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location Sydney, Australia
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visits member for 1 year, 11 months
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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...
Aug
14
comment Determinant of a matrix with $t$ in all off-diagonal entries.
When $n=2$ seems to fail. $1-t^2$ RHS, $-2 t^3-3 t^2+1$ LHS, right?
Aug
12
comment Formal Proof that area of a rectangle is $ab$
What I like about the question and the answer is that it shows that not everything in maths is proving: a good part of any real maths involves finding the right axioms that are useful to the problem at hand (+1 to both)
Jul
30
comment What is $\displaystyle\lim_{h\to 0}\ h\left(\sum\limits_{n=1}^\infty{n}\right)$?
A believe this question illustrates that for limits to be interchangeable, they both need to exist. Although the above does not exists, $\sum\limits_{n=1}^\infty{\left(\displaystyle\lim_{h\to 0}\ h~n\right)}$ does.
Jul
23
comment A set of all rational numbers in $[0, 1]$?
And a set is dense in other (in topology), when its closure is the whole enclosing set. It is mentioned in wikipedia too.
Jul
23
comment A set of all rational numbers in $[0, 1]$?
$\bar{A}$ is the closure operation: adding all limiting points to a set in the topological sense. Some texts use $Cl(A)$ instead, like in wikipedia: en.wikipedia.org/wiki/Closure_(topology). See the link :)