characterized as follows. The field of rational numbers is, up to isomorphism, the smallest field of characteristic $0$. As you say, the rationals can be built up from the integers via a construction which is a special case of the construction known as the field of fractions of an integral domain. In this way the ...

13

Consider the fraction $1/m$. Write $m=2^a 5^b v$ with $\gcd(v,10)=1$. Then the periodic part of $1/m$ has length $e$, where $e$ is the smallest positive number such that $v$ divides $10^e-1$. The non-repeating part has length $f=\max(a,b)$. There are no easy formulas for either $e$ or $f$ in terms of $m$.

12

$$\frac{x}{x^2+1}$$ the inverse of Will Jagy's $x + \frac{1}{x}$. I like $$x - \frac{1}{x} = \frac{x^2-1}{x}$$ because it contains sign information (input magnitude greater or less than one) that you may choose to ignore, and gives a nice zero for $x = \frac{1}{x} = \pm 1$. If you choose to take the absolute value of it (ignoring the sign) it gives you ...

12

Let $a$ and $b$ be distinct irrationals; we lose no generality to suppose $a<b$. Assume they have equal integer parts, since otherwise there is an integer between them and the question is trivial. They have infinite decimal expansions, $.a_1a_2a_3\ldots$ and $.b_1b_2b_3\ldots$. These cannot agree in every position since otherwise $a=b$. So say they ...

12

Suppose that the decimal is $x=a.d_1d_2\ldots d_m\overline{d_{m+1}\dots d_{m+p}}$, where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then $$10^mx=a+d_1d_2\dots d_m.\overline{d_{m+1}\dots d_{m+p}}\;,\tag{1}$$ and 10^{m+p}x=a+d_1d_2\dots d_md_{m+1}\dots ...

12

Euclid's algorithm (originally formulated using repeated subtractions rather than divisions) can be applied to any pair of comparable quantities (two lengths, two masses, two frequencies): just keep subtracting the smaller of a pair from the larger and then replace the larger one by the difference found; stop when the difference becomes $0$, returning the ...

12

Your proof does not work. Indeed, subtracting $1$ from $\frac p q$ will give you a rational number, but it will be negative by assumption, so this doesn't help you (since it doesn't give you a contradiction). A simpler approach: Explicitly state what the infinitely-many positive rationals less than $x$ are. Hint: If $y$ is a positive rational, what can you ...

11

The problem $N=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ with $N,x,y,z \in \mathbb{Z}$ was considered by Andrew Bremner and Richard Guy in "Two more representation problems" published in the Proceedings of the Edinburgh Mathematical Society, vol. 40 pp.1-17 in 1997. An online copy is available here. They showed solutions only occurred for those $N$ where the ...

11

In fact each of $\cos x$, $\sin x$, and $\tan x$ are irrational at non-zero rational values of the arguments. This result is Theorem 2.5 and Corollary 2.7 in Ivan Niven's Irrational Numbers.

11

If $\sin x$ is rational (or even just algebraic), then $\cos x=\pm \sqrt{1-\sin^2 x}$ is algebraic. Therefore $e^{ix}=\cos x+i\sin x$ is algebraic, so by the Lindemann-Weierstrass theorem, $x$ cannot have been nonzero algebraic -- in particular not nonzero rational.

11

My first observation is that you’re getting badly bogged down in symbols. For starters, there is absolutely no reason to replace the clear statement of the problem with the symbolic expression $\forall x \in \mathbb{Q}_{\gt0} \ \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$; that’s just introducing unnecessary obstacles for the reader. The same goes for your ...

11

Please lead me through it step by step. \begin{align*} -100+\frac{1}{2} &= \frac{-200}{2}+\frac{1}{2}\\ &= \frac{-200+1}{2} \\ &=\frac{-199}{2}\\ &=-\frac{199}{2}\\ &= -\frac{198+1}{2}\\ &= -\left(\frac{198}{2}+\frac 12\right)\\ &= -\left(99+\frac{1}{2}\right)\\ &= -99\frac12 \end{align*}

10

A common measure of how "complicated" a (reduced) fraction is is the height: Definition. Let $\frac{r}{s}$ be a rational number, with $\gcd(r,s)=1$. The height of $\frac{r}{s}$ is $\mathrm{ht}\left(\frac{r}{s}\right)=\max\{|r|,|s|\}$. Among those of the same height, you can order them by comparing the minimum. For those with the same minimum, you can ...

10

The decimal expansion of a rational number is always repeating (we can view a finite decimal as a repetition of $0$'s) If $q$ is rational we may write it as an irreducible fraction $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$. Consider the Euclidean division of $a$ by $b:$ At each step, there are only finitely many possible remainders $r\;\;(0\leq r< b)$. ...

10

One thing is in how you construct them. Starting from the natural numbers (and $0$) you construct the integers by saying that $\mathbb{Z}$ is the smallest set that contains the naturals and is a group under addition. Similarly, the rationals $\mathbb{Q}$ is the smallest set containing $\mathbb{Z}$ that forms a group under multiplication (when $0$ is taken ...

Only top voted, non community-wiki answers of a minimum length are eligible