245 reputation
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location Sydney, Australia
age 48
visits member for 2 years
seen Nov 25 at 10:18

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
23
revised How do I figure out the value of a number raised to a fractional power?
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Oct
23
answered How do I figure out the value of a number raised to a fractional power?
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
revised $(x+1)/x = \sqrt{3}$ in form $a+b \sqrt{3}$
edited body
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
12
answered $(x+1)/x = \sqrt{3}$ in form $a+b \sqrt{3}$
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
revised The set of all finite subsets of the natural numbers is countable
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Aug
25
revised The set of all finite subsets of the natural numbers is countable
added 21 characters in body
Aug
25
awarded  Yearling