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12:01 AM
they wanted to distinguish it from E
 
Yeah there are other charts out there that show how Hebrew evolved
(Modern Hebrew is really Aramaic letters)
(not to be confused with Syriac letters… Aramaic actually had about a bajillion alphabets throughout history)
 
the case of the missing vowels
 
Here is my candidate prime counting function $$f(x)= \sum_{n=1}^\infty \exp{\bigg(-\prod_{k=2}^{e^x}} \frac{p_n-1}{\ln(k)} \bigg) $$
where $p_n$ is the nth prime
 
12:22 AM
@geocalc33 there you are, I found you!
On the interweb - I'm your #1 stalker :P
 
oh hi
 
@geocalc33 it's nice to see you working with the primes now as themselves - I.e. you're another victim of the primes and possibly the twin primes - like me :)
 
I'm not doing anymore work with the primes
I promise
 
No matter what approach you take, it always seems to wind up on the analytical side of things, so that's why it's more of an analytic number theory problem rather than algebraic. So in other words I think you're on the right track
Number theory = prime numbers lol
And algebraic number fields of course
 
what do you think
can $f(x)$ play?
 
12:25 AM
Oh, it's a beautiful looking formula
What have you proven about it?
It's nice too to view it as the computer renders it, so let's do that now. I sent you a zoom link
You can also verify it like I did mine in that twin prime thread (did you see my counting formula?)
But you can only verify it up to a certain number of inputs like the first 10,000 say. You still need a math proof
Had to plug in my mike
 
12:56 AM
Hi
If $f:[1,2] ->R$ and $f(1)=1$, $f(2)=3$, whats the supremum of $S = {f(4n-1)/(2n+1), n /in N}$
I can see the inf is 1 but no idea about sup
I think it depends of f
 
yes, it will. presumably you mean f((4n-1)/(2n+1)).
the inf is not determined either.
 
Yeah i mean that
The inf is 1
4n-1/2n+1 is monotone and increasing
 
is f assumed increasing? you didn't mention that.
 
So inf = min = 1
No i mean the subsequence (4n-1)/(2n+1)
 
ok, so this has nothing to do with f? just what's sup and inf of {(4n-1)/(2n+1): n in N}?
if N starts at 1, its inf is 1. the sup is 2 no matter whether N includes 0.
 
1:03 AM
No, it does, we are checking the set S ={ f((4n-1)/(2n+1)) with n in N}
 
if we don't know anything about f, it can act entirely arbitrarily on that sequence. neither the inf nor the sup will be determined.
 
Uhm yeah.. the inf is not determined either
Right
 
if f is assumed increasing and N starts at 1, the inf is f(1) and the sup is f(2). because this connects to the given information, i'm guessing that maybe f being increasing is part of the hypotheses.
but if not, the inf and sup could be anything.
 
We only know that values and that is defined in that compact interval
Ah wait
 
yeah, then anything is possible. S could be any at-most-countable set of real numbers.
 
1:05 AM
We know its monotone, sorry
f is monotone
 
ok, so the inf is f(1) = 1 and the sup is f(2) = 3.
actually i guess i'm also assuming f is continuous. without that the sup could be anything between f(1) and f(2) (inclusive), while the inf would be 1 in any event.
 
Uhm
Why it could be anything between f(1) and f(2) without assuming continuity?
 
pick any value c between f(1) and f(2). draw a straight line from (1, f(1)) to (2,c). define f(x) by using that function as the graph for all 1 <= x < 2 and define f(2) = 3.
will be monotone increasing, and lim_{x to 2-} f(x) will be c, and not f(2) (unless you chose c = f(2))
so that sup will be c also
 
Uhm right thanks :)
That was very clear
 
the key is that f(1) is in that set of values of f, but f(2) doesn't have to be. it has to contain values of f(x) with x arbitrarily close to 2, but need not contain f(2).
and won't, if f has a jump discontinuity there.
 
