Tag Info

Hot answers tagged

19

I applaud your genuine curiosity. There is no general method. To compute the decimal digits for any particular irrational number you start with a definition of the number. For $\pi$ there are lots of methods known. @GregoryGrant 's answer points to one of the oldest and most famous. Archimedes started with inscribed and circumscribed regular hexagons, then ...


19

Here is a thing you can always do but which people don't seem to teach: if you have a proof that you're suspicious of, you can go through the proof with an example. At some point, you'll write down a false statement about your example, and that's probably where the mistake is. (Of course you can't do this if the proof is a proof by contradiction because, ...


18

Here's how Archimedes did it 2,500 years ago. $\pi$ is the circumference of a circle with diameter one. What Archimedes did was to inscribe a polygon with $n$ sides inside the circle. Like this for $n=5$. He then calculated the circumference of the polygon. This gives an approximation to $\pi$ for each $n$ and as $n$ grows it gets closer and closer to ...


17

"Then by the pigeonhole principle, $\{10^rx\}=\frac aq$" is a non-sequitur. The principle says that among too many pigeons, two must be in the same hole, but it does not state thet two pigoens must be in the first hole.


12

You generally need to use arbitrary precision arithmetic to compute large numbers of digits of typical irrational numbers. The exception is oddball things like the Champernowne constant 0.12345678910111213141516… :) There are various arbitrary precision arithmetic packages available. Internally, they use big arrays or lists to hold groups of the digits of ...


10

For some irrational numbers, like $\pi$, there are convenient infinite series that converge to them. So for example $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$ By adding up more and more terms of this series you get closer and closer to $\pi^2/6$. You can take an estimate for some value of $n$, multiply by $6$ and then take the square root. That ...


10

You read the clock wrong (though it's not really your fault). ...


9

No, it's simpler than that. Suppose $y$ is composite, $y=ab$. Then $$ 1+x+x^2+\cdots+x^{ab-1} = (1+x+x^2+\cdots+x^{b-1})(1+x^b+x^{2b}+\cdots +x^{(a-1)b}) $$ and therefore the value of the LHS is composite too.


8

Note the following $$am+bn=1 \implies axm +bxn=x$$ $$abq_1m+baq_2n=x$$ $$ab(q_1m+q_2n)=x$$ i.e., $ab|x$.


6

Well, you need some property. Pick a random number from $1$ to $n$. Given prime divisor $p$ of $n$, what is the probability that your random choice is not divisible by $p$? It is $1-\frac1p$. Then you need to prove that these probabilities are independent - that is, if we pick a random number from $1$ to $n$ not divisible by $p_1,\dots,p_k$, then the ...


5

Construction of the sequence : Denote $p_n$=nth-prime number. Search a number $u$ with $u\equiv 1\ \ mod\ (\ p_1...p_k\ )$ $u\equiv 2\ \ mod\ (\ p_{k+1}...p_{2k})$ ... $u\equiv m\ \ mod\ (\ p_{(m-1)k+1}...p_{mk})$ This is possible because of the chinese remainder theorem. Then $u-m,...,u-1$ have at least $k$ distinct prime factors.


5

Mertens' Theorem says: $$\lim_{n \rightarrow \infty} \ \frac{1}{\log p_n} \prod_{k=1}^{n} \frac{1}{1 - \displaystyle{\frac{1}{p_k}}} = e^{\gamma}.$$ Euler's product formula for the $\zeta$ function and his evaluation of $\zeta(2) = \pi^2/6$ says that $$\zeta(2) = \lim_{n \rightarrow \infty} \ \prod_{k=1}^{n} \frac{1}{1 - \displaystyle{\frac{1}{p^2_k}}} ...


5

Continued fractions are commonly used in order to get the decimal value with enough accuracy. For example, the A001203 sequence in the Sloan's online encyclopedia represents $\pi$ in a form of continued fraction (actually, all infinite, rational numbers can be decomposed into a finite continued fraction, whereas irrational numbers cannot).


5

Hint: Choose any even number to be $y$. Then $1+y^2$ is odd and is therefore the difference between two successive squares.


5

Well, $an\equiv 0\pmod{n}$, so probably you want $b\not\equiv0\pmod{n}$ such that $ab\equiv0\pmod{n}$. Consider $d=\gcd(a,n)$; then $a=cd$ and $n=bd$ for some integers $c$ and $b$; then $$ ab=cdb=cn\equiv0\pmod{n} $$ and $b<n$, so $b\not\equiv0\pmod{n}$.


