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2

Consider the primes, written in order: $p_1, p_2, p_3,\ldots$ That is, $p_1=2, p_2=3, p_3=5, p_4=7, p_5=11, \ldots$. Now consider the sequence $$\frac{p_1-1}{p_1}, \frac{p_2-1}{p_2}, \frac{p_3-1}{p_3},\ldots$$ Much like in the OP, this is a sequence of fractions, all irreducible, with an increasingly larger prime dividing the denominator. Yet the limit ...


6

You proof idea breaks because it could prove that $0\notin\mathbb Q$ by considering the sequence $$a_n = 2^{-n}$$ We have $$\lim_{n\to\infty} a_n = 0$$ but the denominator is exactly $2^n$, wich diverges. Generally, you assume $$\lim_{n\to\infty} \frac{a_n}{b_n} = \frac ab \Rightarrow \lim_{n\to\infty} a_n = a \wedge \lim_{n\to\infty} b_n = b$$ wich is ...


0

The correct answer to the question what the circumference of a circle with diameter $d$ would be $\pi \cdot d $. Of course this is not a satisyfing answer. But since this ridiculous number $\pi$ cannot even be described by the root of a polynomial with coefficients in $\mathbb Q$ we can only approximate $\pi$. This is not a bad thing though. For most, if not ...


16

As I mentioned in the comments, the sizes of the squares are a little more complicated than what you're hoping for. I don't know of a simple expression for the size of each square, but you can get each one by solving a quadratic equation. I wrote a program to do so; it draws all the squares whose sizes are above a small threshold. Here's the result. Hope it ...


1

Your way of filling the corners will only (as far as I can see) a triangular part of them. It's going to be hard to fill them with squares. Chris says the same in his comment except that he actually tries to tell you what you might do (and he caught you miscounting on the number of smaller squares in each step).


2

The integral of any function over a set with measure $0$ is equal to $0$.


3

What will happen if $$(C+4)\pi^2+B\pi+(A-48)=0?$$


0

Ah, 42 and The Hitchhiker's Guide to the Galaxy. Would you like to know how this is connected to the 24-dimensional Leech lattice? :) Given the Ramanujan-type formula, $$\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{An+B}{(C)^{n+1/2}}=\frac1\pi$$ there are relatively simple expressions for $A,C$. Define, $$A(\tau) = ...


2

The formula, $$\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = \frac{1}{\pi}\tag1$$ or its equivalent form, $$\sum_{k=0}^\infty \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^8)^{k+1/2}} = \frac{4}{\pi}\tag2$$ and the similar, $$\sum_{k=0}^\infty (-1)^k \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^9)^{k+1/2}} = \frac{8}{\pi}\tag3$$ belong to the Ramanujan-type ...


1

After you get the angle at or below $2\pi$ the reference angle is just the angle $0\le x\le\frac{\pi}{2}$ that is the distance between your angle and the x axis. So in quadrant I it is just x. Quadrant II it is $\pi -x$ Quadrant III it is $x-\pi$ Quadrant IV it is $2\pi - x$ Some example reference angles in degrees... 30 is 30. 90 is 90. 95 is 85. ...


0

In quadrant III, subtract $\pi$ from the value. In quadrant II, subtract the value from $\pi$. You don't need to subtract anything if it's in quadrant I! Note: I may have misunderstood the question - I'm not sure if this is right.


7

$2\sqrt2$ and $\sqrt2$ are two distinct irrational numbers s.t. the statement: '$2\sqrt2+\sqrt2$ is rational or $2\sqrt2-\sqrt2$ is rational' is not true. $\sqrt2$ and $2-\sqrt2$ are two distinct irrational numbers s.t. the statement: '$\sqrt2+(2-\sqrt2)$ is rational or $\sqrt2-(2-\sqrt2)$ is rational' is true. This illustrates that the statement cannot be ...


12

This seems to be an open problem. It is a conjecture that the statement is false, i.e. that $\pi + e$ and $\pi - e$ are irrational. According to Wikipedia this remains unproven. (Just imagine the impact of the discovery of an equation such as $\pi=e+\frac{4233108252.........3123782}{31238295213.......0591231}$ ... unbelievable!) Remark that at least one of ...


1

$\tau\equiv\dfrac{C}{R}=2\pi$. See tauday.com It is a symbol used by $\tau$ists that define $\tau$ to be the true circle constant.


1

This is probably not the standard way at all but one possible way to derive the value for $ \pi $ is to consider the following diagram. We know the circumference of the circle is $2\pi r$ so we set $|OA|= 1$ Now we get a poor approximation to by saying it is: $2 \cdot |AB| = 2 \cdot \sqrt{2} \approx 2.82$ If we bisect AB at E and draw a segment from O ...


10

Yes, of course there is a 0 in the decimal expansion of $\pi=3.1415926535897932384626433832795\underline{0}2884197...$.


0

A number of different ways of showing this exist. First notice that the area has to be $(\text{constant}\cdot r^2)$ because the area of a region of any shape in a plane must be proportional to the square of the distances. E.g. if you multiply all distances by $3$, then the area is multiplied by $9$. And "constant" in this case means it's the same number ...


2

In around 250 BC, Archimedes expressed $\pi$ as a limit. He constructed sequences of inscribed and circumscribed polygons whose perimeters were lower and upper bounds of the value of $\pi$ respectively, such that the perimeters converged to the same value. I do not know how rigorously the ancient Greeks were capable of proving that the perimeters truly were ...


3

Say you're trying to approximate a number by rational numbers $p/q$. Usually, the bigger $q$ is, the better your chance of approximating the number closely. On the other hand, the smaller $q$ is, the simpler the approximation. In the case of $\pi$, if you want to have a better approximation than $22/7$, you have to go all the way up to $q = 57$. (See this.) ...


2

3.14 is two decimal places. That's the only justification I can give. When doing hand calculations (or sliderule calculations), carrying out more than three significant digits is cumbersome. Remember this: with $N$ digits in a multiplication, you have $N^2$ digits in the final answer before you can round again. In the world of engineering, often you can't ...


4

This is the same as proving that $$\frac{1}{16}(133-37\sqrt{5})>\pi$$ and it follows from the fact that the continued fraction of the LHS is: $$ [3;7,15,1,660,\ldots] $$ while the continued fraction of $\pi$ is: $$ [3;7,15,1,292,\ldots].$$


2

If you were actually generating random real numbers, the probability of any of those falling exactly on the circle line (a set of measure zero) would be zero and thus by the law of large numbers not alter the outcome of the Monte Carlo simulation as the number of points goes to infinity, independent of whether you count those points as "inside", "outside", ...



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