1:20 AM
Yeah the sup can be out of the set
Like it happens in open intervals of R
But the inf must be in the set right?
Ah no
1/n the inf is 0 and is not in the set
Interesting
 
1:39 AM
@leslietownes isnt sup = lim_{n->\inf} f((4n-1)/(2n+1))?
Or is that true only iff f is continuous
 
@Odestheory12 That is $2-\frac3{2n+1}$, so it is increasing.
 
ode: yes, it's true because of the assumption that f is increasing.
[and as robjohn notices because the sequence is too]
 
Uhm
But f is monotone, not increasing
 
if f(1) = 1 and f(2) = 3 there's only one kind of monotone it can be
by "increasing" i mean only x < y implies f(x) <= f(y). i do not assume 'strictly increasing' which would have < in the second inequality
 
1:55 AM
smacks leslie again
There's hardly anything so ugly as <=
 
What i thought was Sup(S) \in ]1,2]
 
note that in the examples i gave above, f is strictly increasing, so this doesn't really come up
ted: i think oo is uglier
 
Yes. Now that you've challenged me, probably you're capable of having ugliness tend to infinity.
I'm going to toss your non-math-Jax math sentences in the trash soon.
I will finally exert my infinite authority here.
 
oddly enough i don't find < and > ugly in plain text. it only looks ugly in tex to me.
when used for langle and rangle, that is.
 
Everything math is ugly in plain text.
I'll settle for 1<2. That's it.
 
2:00 AM
a saving grace of plain text is that its ugliness causes people to write less of it. they focus on the essentials. a lot of old sci.math posts are still very readable.
once mathjax enters the equation, everyone can become the sorceror's apprentice, conjuring up paragraphs of perfectly typeset math babble
 
I'd prefer perfectly typeset any day.
We've been living in the TeX world since 1990 or earlier.
I'm tired of this lazy sh*t.
I'm gaveling this room to order ... or else.
 
well, for most of the 90s people still wrote in plaintext on usenet. i don't know of any news readers that automatically rendered tex.
 
Who's talking news?
 
it sort of stopped existing, which i guess solves the problem.
usenet newsgroups. "google groups". not news news.
 
It's still funny that Joe Harris produced his algebraic geometry book with Machias, the word processor math fonts that Dave Bayer had championed. But it looks so horrible.
 
2:04 AM
sci.math and sci.math.research. most people on those at that time were using some kind of email-like client with plaintext support only. at least i was.
 
I mean, in email exchanges with my students who didn't know TeX, I had to write all sorts of horrible things, but yuck.
 
TeX works in email?
 
i've never seen the harris text, i should check it out.
under: if the recipient can read it. i don't know of a client that renders it.
 
Oh, I see.
 
There are plug-ins somewhere that render it somewhere.
@leslie Springer
practices moving messages to junk
 
2:07 AM
@leslietownes but then sup = lim_{n->\inf} f((4n-1)/(2n+1)) = f(3) = 2
Or where am i wrong
 
ode: you mean f(2) there. and the second to last equality assumes f is continuous from the left at 2, which it need not be.
 
Was writting exactly that :D
You need continuity to evaluate the lim inside of f right
 
Right.
 
Monotone functions can leap!
 
2:09 AM
that's as bad as they get, but they do get that bad.
 
And your example above is why we need continuity
 
Does anyone understand why a Binary Tree traversal would be more efficient for sorting large data sets than other algorithms? Like insertion, for instance?
 
It works not only for the sup but also for this case
 
I'm working on a coding assignment and after the data gets past 500 entries, the tree is much faster than the other algorithms I've implemented.
 
That's beyond my knowledge/pay grade, @UnderMath.
 
2:11 AM
Lol, I figured it would be a long shot asking this in a math chat.
It's my fault. I'm sure my professor explained this at some point, but the effort I put towards my CS work is a fraction of that of my math courses...
 
There are CS-ish people here from time to time who know such things (like @robjohn), but most of us don't.
 
I'm a "CS-ish person" i guess but i promise i'm not good at it
 
We don't know you well enough to judge, @hyper.
 
Anything helps at this point.
 
well insertion sort is really bad
so it's not surprising that any reasonable algorithm would surpass it pretty quickly; insertion is O(n^2)
 
2:15 AM
Oh yeah?
Alright, I have to pull out my time-complexity graph thing really quick.
 
insertion sort takes O(n) to move the next element into place (scan them all, swap min to its appropriate position) and then needs to run that once per element totalling O(n) * O(n)
 
helps to think about worst cases. e.g. what happens if your data is in reverse order.
or extreme cases, even if they aren't worst for a particular method.
 
yeah that's a good way to think about it, thanks leslie
 
my favorite sorting algorithm is 'apply a random transposition and check if the list is sorted, if not, repeat'
 
if by binary tree traversal you mean heap sort, it's fairly straightforward to see why heapification is N log N, which is the best time complexity we have for sorting
@leslietownes ah, the classic BogoSort
 
2:18 AM
@hyper-neutrino isn't it O(log n) if the array is sorted in each pass?
 
