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6

If you allow infinite number of operations, then you can use some algorithm. One easy example is root searching via Newton's method. Here we do the iteration $$x_{n+1} = \frac{a + x_n^2}{2x_n},$$ which eventually converges to $\sqrt{a}$ if $a$ and $x_0$ are positive. See https://en.wikipedia.org/wiki/Methods_of_computing_square_roots for other methods.


4

When you use an infinite ammount of operations, you can use: $$ \!\ \sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}. $$ from here...


11

The set of rational numbers is closed under the elementary arithmetic operations (except for division by zero), i.e. if you have two rational numbers and take their sum, difference, product, or quotient, then the result is again a rational number (again, except for division by zero). Since $\sqrt{a}$ can fail to be a rational number (say, $\sqrt{2}$), there ...


1

First, some thoughts on expanding a rational number in some base, such $s=10/27$ in bases 3 and 10. (Decimal representation is assumed unless otherwise indicated.) If we multiply by a sufficiently large power of 3 then we can turn this into an integer---in this case, $27s=10$---and then we can express the result in base-3 as $9+1=11_3.$ Dividing by $3^3$ ...


1

Any base-$n$-fraction $\frac pq$ will continue WITH a pattern, repeating with a length of at most $q-1$ repeating digits where $q$ is its denominator - no matter what base you are looking at. For example, while $\frac17=0.\overline{142857}_{10}=0.1_7$, $\frac1{10}=0.1_{10}=0.\overline{0462}_7$. If a number repeats or stops in base $10$, it will continue or ...


6

Let our point $P$ be $(\pi, 0)$. If the rational point $(a,b)$ were at a rational distance from $(\pi,0)$, then $\pi$ would be algebraic. But it is not. Detail: If the distance from $(x,0)$ to $(a,b)$ is $r$, then $x^2-2ax +a^2+b^2-r^2=0$. If $a$, $b$, and $r$ are rational, then $x$ is a root of a quadratic equation with rational coefficients. But $\pi$ is ...


-1

You can recursively define the coefficients in order that $$ p_n(x) = a_0 + a_1 x + \ldots + a_n x^n $$ has a root in $x=n$, for instance: $$ a_n = -\frac{1}{n^n}\sum_{j=0}^{n-1}a_j\, n^j.$$ If you meant that every $p_n(x)$ must completely split over $\mathbb{Q}$ for a sequence with infinite non-zero terms, it is believed that there is no chance, but the ...


1

The long division method for polynomials is same as that of numbers, except for the fact that you need to keep a track of the powers of x while performing long divison. I don't know if there exist a way to show long divison like on pen and paper, but I am putting the exact steps below. Okay, so the divisor(d) is $(2 x^2+1)$ and dividend(D) is $4x^4 + 6x^3 ...


0

$$\frac{4x^4 + 6x^3 + 3x - 1}{2x^2 + 1}=\frac{4x^4 -1 +6x^3 + 3x}{2x^2 + 1}=\frac{(2x^2+1)(2x^2-1)+3x(2x^2+1)}{2x^2 + 1}=\frac{(2x^2+1)(2x^2+3x-1)}{2x^2 + 1}=2x^2+3x-1$$


0

I'd use polynomial division, on $\frac{4x^4 + 6x^3 + 3x - 1}{2x^2+1}$, this results in, $2x^2+3x-1$ Since, that is quadratic, it should be easy from here.


0

You ask "can we say $P$ is an approximation of the interval", which is not really a well defined question. In order to talk about approximations, you must have some sort of distance defined. For example, the number $1$ is some approximation of $\sqrt 2$, but the number $1.4$ is a better approximation of the same number, because $d(1,\sqrt 2) > d(1.4, ...


0

If your function is continuous the answer is yes, as $P$ is dense in $[a,b]$. If you start by defining the function on $P$, as long as it is uniformly continuous on $P$, it can be extended to $[a,b]$. Without continuity, in general this doesn't hold.


1

Let $A_n=\{a+k\frac{b-a}{n}|k\in \{0,\ldots,n\}\}$ I assume your function, say $f$ is continuous. Consider $A=\displaystyle\cup_{n=1}^\infty A_n$ It is clear that $A$ is dense in $[a,b]$. Since $f$ is continuous, it is safe to define $f$ only over $A$ and use density to define it elsewhere. If you wish to define $f$ over some $A_N$ for fixed $N$, ...


2

Outside of mathematics, the distinction between the irrational number $\pi$ and the rational number $10^{-10000}\lfloor 10^{10000}\pi\rfloor$ is irrelevant. In everyday life, even the distinction between $\pi$ and $\frac{22}7$ is of little importance.


10

The uncountability of the reals can be used to prove the undecidability of the language $A_{TM}$ consisting of pairs $(M, \omega)$, where $M$ is a Turing machine that accepts input $\omega$. The undecidability of $A_{TM}$ is often used to prove rather applied problems are themselves undecidable. This is practical in the sense that it's quite useful to know ...


1

Everything in "real" world is finite as the universe has finitely many particles. Therefore, if you consider "mathematics" any abstraction out of "real" world then no, there are no applications outside mathematics. Now, solvig for instance PDE's that deals with problems of your day-by-day life is based on models that uses mathematical theories with many ...


1

It has many applications in computer science for example. The fact that $\Bbb{R}$ is separable (meaning it has a countable dense subset: $\Bbb{Q}$), means that many algorithms which work on the real numbers are possible through approximation.


3

See page 64; he says : We shall define the fraction $m/n$ as being that relation which holds between two inductive numbers $x, y$ when $xn = ym$. This definition enables us to prove that $m/n$ is a one-one relation, provided neither $m$ nor $n$ is zero. This symply means that, for each fraction $m/n$, for every "input" $x$ it is univocally defined ...


1

Here's a proof in the opposite direction: Every positive rational number has either a terminating or repeating decimal expansion, and the positive rational numbers $\frac{p}{q}$ (in lowest terms) which have a finite expansion are precisely where $q$ has the form $2^{a}5^{b}$ with $a,b$ non-negative integers. If $q$ has the form $2^{a}5^{b},$ then ...


1

Conjecture: integer fractions always produce terminating or repeating decimal expansions. Divide the denominator into the numerator and write down the whole number result (WNR) if any and the remainder. The remainder is a fraction with the original denominator and a smaller number as the numerator. Multiply the numerator by the first power of the base (eg ...


1

You divide the denominator into the numerator and get a remainder, then you continue to divide the numerator into the remainder (or n times the remainder) to get the decimal expansion. There are only finitely many remainders, so it will repeat or end.


11

Strictly speaking the statement has undefined meaning (because "$1/x$ is rational" has undefined meaning when $x$ can be$~0$) rather than that it is false. To make the statement meaningful, you need to ensure it is not talking about the undefined quantity $1/0$, and depending on how you do that this can make the statement false or true. Since in mathematics ...


24

$x = 0$ is a rational number but $\dfrac{1}{x} = \dfrac{1}{0}$ isn't defined.


4

HINT: What happens in the case that $p=0$?



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