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00:04
thanks...so the division algorithm works, I just have to be consistent with the bases...my textbook did not mention this
no problem
 
1 hour later…
01:27
howde yall
howde do
have i been misspelling howdy my entire life?
well, that's embarrassing
i just googled it
02:31
That really is embarrassing; if you ever build time machine, that is really the first thing you should go back and fix.
Then assassinate archduke Ferdinand; that dude totally needs to be stopped
@BalarkaSen In my investigation of my harmonic crap thing, I think I've found a proof of Hodge's theorem using Moser iteration.
03:27
hello
im so confused on dynamics =/
not sure how to work out what equation to use
what's the quetsion
its in standard units, so the particle moves with constant acceleration -8, at t=0 initial velocity = 10. How many seconds to reach -750m
i think x = vt + 1/2*a*t^2 kinda helps but am not sure how to solve t in that case
03:57
ah i think i got it now
04:23
I am too good
@orbit-stabilizer in that case please reduce that goodness until it's at an acceptable level?
05:02
[Random]
Thinking about what an experiment on the foundation of mathematics be like
Besides codes to check consistency of some user inputted model, can we go one step higher and have codes that allow users to specify elements of a foundation and then run proof checks on it...?
05:15
[The Cult of Infinity]
0. Begin by specifying a formal language which contains all the symbols we are going to use to describe and construct the objects
1. Next, we specify a logic system which provide us the inference rules and syntax
1a. But first, some primitive notions need to be defined:
1b. Propositions P: These are sentence of the formal language which has a fixed truth value
1c. Truth values V: Every proposition is either False (F), True (T) or Null (O)
1d. Conditional propositions P(): These are sentences which its truth value depends on the object specified in the arguement
05:40
(Error: wrong terminology, will be fixed later)
 
1 hour later…
07:08
Let $G = \{z \in \Bbb C \,|\, z^n = 1 \text{ for some} n \in Z^+\}$. How to show that for any fixed integer $k > 1$
the map from $G$ to itself defined by $z\to z^k$ is a surjective ?
@Silent the complex numbers have $k$'th roots of all elements
07:20
@TobiasKildetoft So, suppose $x\in G$, ie $x^n=1$ for some natural no $n$, and take $\sqrt[k]{x}$, then since $x^{nk}=1$, so $\sqrt[k]{x}\in G$ and $\sqrt[k]{x}\to x$ thus it is surjective.
Am io right?
Well, except you should not use that notation since the root is not uniquely determined.
oh. I will try not to use that next time.
07:41
[Random]
One level higher than nonassociativity: Deductive system based on equations:
Consider the following:
a+b=cd
Whenever using the algebraic structure you end up with th above equation, then the following can be done:
(a+b=cd)=>(a+b+k=cd)
Note how unlike a usual rewriting system, it is unidirectional. One can imagine in such systems the notion of putting brackets to an expression extends to the whole equation
The result is that if certain derivation is reached, then the possible derivation in the next step is restricted to certain types of manipulations, thus the whole system form a graph made of implications
The system is deductive since the manipulations that are allowed form a fixed set of inference rules
But nothing stopped us from generalising it such that even the rules themselves can be changed depending the sequence of rewriting thus giving a highly abstract notion of path dependence making it a lot more expressive than n category theory
 
2 hours later…
09:26
This is from Dummit n Foote:
Let $G$ be a finite group which possesses an automorphism $\sigma$ such that
$\sigma(g) = g$ if and only if $g = 1$. If $\sigma^2$ is the identity map from $G$ to $G$, prove that $G$ is
abelian (such an automorphism a is called fixed point free of order 2).
@Silent Is there not a hint with the exercise (there usually is for that one)?
What I can't understand is: do all automorphisms are assumed to satisfy above condition in this problem? What does 'possess' mean?
I dont think $\sigma\in G$
No, it is just assumed that the group has an automorphism with that property
"An" automorhism. OK, so there may be others that do not satisfy this.
09:30
Oh, but, isn't $\sigma=1$ satisfy the requirement trivially, meaning every finite group abelian?
no, because that does not satisfy the first part
Sorry!
Guys any constructive feedback?
1
Q: Is this evidence that $g(x)$ can be analytically continued?