@LeakyNun wdym sorted in each pass?
 
I remember my prof explaining that the tree is O(logn)
 
heaps take O(log N) to insert and remove elements
 
@hyper-neutrino inserting "4" into "1 2 5 7" is log n
 
what does that have to do with insertion sort tho
like
 
2:19 AM
Oooh, ok. I'm starting to see it now.
 
insertion sort scans the whole unsorted section and finds the minimum, which takes O(n)
and then insertion is O(1) because you know where the element should go
but you need to do N scans
 
lots of clever stuff in sorting.
 
I'll need to work on understanding this stuff more thoroughly, in case there's a complexity table on the final.
 
i like watching bubble sort. its sortof soothing as such
 
2:34 AM
lava lamp sort
 
Understanding is good for its own sake, not just for tests.
 
True, true. I try not to fall back into that habit of studying for tests, rather than understanding. It comes to the forefront when I'm stressed lol.
 
That’s partly why you stress. This bad habit of cramming for tests.
 
It's a vicious cycle.
3
 
Only if you are a procrastinator and let it become one.
 
2:40 AM
the fundamentals should be in your heart. it is not a head thing. insertion sort should be in your heart.
michigan is beating iowa. :(
 
A disciplined student barely studies for finals.
 
;-;
Next semester is going to be different. I'm finally taking care of some stuff I was dealing with. So I'll be more focused.
But for now, the next few weeks are gonna suck.
 
An undisciplined student also barely studies for finals :)
2
 
hyper :)
 
2:44 AM
i wish i had worse study hygiene in law school. i could have had one of those 'paper chase' moments around finals. missed out on that experience.
 
not a particularly bad thing to miss in all fairness
 
yeah, it didn't make it look like a lot of fun.
i had a 'study group' of sorts but we realized early in the first semester there was really no point to it. we would get together for dinner and talk about anything other than law school.
 
Character assassination of profs out of bounds?
 
well, they had no character to assassinate in the first place. but basically, no.
i don't think criticism of profs or other students would have broken our informal rule against talking about law school.
i guess the rule might have been, as long as it isn't studying, it's OK.
 
My tangential point.
 
2:59 AM
@TedShifrin this was a nice thing to discover some time ago
also, spaced repetition programs
 
3:13 AM
i replaced my wheel and went north on the bay trail, ended up getting a puncture near rosie the riveters! today was not meant to be a cycling day. the ghost of willamete...
loved paper chase
 
3:51 AM
Wait, no blaming Willamette!
 
i went outside once today, to check the mail and look at the neighbors' holiday lights.
 
exercise is my drug. that is why i am procrastinating on my hip.
nice balmy 50f outside at the moment.
paid my property taxes. truly exciting day.
 
ooh! i did that a week or two ago. felt great.
55F here. the heat just came on.
 
my dad was an accountant, so impressed on me the importance of delaying payment as long as possible, but the tension gets to me :-)
i am not sure why my typing is so bad. the mistyped letters are often far away from the intended, so its a deeper problem.
 
same. i hate giving them interest-free loans but i also hate having it on my to-do list.
there's probably a generation of accountants for whom the importance of timing is something of a lost art or a textbook thing. i should ask my accountant friend about that.
with interest rates doing who knows what and inflation, people are going to get good at this again.
 
4:03 AM
i suspect our inflation is a direct result of stimulus.
the need to accumulate toiler paper.
 
You can try a grammar checker like Grammarly.
 
are you an ad bot?
 
No. I am real human.
 
i get lots of ads for grummerly
 
I use DNS blocker, so I dont get ads.
 
4:07 AM
gremmer an spuling would npot by my stroing pints
 
Youre are now forcefully writing bad.
 
thankfully i am pretty regular
writing badly, i suspect grammarly would point out...
 
Whats wrong with that?
 
writing bad is the act of writing the letters b a d.
 
i wouldn't mind a few strong pints.
 
