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Even more is true: the leading digits of $p^n$ can be any sequence you want. Note how $$\text{the leading digits of }p^n\text{ are }\overline m$$ is the same as saying $$\overline m\cdot 10^k\leq p^n<(\overline m+1)\cdot 10^k\text{ for some }k\geq0.$$ Equivalently, $$\log_{10}(\overline m)\leq n\log_{10}(p)-k<\log_{10}(\overline m+1).$$ Indeed there ...

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You get this because the golden ratio is a quadratic algebraic integer. Let $\tau=(1+\sqrt5)/2$ and $\overline{\tau}=(1-\sqrt5)/2$. These are the zeros of the polynomial $$p(x)=(x-\tau)(x-\overline{\tau})=x^2-x-1.$$ Let then $a/b$ be a rational number (in reduced form). Because $\tau$ is irrational we get that $p(a/b)\neq0$. But $$p(a/b)=(\frac ... 2 Fleshing out my comment into an answer: we know that e^9\approx 8100, so that the integer part of e^{e^9} is roughly 12,000 bits. This means that we can get a value to sufficient accuracy by: Computing e to 'sufficient' accuracy (being conservative, 2^{16} bits should be enough here; it's not hard to do a more accurate error analysis, but the ... 4 We could prove it is not an integer by providing explicit bounds. If you really wanted to, you could note the following common inequality, holding for all positive n:$$\left(1+\frac{1}n\right)^n<e<\left(1+\frac{1}n\right)^{n+1}.$$Which yields a series of rational bounds for e, where both sides converge to e as n heads to infinity. The point ... 1 ((x-a)(x-b)(a-b))^2 = -23 which is not a square in \Bbb Q. And so \sqrt{-23} \in \Bbb Q(x,a). This extension must have even degree over \Bbb Q. If \Bbb Q(x,a) were equal to \Bbb Q(x) it would have degree 3, which is impossible. Hence those fields are different, and so a \notin \Bbb Q(x). d cannot be a square in \Bbb Q(x). 1 Presumably you're not allowing a_n = b_n = c_n = 0. Clearly yes, in some sense: if you restrict the possible (a, b, c) in some way (e.g. a bound on \max(|a|,|b|,|c|)) that leaves finitely many possible choices, one of them must have the smallest value of |at^2 + b t + c |. If, for example, you impose a = b, you're essentially approximating t^2 + ... 0 the following series is used for \sqrt{3}$$\sum_{n=0}^{\infty }\frac{(2n-1)!!}{n!3^n}=1+1/3+1/6+5/54+35/684+..$$1 I outlined a really easy to use method originally created by the Babylonians to approximate the square root of a number N in this answer. Here's the idea: make an initial guess as to the square root, and let this initial guess be given by r_1. Now, continue improving your guesses by using$$ r_{n+1}=\frac{1}{2}\left(r_n+\frac{N}{r_n}\right),\tag{1} $$... 1 Brahmagupta's equation gives such approximations. An integer solution for X^2-dY^2=1 leads to the rational number X/Y an approximation for \surd d.$$49-3\times16=17^2-(4\surd3 )^2=1(7-4\surd3)(7+4\surd3)=1$$Now raising to an arbitrary power$$ (7-4\surd3)^n(7+4\surd3)^n=1$$Expressing (7+4\surd3)^n in the form a_n+b_n\surd3 you can ... 2 To test if x>\sqrt3, it suffices to check if x^2>3, since these are equivalent when x is positive. With that in mind, I then see if 1.00>\sqrt3, if 1.01>\sqrt3, if 1.02>\sqrt3, if 1.03>\sqrt3, in order, until I find the smallest one that's bigger than \sqrt3. I then take the one immediately before that. Hey, nobody said ... 4 May be, we could consider solving for x$$f(x)=x^2-3=0$$and use Newton method which will give$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{x_n^2+3}{2 x_n}$$Starting with x_0=1, we should get as iterates 2, \frac{7}{4}, \frac{97}{56}, \frac{18817}{10864}, \frac{708158977}{408855776}. Using Halley method instead, which will give ... 8 It is an overkill for sure, but since the continued fraction of \sqrt{3} is given by:$$ \sqrt{3}=[1;\overline{1,2}],\tag{1}$$we have that:$$ [1,1,2,1,2,1,2] = \frac{71}{41} \tag{2} $$is an accurate approximation, and:$$\left|\sqrt{3}-\frac{71}{41}\right|\leq\frac{1}{41^2}<\frac{1}{1000}.\tag{3}

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There are $100$ distinct rational numbers between $1$ and $2$ of the form $1.XY5$. Square them and find where the square changes from below $3$ to above $3$. Take the average of the largest number whose square is below $3$ and the smallest number whose square is above $3$.

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One way is: use decimal search to find the first 2 decimal places, then multiply by 100, truncate to an integer, and use this value over 100. Decimal search is: Find each digit by seeing if a certain value is two high or two low. Eg: calculate $1.5^2$. If this is < 3, try $1.6^2$, $1.7^2$ until we find a value which squares to more than 3. Then do the ...

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