5

Write $2014 = 2N$, where $N = 1007$ is odd. I claim that $$f(x) = N x^2 + 2 x = 1007 x^2 + 2x$$ has the required property. Note that $f(x)$ is odd if $x$ is odd and even if $x$ is even. Hence, if $f(i)$ and $f(j)$ have the same remainder when divided by $2014 = 2N$, then $i$ and $j$ have the same parity, and so $i-j$ is divisible by $2$. Suppose that ...


4

Hint Let $y=2k, z=2k^2$ and then complete the square on right hand side.


4

Not only are the answers from the Mathematics StackExchange community "more intuitive and instructive," they're much more valid than anything you will find on Wikipedia. Although it's true that a lot of the "community" here are also active on Wikipedia, their talents and insights are mostly wasted over there. There is much tighter control here than there. ...


4

There is no general way to compute the decimal expansion for an arbitrary member of the set of irrational numbers, I. :( --Proof Start-- Here's a wishy washy proof (The statements that most obviously lack support are marked with a "*", so if anyone's interested, you should look into them yourself): Consider R to be the set of real numbers. I is a proper ...


4

By the Prime Number Theorem $$ \pi(x)\sim\frac{x}{\log x}. $$ Since $(\sqrt k-1)^2=k-2\,\sqrt k+1<k$ we have $$ \pi(2\,k)-\pi((\sqrt k-1)^2)\ge\pi(2\,k)-\pi(k)\sim\frac{2\,k}{\log k+\log2}-\frac{k}{\log k}\sim\frac{k}{\log k}\quad\text{as }k\to\infty. $$


4

To help you get you started: If $n \le x < n+\dfrac{1}{3}$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = 2n \cdot 3n = 6n^2$. If $n+\dfrac{1}{3} \le x < n+\dfrac{1}{2}$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = 2n(3n+1) = 6n^2+2n$. If $n+\dfrac{1}{2} \le x < n+\dfrac{2}{3}$ for some integer $n$, ...


3

The following are some bits and pieces from Section 2.6 of my book. Here I will demonstrate what the OP is trying to do, but with another elliptic curve. Hopefully the OP can replicate this for the elliptic curve above. Let $E/\mathbb{Q}$ be the elliptic curve $y^2=x^3+3$. Its minimal discriminant is $\Delta_E=-3888=-2^4\cdot 3^5$. Thus, the only primes of ...


3

Let $n=\prod_{i=1}^{k}p_{i}^{a_i}$ where the $p_i$'s are distinct primes and the $a_i$'s are all $\geq1$. Let $A_i$ denote the subset of $\{1,2,\ldots, n\}$, all of whose elements are divisible by $p_i$. Then the set $A := \{m|1 \leq m \leq n, {\rm gcd}(m,n) \gt 1\}$ is precisely the union of the $A_i$'s and we have, from the principle of ...


3

Yes this can be shown via inclusion-exclusion but is a bit cumbersome. I will spell it out up to three distinct prime-divisors only. First recall that $1\le a \le n$ is coprime to $n$ when it is not divisible by any prime $p$ that divides $n$. Let $p \mid n$. The number of $1\le a \le n$ such that $p \mid a$ is $n/p$ so the number not divisible by $p$ ...


3

The theorem itself is partially hinting on the multiplicative property of $\varphi$ since it is asking you to separate in the primes. I don't know if the following argument is useful. Let $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_r^{\alpha_r}$ then a number $a$ is coprime to $n$ if and only if $a$ is congruent to a number that is not a multiple of $p_i\bmod ...


3

Let $(x_0,y_0)$ be a particular solution, and let $(x,y)$ be any solution. Then $ax_0+by_0=ax+by=N$. Subtracting, we find that $a(x-x_0)=-b(y-y_0)$. Let $a=da'$ and $b=db'$. Then $a'$ and $b'$ are relatively prime, and $$a'(x-x_0)=-b'(y-y_0).$$ Note that $b'$ divides $a'(x-x_0)$. Since $a'$ and $b'$ are relatively prime, we conclude that $b'$ divides ...


3

I think there are two ways to answer this question. I am atleast not sure that there are more in number theory in other fields. However as number theory is close to basic arithmetic at times the conjectures are easier to understand and thus they get more famous, so it simply appears as there are more conjectures than in other fields. If 1. is false i.e. ...


3

It should be $$\prod_{m\geq 0}(1+z^{2^m}),$$ since: $$ (1-z)(1+z)(1+z^2)\cdot\ldots\cdot(1+z^{2^N}) = 1-z^{2^{N+1}}.$$


3

Putting $ x = y = 0 $, we get $ f(0) = 0 $. Putting $ y = 0 $, we have $ f(x^2) = x^4 $. Putting $ z = x^2 $, we have $ f(z) = z^2 $ for all $ z \geq 0 $. $$\therefore f(2015) = 2015^2= 4060225$$



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