4:09 AM
or somer netflix production about who knows what
i am tempted to break out a bottle of red.
i just texted a friend to see if he wants to open a bottle of something red
@Osmium i am not usually pedantic.
interesting choice of username, a bit dense i might add
 
Yes it is. I use this type of naming everywhere.
Do you like chemistry?
 
i like it but am very much at the pauling general chemistry level.
my favourite parts are related to thermodynamics.
 
I love it too.
Do you know about chemical kinetics?
 
i like some of the semiconductor stuff as well.
rates of reactions?
yes.
 
I have asked this question related to it on ChemSE but it got closed, can you answer it. chemistry.stackexchange.com/questions/160166/…
 
4:44 AM
@Osmium it has been quite a while since i dealt with such things but (assuming constant pressure) the change in enthalpy is just the heat given off or absorbed by the system?
 
Yes, you are correct.
 
Are you dealing with solids?
 
No, it is part of transition state theory; it describes what happens at molecular level when a reaction happens.
Maybe you meant to ask whether I am dealing with a system consisting of solids, in that case, no.
At constant pressure for any system change in enthalphy equals the heat exhcnage.
 
And your question is asking why the heat exchange is only potential energy exchange?
 
Yes.
 
4:52 AM
constant temperature?
 
There is no need for constant temperature; at least the book didn't mention it.
 
just guessing that constant temperature would make the kinetic energy the same. the quick answer it i don't know.
most of my interaction with enthalpies were with steam :-)
 
You're right, for any two ideal gases at the same temperature they have the same average kinetic energy. But here we are dealing with real compounds.
When I study something deeply after 1-2 months I forget almost everything I learnt.
But you seem to not forget what you study.
How do you study?
 
5:07 AM
I only remember things for which I have some intuition for.
I regularly have to consult texts to remember the details.
 
But what about tests?
 
I used to have a strong short term memory :-)
but the topics had to have some structure/connectivity. I am unable to remember collections of unrelated facts.
 
I don't have both. Neither short nor long term memory.
 
i remember enthalpy for a few reasons, one of which is that gibbs found it easier to work with than energy. for some reason this stuck in my head.
so, sometimes a nice narrative helps.
 
I have noticed that when I visualise something or take a very concrete example of a statement I remember it for months.
Old Russian books include many such examples.
 
5:19 AM
I like some of the old Russian texts, I found them more concrete and less abstract. Suited me.
 
If you can't prove that a Turing machine doesn't halt (and can't prove that it does), then there's a model of your axioms in which it does halt. This model looks like the set $\Bbb N$ union many "extra" numbers that are larger than everything in $\Bbb N$. In a sense, this means that, though the Turing machine doesn't halt in the "standard model" $\Bbb N$, you just haven't waited long enough!
Nonstandard models of arithmetic are weird.
In fact, there's a "universal" Turing machine such that, for any partial function $\subseteq\Bbb N\to\Bbb N$, there exists a model of arithmetic in which it will compute exactly that function
And if the domain of the partial function isn't all of $\Bbb N$, for any partial function extending the domain of that function to more inputs, there's an extension of the model in which it computes the extended function
 
Most of the books were published by MIR publishers. You can find their PDF versions here: mirtitles.org
 
Oh, and PA (or whatever axiom system you're using) can prove that the function is only defined on finitely many inputs
 
i have a question about why vakil's "magic diagram" is cartesian - actually my concern is found in this comment (i am not convinced by the response): math.stackexchange.com/questions/778186/…
i think the issue i'm having is that if we have a morphism $Y\rightarrow Z$ and we draw out the big diagram involving $X_1\times_{Y}X_2\rightarrow X_1\times_{Z}X_2$, i don't know how to show this diagram actually commutes (and maybe it doesn't, but it seems like we need this for the argument given in the above post to work)
more specifically i don't see any reason why the compositions $X_1\times_{Z}X_2\rightarrow X_i\rightarrow Y$ should be equal, if they even necessarily are - all we know is that they are equal after composing with $Y\rightarrow Z$
 
Thank you for your time, Joe Higgins. Thank you very much.
Your chat is unreadable P-addict.
 
5:30 AM
@Osmium it was nice chatting :-), no need for thanks!
@Osmium you need to activate the chatjax feature to see the math. see the link at the left of the page near 'Latex in chat"
 
 
5 hours later…
Jam
10:09 AM
the boundary of $\mathbb{R^3}$ is empty right?
 
