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location Sydney, Australia
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Nov
23
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.
Nov
23
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.
Nov
23
comment On contemporary mathematics education
What is the difference between learning to swim with floating tubes or without them? You bet, selfconfidence!
Nov
6
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 $P$.
Nov
3
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.
Oct
16
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?
Oct
14
comment Is this true: $\lim_{\lambda \rightarrow 0}E\left[ e^{-\lambda X} \right] = P\left(X < \infty \right)$?
I bet the OP meant $\lim_{\lambda \to 0}$
Oct
12
comment $(x+1)/x = \sqrt{3}$ in form $a+b \sqrt{3}$
@columbus8myhw, thanks. Yes, even simple questions in maths can have deeper answers.
Oct
2
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.
Aug
27
comment Proof that group is commutative if every element is its inverse (feedback wanted)
Seems fine to me :)
Aug
26
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.
Aug
26
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.
Aug
26
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?
Aug
25
comment The set of all finite subsets of the natural numbers is countable
@GuilhermeD, I posted as answer comments of your proof. Hope helps.
Aug
25
comment The set of all finite subsets of the natural numbers is countable
@GuilhermeD - what is $A_n$ ?!
Aug
25
comment The set of all finite subsets of the natural numbers is countable
@GuilhermeD, I believe is easier if you invoke the binary representation of every natural number... but that is not approach.
Aug
25
comment The set of all finite subsets of the natural numbers is countable
@GuilhermeD, I may have a mistake. One can inject X into N, but turning that into a bijection may require more steps :(
Aug
25
comment The set of all finite subsets of the natural numbers is countable
You invoked the fundamental theorem of arithmetic. It actually gives you a bijection between $N$ and $X$. Do you see why?
Aug
18
comment How does exponentiation relate to multiplication?
You need to prove that $ln(x~y) = ln(x) + ln(y)$, which follows from the definition by a change of variable. Then it should not be hard as $e^x$ is defined by the inverse (in your text).
Aug
14
comment Determinant of a matrix with $t$ in all off-diagonal entries.
So, there is a typo in the post...