More AnonymousArgument Let, $f(x) = \sum_{r=1}^\infty \mu(r) x^r$ where $\mu(r)$ is the mobius function. Hence, $$ f(x) + f(x^2) + \dots = x$$ Now, let $x \to x^2$: $$ 0+f(x^2) + 0+\dots = x^2$$ Similarly $x \to x^3$ : $$ 0 + 0 + f(x^3) + \dots = x^3 $$ Adding the above (vertically) we get defining $\...

(hey)
@Silent Hint: Define a new map $\varphi: G\to G$ by $\varphi(g) = \sigma(g)g^{-1}$. Show that this is surjective.
@TobiasKildetoft thank u so much.
a similar hint is given in text.
10:22
How to find the degree of the remainder $r(x)$?
$f(x)= q(x)g(x)+ r(x)$
@Abcd That depends on the polynomials
@TobiasKildetoft Intuitively, I just subtract the degree of the divisor from that of the dividend, and I get the degree of the remainder, what would be the rigorous proof of this?
which one do you call the dividend and which the divisor?
@TobiasKildetoft FIxed
doesn't answer my question
10:26
3/2 , 3 is the dividend 2 the divisior
but you are not dividing stuff here
anyway, there is no way to tell the degree of $r$ from that expression
@TobiasKildetoft This one: $\dfrac{x^5+x^2+3}{x^2+2}$
What would be the degree of $r$?
3?
Just do the long division to find out
it depends on the actual polynomials, not just their degrees
@TobiasKildetoft I am talking about the maximum possible degree
certainly not $3$ since we always require that the remainder has smaller degree than what we divide by
10:31
So that I can utilise the remainder theorem
Not sure what you mean by maximum. With respect to what choices?
Like what function would we assume for r in this:
$f(x)= g(x)h(x)q(x)+r(x)$
where $g(x),h(x)$ are divisors
where did the extra factor some from?
We have a quadratic at the bottom
anyway, we don't "assume" a function for $r$. We pick $r$ as the unique polynomial with certain properties
10:34
Thats why two linear factors
to utilise the remainder theorem
but that quadratic does not factor
@TobiasKildetoft Should I share a question?
We don't need linear factors to use the remainder theorem.
Dividing $f(z)$ by $z-i$, we obtain the remainder $i$ and dividing it by $z+i$ we obtain the remainder $1+i$, then the remainder upon the division of $f(z)$ by $z^2+1$ is @TobiasKildetoft?
so it should be a polynomial which has remainder $i$ when divided by $z-i$ and $1+i$ when divided by $z+i$, and it should have degree at most $1$.
10:39
why degree at most 1?
because when we divide we require the remainder to have smaller degree than what we divide by
And why is that?
because otherwise "the" remainder is not well-defined
@abcd division algorithm: let $P$ and $D$ be polynomials. Then, there is unique polynomials $Q$ and $R$ such that: 1. $P=DQ+R$, and 2. $\deg R < \deg D$
and in that case, $R$ is called the remainder
@LeakyNun Silly question: What does $x^3+ 3x^2+ x $ divided by $x^2+2$ mean intuitively?
10:48
the same as what it means 26 divided by 3
Okay.
division algorithm for integers: let P and D be integers, where D>0. Then, there is unique Q and R such that: 1. P=DQ+R, and 2. 0<=R<D
and in that case, R is called the remainder
11:32
Sometimes it’s hard to figure out where to even start with an OP’s misconceptions: math.stackexchange.com/q/2692148/137524
I guess “the kernel and range of a matrix are subspaces not numbers”
12:05
[The Cult of Infinity]
0. Begin by specifying an alphabet $\mathcal{A}$. This give us the symbols to be used
1. Next, define a formal grammar $\mathcal{G}$ which provide the rules in producing new strings and how they are concatenated, truncated etc.
2. Together, these form the formal language $\mathcal{L}$ where all the details of the foundation $\mathfrak{F}$ is writtened and expressed
3. Next specify a logic $L$ with its inference rules to provide the syntax of reasoning in $\mathfrak{F}$
But before we begin, we need to define a couple of primitive notions:
3a. Truth values: These can be True (T), False (F) or Null (O)
3b. Expression: A string or sentence constructed using $\mathcal{L}$ such that it is well formed (satisfy the syntax given by $\mathcal{L}$). They don't necessary have a truth value.
3c. Proposition P: An expression that has a fixed truth value. Can be though as a 0th argument predicate.
12:24
In referring to "the surjection" $\mathcal{O}_K/\mathfrak{a}_1 \to \mathcal{O}_K/\mathfrak{a}_2$ with $\mathfrak{a}_1$ and $\mathfrak{a}_2$ ideals of $\mathcal{O}_K$ is it meant that a coset $k + \mathfrak{a}_1$ is mapped to $(k + \mathfrak{a}_1) + \mathfrak{a}_2$?