10:30 AM
Yes, that's the case for any clopen set
 
11:18 AM
in fact they're equivalent
 
11:32 AM
What is the difference between pointwise and elementwise inequalities?
 
12:00 PM
in what context?
 
12:19 PM
Its in relation to a description of a polyhedral which I believe it makes sense to talk about equalities being elementwise, but I dont have a firm grasp of the difference between them.
 
12:56 PM
help me
in solving
lim u tends to infinity
f(u) = u-1/root(u^2-2u+2)
help ?
 
1:22 PM
@P-addict they certainly aren't in general, but the argument doesn't need that either
like, think you are in Sets and take $Z$ to be a singleton. then the map between the fiber products becomes the canonical inclusion $X_1\times_YX_2\subseteq X_1\times X_2$. of course, the projections from the product to the factors followed by the maps to $Y$ are not equal, unless the inclusion is an equality
in fact, generally, if they were equal, you'd obtain a map between the fiber products the other way round and this would be inverse to the map already given, since the composites are just canonical projections in each component
 
 
5 hours later…
6:04 PM
@Thorgott right, this too... i knew something was up when i was getting results like "all fibered products of $X_1,X_2$ are isomorphic"
any idea on how to show the diagram is cartesian? i think the yoneda lemma is one option but as was mentioned in the post this exercise shows up in vakil's notes before he mentions yoneda so i was looking for another proof
 
Najib's answer is how I would do it
 
6:33 PM
@Thorgott oh! is it that the blue and red paths are equal because once we get to $Y\times_{Z}Y$ we can compose with the map $Y\times_{Z}Y\rightarrow Y$, and since the last map in both the blue and red paths to $Y\times_{Z}Y$ is $Y\rightarrow Y\times_{Z}Y$, we get the composition $Y\rightarrow Y\times_{Z}Y\rightarrow Y=\text{id}_Y$?
(sorry, that really wasn't very clear)
 
I'm trying to prove that $|\mathbb Z| = |\mathbb Z - \{2\}|$. I know one way to do this is to show that there is a bijection between the sets. Can I do this by forming the function $f: \mathbb Z - \{2\} \rightarrow \mathbb Z$, such that $f(z) = \frac{1}{2-z}$ and then showing the bijection?

Another option I could think of: Since $\mathbb Z$ is denumerable, it forms a bijection with the natural numbers. So I can just claim this, and then show that $\mathbb Z - \{2\}$ also has a bijection with the natural numbers and therefore $|\mathbb Z| = |\mathbb Z - \{2\}|$.
 
actually, is the argument i just made really just saying $Y\rightarrow Y\times_{Z}Y$ is mono because it has a left inverse?
 
under: your rule for f doesn't produce integer outputs for most z, but you could fix it so that it does. the second approach doesn't seem much simpler than the first (although sometimes with other sets and contexts it might be)
under: e.g. it should be possible to prove very quickly that f(z) = z if z < 2 and z - 1 if z > 2 defines a bijection from Z - {2} to Z
 
Hm, ok. I'll see what I can do.
 
@P-addict this is true, but I don't see how this is relevant. the blue and the red path are both the composition $X_1\times_YX_2\rightarrow X_1\rightarrow Y$.
 
6:45 PM
by the second approach, i generally mean, showing that A and B have the same cardinality via some chain of bijections A -> C_1 -> C_2 -> ... -> C_n -> B which are known to exist, or are more simply described or proved to be bijections, than a 'directly' constructed map from A to B (e.g. because an explicit formula would be annoying to compute)
i don't think we're in that world here
 
Ooh, ok. I think I see it more clearly now.
 
