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Jan 22, 2019 12:52
Yeah I effectively did that by not doing anything :)
I answered it for him and left him a comment as well with suggestions for the future.
Jan 22, 2019 12:41
@dessert Thanks, I'm only recently started got 1K to be reviewing posts and I'm trying to do a good job and not be rude to his first post....
Jan 22, 2019 09:36
I mean, it's clear... and there is a definitive answer ("Not possible. Yes. No.")
Jan 22, 2019 09:34
Hey all, what's that best approach to reviewing questions like this: askubuntu.com/questions/1111860/…
 

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
Mar 16, 2017 23:21
Do you think the || supposed to be logical or?
Mar 16, 2017 23:21
Mar 16, 2017 23:20
I'm trying to read this formula
Mar 16, 2017 23:20
Hi everyone
Jan 2, 2017 01:36
@Fargle I hope one day to achieve enlightenment and become an Algebra, myself. I'll float through the universe.
Jan 2, 2017 01:35
@AlessandroCodenotti Thanks. I always think of $| |$ in general terms being particular to vectors, rather than equivalent representation of sqrt, though. But at least my algebra wasn't wrong, it's just a representation thing.
Jan 1, 2017 23:54
(why is the formatting syntax different in chat from SE posts? that's dumb)
Jan 1, 2017 23:51
In *Introduction to Abstract Algebra* (2014) by Fine, Gaglione, Rosenberger, p.459, we have a very clear definition of the properties of the **norm** on a vector space such that *"Any inner product space $V$ is a normed linear space with |v| = $\sqrt{<v,v>}\ $ for* [all] *v $\in V$."*

Where *A* **norm** *on a vector space is a function $|\ |\ :\ V \rightarrow \mathbb{R}\ \ $satisfying:*

$(1)\ \ |v| \geq 0 \ \ for\ all\ v \in V$

$(2)\ \ |v| = 0 \ \ if\ and\ only\ if\ \ v = 0_{v}$

$(3)\ \ |tv| = |t||v|\ \ for\ \text{[all]}\ v\ \in V\ and\ t\ \in \mathbb{R}$
Jan 1, 2017 23:51
Quick question, probably not worthy of posting:
Jan 1, 2017 23:51
Happy New Year to y'all!
Jan 1, 2017 23:51
Hi Beautiful Math Geniuses.
Nov 4, 2016 00:06
I should have said that they are not easily layerable/stackable
Nov 4, 2016 00:05
I feel much more comfortable with a statement saying: "Given a group $G$, $\exists$ $g \in G$ $s.t.$ (etc etc)"
Nov 4, 2016 00:03
To show that a statement about existence (one with) is true, all we need is to find an element (an example) that satisfies the requirement(s) set by the statement.
To show it is false, we need to go through every single element in our consideration and confirm that no element satisfies all the requirements set by the statement.
Nov 4, 2016 00:03
To show it is false, we only need to provide an example that doesn’t satisfy at least one of the requirements set by the statement. Such an example is called a counterexample.
Nov 4, 2016 00:02
To show that a statement about universal truth (one with) is true, we need to go through every single element in our consideration and confirm that each element satisfies the requirement(s) set by the statement.
Nov 4, 2016 00:02
I will just say this:
Nov 4, 2016 00:01
I found my notes
Nov 4, 2016 00:01
@DHMO Yep sorry I am wrong
Nov 3, 2016 23:59
lemme find you what my Discrete Math prof said about it
Nov 3, 2016 23:58
One is a symbol for universality, the other for existence
Nov 3, 2016 23:57
In fact we learnt that you never use them in the same statement
Nov 3, 2016 23:57
Hi Devillus, you need to remember that $\forall$ and $\exists$ are not layerable.
Nov 3, 2016 23:50
For example I could state:

Let $v_i \sim v_j$ denote that vertex $i$ is adjancent to vertex $j$ in some graph $\text{G}$.
Nov 3, 2016 23:49
more info: Consider the following statement from "Pearls in Graph Theory" (Harstfield, Ringel):

"Two graphs $\text{G}_1, \text{G}_2$ with $p$ vertices are said to be $isomorphic$ if the vertices of $\text{G}_1$ and $\text{G}_2$ can be labeled with the numbers from 1 to $p$ such that whenever vertex $i$ is adjacent to vertex $j$ in $\text{G}_1$, then vertex $i$ is adjacent to vertex $j$ in $\text{G}_2$ and conversely.

Such a labeliing is the same as a one-to-one correspondence between $\text{V(G}_1)$ and $\text{V(G}_2)$ that preserves adjancency."
Nov 3, 2016 23:32
$\Phi$ is for incidence
Nov 3, 2016 23:32
(lemme find it)
Nov 3, 2016 23:32
for example, we have a symbol for incidence
Nov 3, 2016 23:31
is there a symbol for vertex adjacency in Graph theory?
Nov 3, 2016 23:30
Hi guys
Jun 23, 2016 08:56
(fyi – this is about quickly getting the fellow integer generators of some cyclic group $\langle a \rangle$.
$\langle 21 \rangle \in \mathbb{Z}_{30}$ was the one in the exam, which, admittedly is quite easy, but in past exams it's been $\langle 3 \rangle = U(31)$ for example which (at least for me) is a bit trickier just to do in my head)
Jun 23, 2016 08:49
Interestingly, the example sequential set of numbers I wanted to test for being coprime was exactly what was needed in the exam, so all I had to do was press the up arrow on the TI-89
Jun 23, 2016 08:48
@TedShifrin They exam didn't have any restrictions on calculators. I really just used it for one thing, which was to find the LCM or GCD of two numbers faster than in my head and to quickly generate list of natural numbers that are less than x and coprime to x
Jun 23, 2016 08:46
congratulations @TobiasKildetoft !
Jun 23, 2016 00:54
thanks guys.
Jun 23, 2016 00:54
but too verbose for my needs. Eh, I'll just type it in manually.
Jun 23, 2016 00:53
I think the seq() function is the ticket though
Jun 23, 2016 00:50
Algebra III – i.e. Abstract Algebra
Jun 23, 2016 00:49
Ali – too much time for what I need to do in the exam, easier just to enter it manually
Jun 23, 2016 00:49
I am just gonna try and live chat with someone on the TI.com website
Jun 23, 2016 00:48
haha yep found it too
Jun 23, 2016 00:46
basically I can't see anything on the syntax of range operators
Jun 23, 2016 00:46
I want something like [1..10] or [1:10]
Jun 23, 2016 00:46
yes but how, Ali?