@ÍgjøgnumMeg I assume one ideal is contained in the other?
@TobiasKildetoft Yeah that's something I should've mentioned
$\mathfrak{a}_1 \subseteq \mathfrak{a}_2$
So yes (that map is well defined)
Wonderful, thanks :)
and it just gives you the usual map from the full ring
note that $(k + a_1) + a_2 = k + a_2$
12:30
I thought as much
Thank you
If I have $2n$ balls in a connected manifold
Is it possible to find a path between each possible pairs such that none of those curves intersect
@Slereah is $n$ the dimension of the manifold?
No, it is arbitrary
Dimension of the manifold is at least 2 tho
then no, since there are non-planar graphs in the plane
unfortunate
What about the case where the manifold is at least of dimension 3?
12:35
probably the answer is yes in dimension 3 and up
Good to know
but that is only a guess
certainly it is true in $\mathbb{R}^3$
It is true in dimension > 2
If two paths intersect make them transverse by perturbing a little
By transverse 1-dimensional submanifolds of a manifold of dimension > 2 intersect trivially
12:55
@BalarkaSen so this is actually an exercise in a riem geo book
bit of a hard exercise lol
13:10
where has this been my whole life tikzcd.yichuanshen.de
what is this
oh, dam
I'll never use this, but cool
What I'd like is a decent program to draw TeX diagrams
pstricks or DEATH!
13:21
I tried TikzEdt but it's not great
Given the parameters in unusual form, how to find the vertex of the parabola?
Unusual form: $y = t^2+ t+1$, $x = t^2-t+1$
The usual form of parameters is $y = 2at$, $x= at^2$
The most i've used so far is Geogebra's tikz export but it's awful
@Abcd I guess what I notice is what happens when you add/subtract x and y
@Semiclassical Yeah even I noticed that $x+y =2t^2+ 2$
But they want vertex of the parabola
Uh, no
If it was x+y=2 that’s a line
13:24
@BalarkaSen Offset's "I can't vibe with queers" line made it onto YFN Luci's album. You'd think they'd edit that out.
Ok. What about x-y?
$x- y = 2t$
Right. Can you eliminate t now?
sure
And that would be the equation of the parabola, I think.
yeah. You’re not necessarily done at that point, but it’s the right place to start
13:28
I can find the vertex easily after that
y^2 = 4ax
comparison.
@Semiclassical One thing I never get is why elimination of parameter yields the equation?
Well, let’s find out. What do you get for the equation?
$4x+ 4y = x^2 + y^2 - 2xy + 8$
@Abcd not sure what a good answer to that is tbh
The method "just works".
They do it so many times.
@Abcd that doesn’t look quite right
13:31
This sort of elimination
Let me recheck
Should be 2(x+y)=4t^2+4=(x-y)^2+4
$2x+2y = x^2 +y^2 - 2xy + 4$
Yeah
The thing to see here is that this isn’t of the form y^2=4ax, at least not in terms of x and y
The point is that while this is a parabola, it’s not one whose axis of symmetry is parallel to the y-axis
It’s a rotated parabola
@Semiclassical its symmetric about the line x=y
I just have to see its intersection with that.
Good call
13:38
Hmm, done.
equally well, you can note that x=t^2+t+1=y=t^2-t+1 has one solution
If $f_n : X \to \Bbb{R}$ is uniformly continuous for every $n$, where $X$ is a metric space, is $\inf \{f_n \mid n \in \Bbb{N} \}$ uniformly continuous?
But then:
--> I don't understand the working of parameter elimination
Well, here’s a brute force statement
Suppose x, y are both quadratic functions of t.
Then I can choose some linear combination of x,y such that the t^2 term drops out
And I can pick another linear combo such that the t term drops out
First one has only t in it, and second has only t^2, so you can eliminate t from the second
That’ll mean you have something like (ax+by+c)^2 = dx+ey+f
Which is a conic section
So having x,y being quadratic functions of t always yields a conic section
The thing I’d point out here is that, had the polynomials been cubic, this wouldn’t have worked
You could pick a linear combination that killed off t^3, but you’d still have t and t^2 dependence
So you’d not be able to get an equation involving just x,y in this manner
So the case of two quadratic functions is rather special
I guess the short version of this is that, while the quadratic case may look complicated, it’s still simple enough that the “eliminate t” advice works
I’m sure there’s more that could be said about this, but that’s all I can say myself
(If there are other thoughts, I invite people to give them!)