@Thorgott well, i guess what i'm saying is when trying to prove the universal property, we start off with two maps through $Y$ and $X_1\times_{Z}X_2$ from an object $A$ which are equal when they go to $Y\times_{Z}Y$. it turns out both compositions actually come from mpas which first go through the $X_i$ then $Y$ (namely the blue and red paths) and end with $Y\rightarrow Y\times_{Z}Y$, so since this map is mono we actually have maps into $Y$ which play nicely with the $X_i\rightarrow Y$...
...and that is how we get our map $A\rightarrow X_1\times_{Y}X_2$, by the universal property
in other words i was stuck on showing the red and blue paths were equal, and this follows from the fact that they are once we continue them to $Y\times_{Z}Y$ via a mono map
hopefully this argument works
 
7:36 PM
I found an even simpler solution: $\mathbb Z - \{2\}$ is an infinite subset of a denumerable set, $\mathbb Z$. There is a theorem that states that any such subset is also denumerable, and as such, $|\mathbb Z - \{2\}| = |\mathbb Z|$.
 
the morphism $A\rightarrow X_1\times_ZX_2$ already determines the two morphisms $A\rightarrow X_1$ and $A\rightarrow X_2$ that will determine the morphism $A\rightarrow X_1\times_YX_2$, the additional data merely specifies the compatibility necessary for this morphism to be induced
I don't quite follow you, but the fact that that map is a mono is not particularly relevant here
 
@UnderMathUate you can just define $\phi(n) = n$ for $n \ge 1$ and $\phi(n) = n-1$ for $n \ge 3$. show that $\phi$ is a bijection.
 
i don't agree that this is simpler. you are detouring through two facts (that Z - 2 is infinite, and that Z is denumerable) which are useful generalities but have their own proofs somewhere in terms of bijecitons, when you could just be exhibiting one function and proving it is a bijection.
copper, this has been proposed above
 
its like a committee meeting, everyone has got to say their bit even though it has been said uncountably many times before.
 
a lot of general facts about cardinalities are proved via cardinal arithmetic, which is basically what you are doing here. but cardinal arithmetic can be subtle with infinite sets, and it's arguably overkill when you can just write a map down.
 
7:43 PM
@Thorgott i think the thing i wasn't sure on was why the two morphisms $A\rightarrow X_i$ we got agreed when composed with the $X_i\rightarrow Y$. they agree when you then compose this with $Y\rightarrow Z$ (obviously), but that they do so before was what i needed to show
 
then there's the shredder-bernie theorem
 
Oof, major oversight on my part. I was essentially just going to assume that Z-2 is infinite. We sketched out that Z is denumerable in class.
 
because only then could i invoke the universal property of $X_1\times_{Y}X_2$
 
But I did manage to work out the other solution you proposed.
 
all statements about cardinal arithmetic (e.g. |A| = |B| because |A| = |C| and |C| = |B|) are encoding statements about bijections that you aren't bothering to write down. this is fine when that's the shortest route and the maps you would be writing down have complicated formulas or involve tons of arbitrary choices.
i mean, it's generally fine, but as a matter of aesthetics or simplicity, explicit bijections beat cardinal arithmetic
you see this a whole lot even in finite sets in combinatorics. often it is far more informative to have a bijection between two sets of things than simply to know those sets have the same size
 
7:45 PM
yeah, you have to use that the maps to $Y\times_ZY$ agree and then look at them component-wise (because that's how we understand maps into fiber products usually)
that's also implicit in your observation that $Y\rightarrow Y\times_ZY$ is a mono
 
Alright, I can see where you're coming from.
I think in the case of learning the concept, it's probably best to lean more towards explicit descriptions than not anyhow.
 
fighting the urge to answer a convex psq
for poetry implicit is good, for maths explicit is usually better
 
e.g. there are the same number of k-element subsets of X = {1,2,...,n} as there are (n-k)-element subsets of {1,2,...,n}. the simplest proof is to show that S -> X \ S defines a bijection between those collections of sets. you don't need to compute what that number is to do that, as other routes to establishing this fact might do
 
How does one complete the character table of a finite group knowing only data of conjugacy classes? Only way I could think of is via induced representations
is there another way?
 
i'd almost say that in some contexts it would be missing the point to bring binomial coefficients (i.e. counting, i.e. bijections with {1,2,...,size of set}) into it
 
7:49 PM
@Thorgott cool, this was easier than i thought it was. sometimes all one needs is a bit of sleep i guess. thanks as always Thorgott!
 
8:24 PM
hey guys
 
8:39 PM
hi @Allie
 
8:53 PM
can i list a short proof to see if i did a problem right
 
Sure. I'm not going to be here too long. Have to interview a potential college student.
 
gotcha
so the question is
suppose A is an n x n matrix
and let v_1, ..., v_n are vectors in R^n
suppose Av_1, Av_2, ..., Av_n are linearly independent
prove that A is nonsingular
so what i did was
we know that
 
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