/sci/ - Science & Math

>Consider the subgroup [math]G=\left\langle \sigma,\pi \right\rangle [/math] of [math]S_{4}[/math]...
can someone clarify what this notation means?

>>9741634
>The letter e = exponential function you know?
yes i fucking know e is the fucking exponential function, but it just so fucking happens that you dont know enough fucking algebra to be at the level to define this function through calculus

i would suggest either start again with the basics, or a bullet to the head

>>9419046
>his brain is too tiny to understand the different schools of mathematical thought that developed historically and continue into today
Even mathematicians of Cantor's time were dubious of the notion of "infinite" sets. Read a book.

>Exercise: (recommended only for math olympiad students)

>He organizes his study hours and makes adjustments in order to make up for unforeseeable events that disrupted his study, these changes are make weeks in advance if possible. All in order to maintain a constant and rigorous study habit that will result in good grades and a fulfilling life

>>9255810
>tfw it's too trivial for me to see

>>9252615
>actual explanation and exposition is wrong
>what reaffirms my feelings is correct

>>9242887
any takers?

>>9238771
well done sir. you are amazing

>>9229030
oh well, at least the most important part is displaying correctly in my first post.

Brainlet coming through.

Prove: Given a non-empty subset [math] A \subseteq \mathbb{R}[/math], suppose [math]b[/math] is an upper bound for [math]A[/math] and [math]\epsilon \in \mathbb{R}[/math] Prove that
[eqn] b = \sup A \implies \forall \epsilon > 0 \hspace{0.2cm} \exists\hspace{0.2cm} x \in A ,\hspace{0.2cm} \text{such that}\hspace{0.2cm} x > b-\epsilon [/eqn]

Proof:

Suppose this was not true. Then [math] \exists \epsilon > 0 \hspace{0.2cm} \forall \hspace{0.2cm} x \in A ,\hspace{0.2cm} \text{such that}\hspace{0.2cm} x \leq b - \epsilon[/math] , which is equivalent to [math] x + \epsilon \leq b [/math] . Can I choose epislon to be [math]n\epsilon[/math], for [math]n\in \mathbb{N}[/math], by the archemidein principle? Hence, [math]x+n\epsilon \leq b[/math] which contradicts [math]b[/math] being the supremum?

>>9194893
right on. sucks being a brainlet. many thanks.

>betwixt