13:57
@TedShifrin do you have any lectures on an intro to manifolds? I didn't find them in your course.
14:17
Hi,
So I am struggling a bit with a percentage question now
"in 2016 B was = to 55% of A. Suppose A Grows by 4% annually, and B grows by 6% annually. When will B = A?
@Semiclassical Could you also tell me how does symmetry about the line $x=y$ work? I just know how to test it... by replacing xs by ys and ys by xs and if we get the same equation, the graph is symmetric about $x=y$
What you should notice is that swapping x and y is the same as changing t to -t
But you’re considering all real values of t, positive and negative
So for instance (x(1),y(1))=(1,3) while (x(-1),y(-1))=(3,1)
Why does $\sigma(S) \ne S$ imply the action is faithful here?
You could also imagine a scenario where y(t)=x(1-2t). In that case, you’d still end with symmetry but it’d be a lot harder to see in the algebra
Nevertheless, you’d still see the symmetry once you eliminate t from the equations
That’s one reason why that form can be more convenient
14:33
Guys , I have a doubt
This is unusual , I have no idea how to begin with a limit like this.
As discussed in another room, guys, this is a very very very strange limit question that you might be interested to have a look at it, cause it has pontentially unbounded number of terms in it, thus even if the limit for each term exists, the formula $\lim_{n\to a} (f+g) = \lim_{n \to a} f + \lim_{b\to a} g$ may not be valid
@Tanuj why don't you post on main
@Tanuj Recall that if $f : [a, b] \to \Bbb R$ is a (continuous) function then $$\lim_{n \to \infty} \frac{b - a}{n} \sum_{k = 1}^n f\left (a + \frac{k}{n}(b - a)\right) = \int_a^b f(x) dx$$
@Abcd I thought of giving it a try and ask it here rather on main.
What do you have in mind for p here? If p>1 then n^p blows up as n goes to infinity whereas for 0<p<1 this goes to zero insteaf
14:37
@BalarkaSen yes , I was thinking on the same lines
Try out your limit for $f(x) = x^p$ on the interval $[0, 1]$.
There will be conditions on $p$ for which this converges/converges to whatever it does like @Semiclassical says of course
o wow, that's a riemann sum in disguise, I never expected that...
@Semiclassical But symmetry is related with reflection
If p is a positive integer then the limit can be checkef using Faulhaber’s formula
@BalarkaSen I have done something like $\sum_{r = 1}^n \dfrac{r^p}{n^{p+1}}$
What do I do now , how do I convert it into an integral from here ?
14:39
(Actually deriving Faulhaber isn’t easy)
It's $1/n \sum_{k = 1}^n (k/n)^p$
@Abcd ok?
@BalarkaSen yes
@Semiclassical wrong link, it links to some matrix weirdness?
@Semiclassical I wanted to understand it in the reflection sense...why does this way work: swap x and y , you get the same equation \implies your equation is symmetric about x=y
14:40
Doh
@BalarkaSen ahh got it . One doubt though , how did you decide the limits of the integral ?
@0celo7 lol
I heard about that controversy
@Tanuj I mean it's $1/n \sum_{k = 1}^n (0 + k/n \cdot (1 - 0))^p$
So $a = 0$ and $b = 1$ seems like the right choice
@BalarkaSen umm , I would never find that out myself , isn't there a more intuitive way to find limits of the integral ?
14:43
@Abcd well, the values of t you consider run from -infinity to infinity
Which means that the values of -t run from -infinity to infinity as well
@Semiclassical Lets forget about that question (the $t$)...I am talking about a general $x-y$ equation
So therefore (x(t),y(t)) and (x(-t),y(-t)) trace out the same curve
like $2x^2 + 2y^2 = x+ y $, swap x and y, you get same equation.
Right. So if (x0,y0) solves the equation then so does (y0,x0)
And the latter is the mirror image of the former across the line x=y
So the mirror image of every point on the curve is also on the curve. But that means the curve is symmetric across that line
I see. Got it.
14:48
Things are more tedious if you wanted to reflect across a line like y=1-2x
The same idea is there, though
@BalarkaSen Thanks man :)
@Tanuj Well, look up Riemann sums. $a + (b-a) \cdot k/n$ are the "partition points" of the interval [a, b] over which you integrate $f$
(A good exercise for this would perhaps be to show that every line through the center of a circle is a line of symmetry for that circle)
So it shouldn't be too hard to conceptualize
@Tanuj No problem
15:26
is $\frac{-x^2}{y}$ convex?
on $\Bbb R \times (-\infty,0)$
Hello!!

