# Tag Info

0

Leaving aside the definition and use of the term “infinitesimals”, you can see that real numbers can be constructed from rational numbers as the collection of all limits of convergent series made up by rationals. Many series, e.g. Euler’s limit of the sum of $1/n^2$ can be used to define pi as a limit. These reals can be constructed by a process defined by a ...

2

We will show more: that if $x \ne 0, 1$ is algebraic (possibly non-real), then $\sin(\log(x))$ is transcendental (regardless of which value of the logarithm we take). To do this, we apply the Gelfond-Schneider Theorem, which states that if $\alpha, \beta$ are algebraic numbers with $\alpha \ne 0,1$ and $\beta \notin \mathbb{Q}$, then any value of ...

0

It is possible to find the integer part of any number divided by two using basic arithmetic: $$\text{Floor}\left(\frac{n}{2}\right) = \frac{n}{2} - \frac{1}{4} + \frac{(-1)^n}{4}.$$ I'm not aware of a solution for division by other numbers.

2

For any three integers $m,n,b$ such that $$m \ge 1,\quad n, b > 1\quad\text{ and }\quad\gcd(m,n) = \gcd(m,b) = \gcd(n,b) = 1$$ Let $\displaystyle\;\tau\left(\frac{m}{n}, b\right)\;$ be the "period" of the expansion of $\displaystyle\;\frac{m}{n}\;$ in base $b$. Let $\lambda = \lambda(n)$ be the value of Carmichael function for $n$, we have ...

6

Let $x^2-ax=c$ and $x^3-ax=d$. Let $P(y)$ be the minimal polynomial of $x$ over $Q$. Then $P(Y)| Y^2-aY-c$ and $P(y) | Y^3-aY-d$. In particular the degree of $P$ is at most two. Case 1 $\deg P=1$, it follows immediately that $x \in \mathbb Q$. Case 2 $\deg P=2$, then $P(Y)=Y^2-aY-c$ and hence $$Y^2-aY-c| Y^3-aY-d \,.$$ This shows that $$Y^3-aY-d = ... 15 Put x^2-ax=b and x^3-ax=c with b,c\in \mathbb{Q}. As x\in \mathbb{R} is a real solution of y^2-ay-b=0, we have D=a^2+4b\geq 0. Now:$$x^3=xx^2=x(ax+b)=ax^2+bx=a^2x+ab+bx=(a^2+b)x+ab$$This is also ax+c. If we suppose that x\not \in \mathbb{Q}, we get that a=a^2+b. Hence D=a^2+4b=4a-3a^2<0 as \displaystyle a>\frac{4}{3}, a ... 0 There's already a ton of useful answers here, but I figured I throw in another stimulus for reflection. From the way you ask your question (which is very legitimate), it looks to me as you consider numerical (discrete) solutions as if they had nothing to do with the real (continuous) problem. Or, in other words, that there's no need of real numbers and ... 1 I know similar ideas already came up here, but the first thing I thought of was to construct the bijection f:\mathbb R\longrightarrow(-2,2) given by$$ f(x)= \begin{cases} -2-\frac{1}{x}&x\in(-\infty,-1)\\ x&x\in[-1,1]\\ 2-\frac{1}{x}&x\in(1,\infty) \end{cases} $$with explicit inverse$$ f^{-1}(x)= \begin{cases} ...

6

The answer to this question: (Is there a bijection between $(0,1)$ and $\mathbb{R}$ that preserves rationality?) -- that is, $f(x) = \frac{1}{x} - 2$ for $(0,\frac{1}{2})$ and $f(x) = 2 - \frac{1}{x - 1}$ on $[\frac{1}{2},1)$ -- is a bijection from $(0,1)$ to $\Bbb R$ that preserves rationality; then $f|_{\Bbb Q}$ is a bijection from $(0,1)_{\Bbb Q}$ to ...

