# caya

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bio website location Sydney, Australia age 48 member for 2 years seen Nov 25 at 10:18 profile views 26

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 Nov23 comment $x,y,z$ are positive real numbers and $x+y+z=1$ $\implies$ $\bigg(1+\dfrac 1x\bigg)\bigg(1+\dfrac 1y \bigg)\bigg(1+\dfrac 1z \bigg)\ge 64$? Hum, Jensen's inequality seems to give an smaller bound ... sorry. Nov23 comment $x,y,z$ are positive real numbers and $x+y+z=1$ $\implies$ $\bigg(1+\dfrac 1x\bigg)\bigg(1+\dfrac 1y \bigg)\bigg(1+\dfrac 1z \bigg)\ge 64$? It smells like Jensen's inequality. Nov23 comment On contemporary mathematics education What is the difference between learning to swim with floating tubes or without them? You bet, selfconfidence! Nov6 comment Explain proof of irreducibility of $x^{p-1} + 2x^{p-2} \dots (p-1)x + p$ Isn't $Q(0) = p$? as if all zeros are there, then $Q$ = original $P$. Nov3 comment Prove that the set of limit points of a function is all $R$ A limit cannot be whole real line. I bet you need to restate what you want to prove. Oct23 revised How do I figure out the value of a number raised to a fractional power? added 82 characters in body Oct23 answered How do I figure out the value of a number raised to a fractional power? Oct16 comment Propositional Calculus: Stating and proving the unique readability theorem in Polish notation "an initial segment of one another, which we know is not possible" I bet, that had to be proved as a lemma, ah? Oct14 comment Is this true: $\lim_{\lambda \rightarrow 0}E\left[ e^{-\lambda X} \right] = P\left(X < \infty \right)$? I bet the OP meant $\lim_{\lambda \to 0}$ Oct12 revised $(x+1)/x = \sqrt{3}$ in form $a+b \sqrt{3}$ edited body Oct12 comment $(x+1)/x = \sqrt{3}$ in form $a+b \sqrt{3}$ @columbus8myhw, thanks. Yes, even simple questions in maths can have deeper answers. Oct12 answered $(x+1)/x = \sqrt{3}$ in form $a+b \sqrt{3}$ Oct2 comment Formula for the terms of the sequence defined by $a_0 = 1$, $a_1 = -2$ and $a_{n}=-4 a_{n-1}-4 a_{n-2}$ @Mark, induction is: show that the first domino falls, and that every domino that falls knocks the next one; then you can be sure all dominoes fall. Strong induction is, show that the first domino falls, and that if all dominoes up to a certain point fell, then the next one falls too. So, strong induction -although equivalent- allows you to use as argument a stronger assumption ("all dominoes up to a point") to prove that the next one also falls. Stronger assumptions make proofs easier. Aug27 comment Proof that group is commutative if every element is its inverse (feedback wanted) Seems fine to me :) Aug26 comment Verify my proof: If $X$ is infinite, then there exists $f: \mathbb{N} \rightarrow X$ such that $f$ is injective. Your definition is different. Aug26 comment Verify my proof: If $X$ is infinite, then there exists $f: \mathbb{N} \rightarrow X$ such that $f$ is injective. If you check en.m.wikipedia.org/wiki/Infinite_set, what you want to prove is mentioned there and requires the axiom of choice. Aug26 comment Verify my proof: If $X$ is infinite, then there exists $f: \mathbb{N} \rightarrow X$ such that $f$ is injective. What is your definition of infinite set? Aug25 revised The set of all finite subsets of the natural numbers is countable added 3 characters in body Aug25 revised The set of all finite subsets of the natural numbers is countable added 21 characters in body Aug25 awarded Yearling