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Let $a = m/n$ and $b = p/q$ where $(m,n) = 1 = (p,q)$. The equation translates to: $(mq)^2 + (np)^2 = (3nq)^2$. So these are Pythagorean triples and the general solution can be found ( H.Stark book ) to be: $mq = x^2 - y^2$, $np = 2xy$, and $3nq = x^2 + y^2$ where $x > y > 0$ and $x, y$ are naturals. So this gives: $p/q = 6xy/(x^2 + y^2)$, and $m/n = ... 1 For any rational number$b$we want to find$a^2+b^2=C \rightarrow a^2=C-b^2\rightarrow |a|=\sqrt{C-b^2}$For what values of$b$is$\sqrt{C-b^2}$rational ? Let$b=\frac{p}q$where$p$and$q$are relatively prime.$|a|=\sqrt{C-b^2}=\sqrt{\dfrac{Cq^2}{q^2}-\dfrac{p^2}{q^2}}=\dfrac{\sqrt{Cq^2-p^2}}{|q|}\in\mathbb{Q}$Therefore$Cq^2-p^2$must be a ... 2 Hint:$3^2 + 4^2 = 5^2$. Scale... 3 Clearly, and almost vacuously$b=\pi+q$for some rational number$q$. You can easily show that these are the only ones. 1 Your idea about prime numbers is a good one! Let$T$be the set of primes which appear in either the numerator or denominator of some element of$S$. Then$T$is finite (why?) so there is some prime$p \not \in T$. Show by unique factorization that$p$cannot be the product of powers of elements of$S$. 3 It is$\dfrac1{789}$in decimal form. A better approximation for$\dfrac{22}7-\pi$would be$\dfrac1{790\frac56}=\dfrac6{4745}$. But the best is Ramanujan's:$\pi\simeq\dfrac{\ln(640,320^3+744)}{\sqrt{163}}$, with a precision of$30$decimals. Hope this helps! As an aisde, I know a short one for$\gamma\simeq\dfrac1{\sqrt3}$, with an error of about ... 1 No set of more than one rational number is independent, so there is no basis, the rationals are not a free module over the integers. However, your argument that there is no finite generating set is correct. 1$\mathbb{F}_2[x]$as an additive group is countable. It has a number of elements of order$2$, which shows it is not isomorphic to$\mathbb{Q}^+$. Construct a bijection between these sets and you're good. 2 Short answer: Integers have a finite length, so the question is meaningless. An (non-negative) integer is, by definition, a cardinality of a finite set, and this can always be represented by finitely many digits. There are several things in mathematics which look a bit like an infinite integer, but none of them help save the idea. You could think of ... 4 There is no such thing as an "infinitely long integer". -1 The set of rationals is dense, which means there is a rational number between any two rational numbers. let S be sum of all rationals in the interval (1,10) => Every rational number in the sum is greater than 1 => S is sum of infinite rational numbers each of which is greater than 1 Hence, sum of rationals over (0,10) cannot be finite 6 Isn't $$f(x)=\frac{1}{x^3+3}$$ a counter example? A "non-trivial" example would be $$f(x)=\frac{ax^3}{x^3+3}$$ It is easy to see that this is injective when$a \neq 0$since it can be written as $$f(x)=a-\frac{3a}{x^3+3}$$ Moreover, as$f(x)$is decreasing as a real valued function, if$P(x)$is any polinomial with$P'<0$then ... 1 It works for any function$f(x) = g(x, \frac{1}{x})$where$g$is symmetric (i.e.,$g(x, y) = g(y, x)$). 2 Let$f(x)$be arbitrary for$-1\leqslant x\leqslant 1,$and define$f(x):=f(1/x)$otherwise. 2 For lack of anything worse than this,$f(x)=(x - \frac{1}{x})^2$. 9 For lack of a better idea,$f(x)=|\ln x|$. 12 $$\frac{x}{x^2+1}$$ the inverse of Will Jagy's$x + \frac{1}{x}$. I like $$x - \frac{1}{x} = \frac{x^2-1}{x}$$ because it contains sign information (input magnitude greater or less than one) that you may choose to ignore, and gives a nice zero for$x = \frac{1}{x} = \pm 1$. If you choose to take the absolute value of it (ignoring the sign) it gives you ... 8 $$\frac{1}{1 + x + \frac{1}{x}} = \frac{x}{x^2 + x + 1}$$ extends to the real analytic function written on the right. Notice that$x^2 + x + 1 = (x + \frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4}. $1 Yes, of course, but, due to the Gelfond-Schneider theorem, they would have to be transcendental. 1 For cooking, I only see denominators$2,3,4,8$, so I would convert to the closest of those. In fact, thirds only come with cups and eighths only with teaspoons, so you could do that. A fraction of eleventh a doesn't seem useful. For other applications, I would still be strongly biased toward small denominators. You could make a score from the error plus a ... 0 The biggest thing about recipes is "the proportions of quantities relative to each other." The best thing (IMO) you can do, is move as much of all quantities to integers as possible by multiplying all quantities by some integer constant$c$so that the smallest is$1$, then rationalizing all the non-integers to "reasonable" fractions (with no denominators ... 1 (Beware, numbers below are base$6$) Fractions can be reduced by dividing the greatest common divisor from both the numerator and denominator. And you can find GCD by Euclidean algorithm, but with senary (heximal) multiplication and division. E.g. for$\frac{4535}{1555}$, (it took me some time)$\begin{align*} 4535 =& 1555 \times 2 + 541 = 3554 + ... 1 First, note that if1/b$terminates then$a/b$terminates, so we really only need to consider reciprocals. If$1/b=(0.a_{-1}a_{-2}a_{-3}\ldots)_6$, then$6/b=(a_{-1}.a_{-2}a_{-3}a_{-4}\ldots)_6$. If$6$is coprime to$b$, then the multiplicative order$n$of$6\pmod{b}$, which must divide$\phi(b)$, is the least number such that$6^n\equiv1\pmod{b}\$. ...