4

Start: Pick any monotonic decreasing sequence $l_i$ of positive rationals converging to zero and any monotonic increasing sequence $r_i$ of positive rationals converging to $1$, with $l_1<r_1$. Consider $$\begin{array}{lllll} \Bbb Q= & \cdots & \sqcup & [2,-1) & \sqcup & [-1,0) & \sqcup & [0,1) & \sqcup & [1,2) & ... 4 Cantor showed that any two countable densely ordered sets with no first or last element are order isomorphic. One can build the isomorphism using his Back and Forth Method. If all we want is a bijection, we can more simply note that both sets are countably infinite. 1 "Absurd" is not quite enough. You should say that if q=\pm1 and p=\pm1 or p=\pm2 then \left(\frac{p}{q}\right)^2=1 or 4. But \sqrt{2} ^2=2, so \sqrt{2} cannot be written as \frac{p}{q}. 0 All nice and well, but people tend to forget that eventually symbols are numbers or numbers are symbols (what is called? ahh Goedel encoding). So numbers/symbols are interchangeable tokens. One can very well do math using just the "symbols" from 0 to 9 plus the "inference rules" of numerical analysis. Now the fact remains that either with "real symbols" ... 1 Without symbolic math it would be impossible to solve for instance Einstein equations for some more complex metrics that Minkowski's. Chaos theory is all about nonlinear equations. Mostly all of the physics laws are too complex to be solved exactly. Therefore we need approximations (I don't mean numerical approximations) and symbolic math to solve them. 3 Symbolic algebra is far newer than numerical approximations, and developed in Europe during the Enlightenment (ref. Florian Cajori, A History of Mathematical Notation). Even the Eighth-century Arabic Algebra of Al-Khowarizmi described its operations in words (it didn't even use numerals). This older "algebra" is in many ways more of a strategy guide for ... 2 If you want \frac{a^2}{b^2}+\frac{a}{b}+3=\frac{c^2}{d^2} with a,b coprime integers and c,d integers, then you want$$a^2+ab+3b^2=z^2$$where z is rational. Dividing through by z^2, and letting u and v be the rational numbers \frac{a}{z} and \frac{b}{z} respectively, you have the equation$$u^2+uv+3v^2=1$$On the uv-plane, this defines a ... 1 First, you need a well-defined notion of nonzero infinitesimals. Nonstandard analysis gives us such a notion, and there is a construction of the real numbers using them. The basic idea is: Construct the nonstandard version of the rational numbers. Throw away every infinite nonstandard rational number. (that is, all q such that |q| > n for every ... 0 The questions are a bit vague but let me try to answer my interpretation of one of them. You write that "infinitesimals give infinite precision", and furthermore "the real numbers require infinite precision", so why can't infinitesimals be used to construct the real numbers? Now the usual construction of the reals passes via equivalence classes of ... 0 Understand so. It is necessary to solve the equation:$$x^2+xy+3y^2=z^2$$You can easily write the formula of the solution.$$x=3s^2-p^2y=(2p-s)sz=p^2-ps+3s^2$$To write all the decisions. You should check out the rest of the formula. Their output is to use the formula for the solutions of this equation in the General form. The formula you ... 1 Symbolic mathematics has helped in discovering new physics. For example, Dirac (I think) was looking at the symbolic solutions for electrons & couldn't just ignore a sqrt(-1) issue. His solution was to postulate that electrons can have both positive & negative charge & hence we end up with a description for the positron. i.e. antimatter. 1 The middle term is a generalized continued fraction, with \{a_1,a_2,\ldots\}=\{1,1,2^2,3^2,\ldots\} and \{b_0,b_1,\ldots\}=\{0,1,1,1,\ldots\}. The recurrence relation for the numerator and denominator of the convergents gives$$\begin{align} x_n&=\frac{A_n}{B_n}\\ &=\frac{\mbox{[A024167](https://oeis.org/A024167)}}{n!}\\ ...

1

Here is a brute force method to prove the first inequality. We will see the calculations for $n=2,3,4,5$ and that will be sufficient to prove for general $n$. Here goes: $n=2$ $$\frac{7}{12} \leq \cfrac{1}{1+\cfrac{1^2}{1 + 2^2}} \iff \frac{12}{7} \geq 1+\cfrac{1^2}{1 + 2^2} \iff \frac{5}{7} \geq \cfrac{1^2}{1 + 2^2}$$ which is true. For $n=3$, we'll ...

Top 50 recent answers are included