Let $v_1, v_2, v_3, v_4$ be elements of a vector space.

Does it hold that $\langle 2v_1, 3v_2, \frac{1}{2}v_2+v_3, -3v_4\rangle=\langle v_1, v_2, v_3, v_3+v_4\rangle$ ?
15:42
If $f : \Bbb{R} \to \Bbb{R}$ is continuous at a point, then is $f$ continuous on some nbhd of that point?
16:12
Hi guys! In this question, I know that $f(x + 2\pi) = f(x)$ and $f(x - 2\pi) = f(x)$ but don't know how to deal with the bounds of the integrals
make a change of variables
Does it not simple follow up from the property?
0
Q: Find product of matrix with mapping

EvindaWe consider in $\mathbb{R}^2$ the basis $B=\{ (2,1), (5,3)\}$ and $C=\{ (1,1), (2,3)\}$ and the mapping $f: \mathbb{R}^2 \to \mathbb{R}^2$ such that $f(x,y)=(x+y,-2x+y)$. Let the matrix $Cf_B=\begin{pmatrix} \alpha_{11} & \alpha_{12}\\ \alpha_{21} & \alpha_{22} \end{pmatrix}$. I want to compute ...

Hey... do you have an idea?
Is anyone familiar with Python ?
A little
Familiar with Py
16:16
Do you know how to model a hexagonal network?
Hey @Semiclassical
Did you see my question?
It is to model what is represented by the image. For squares, it's easy just take a matrix. Almost every square has 4 neighboring squares, for a hexagonal network it would take 6 neighboors for each box.
Well . You can use custom data type for this
Define an Entity type
(Where Entity denotes a Hexagonal element)
Define another type NetworkManager
Ok. I will look at this. Thanks
16:26
Now NetworkManager has a connect Method() which takes two params and keeps this params in its Cache
That is a Relation between Entity is established
And kept in Cache
Ok
Which will be later retrieved through get Neighborhood method of the Manager
@Lucas $u=x-2\pi $
$x=u+2\pi $
$dx=du$ and $u_1=a-2\pi, u_2=b-2\pi$
$\int_{a-2\pi}^{b-2\pi}f(u+2\pi)du=\int_{a-2\pi}^{b-2\pi}f(u)du=?$ And then ?
Did you use the fact that $f$ is $2 \pi$ - periodic ?
just apply the variable change correctly and use the periodicity
Yes, this is how I wrote $f(u+2\pi)=f(u)$
16:39
you started from the first integral (in the picture) so you can conclude the first and the third are equal
Then do the same with u= x + 2 pi instead of u= x - 2 pi
16:50
@Lucas How to get $\int_{a-2\pi}^{b-2\pi}f(x)dx$ from $\int_{a-2\pi}^{b-2\pi}f(u)du$ ?
But that's the same
whatever the name of the variable is
So there is no harm to conclude this $\int_{a-2\pi}^{b-2\pi}f(u)du=\int_{a-2\pi}^{b-2\pi}f(x)dx$ I mean is it proper to write that way?
17:06
yes
@user193319 no
17:19
Non brute force method for:
$a^2 + 19a + 92 ---> \text{perfect square}$
0
Q: Error in Problem Statement of Prob. 2 Chap. 2 in Rudin's RCA?

user193319 Let $f$ be an arbitrary complex function on $\Bbb{R}$, and define $\varphi(x,\delta) = \sup \{|f(s)-f(t)| : s,t \in (x-\delta, x+ \delta) \}$ and $\varphi (x) = \inf \{\varphi(x,\delta) : \delta > 0 \}$. Prove that $\varphi$ is upper semicontinuous, that $f$ is continuous at a point if and onl...

@Abcd for integer $a$?
Testing it in mathematica, I find that $a^2+19a+92$ isn't a perfect square for any integer $a$ from 0 to 10000
@Semiclassical O_O
Which is something mathematica can do surprisingly fast...
Okay, the actual question is:
17:24
lol, it does a up to 100000 in 4.625 seconds
god bless mathematica
$a^2 + 19a + 92 = (a + 19/2)^2 + (92-(19/2)^2)$. Multiply out by $4$ to get $(2a + 19)^2 - 7$
If $a^2 + 19a + 92$ was a perfect square this would be too.
But then not many situations where (2a + 19)^2 - (2a + 18)^2 is less than or equal to 7....
yeah, especially since that simplifies a bit
Mhm.
(n+1)^2 - n^2 is 2n + 1 which is greater than 7 whenever n is greater than 3.
17:29
actually, there are integer solutions if you pick $a<0$
But I"m not sure that's what the question wanted.
Ya.
You can use the same technique to count the solutions even if a < 0
hmm
@BalarkaSen I'm not sure this is enough for $a<0$, actually; it gives an upper bound on $a$, but not a lower bound.
9 mins ago, by Abcd
Okay, the actual question is:
Actually, why (2a+18)^2?
If roots of equation $x^2 - (2n+18)x - n- 11= 0$, $n \in \mathbb{Z}$ are rational numbers for $n=a$ and $n=b$, then $|a-b|$ is: ?
17:36
I'd use the rational root theorem there
@Semiclassical It is enough for sure. You're asking when n^2 - 7 is a perfect square. That forces n^2 - 7 to be (n-k)^2 for some k. If n > k, (n-k)^2 is less or equal to (n-1)^2, which is why you study the gap n^2 - (n-1)^2.
You need to think about the various cases which I can't be bothered to do right now
Fair enough.
If n becomes negative (n-k)^2, for k > 1, becomes greater than (n-1)^2, being the idea.
Right.
@Abcd If you make use of the rational root theorem, I don't think you have to think about when $a^2+19a+92$ is a perfect square.
@Abcd Presumably there's some clause that $a\neq b$ as well
@Semiclassical maybe we have to assume that
@Semiclassical {-n-11} is a root?
17:41
otherwise, you could just find one value of $a$ that works and do $b=a$
That's one (possible) condition. But if $-n-11$ is a root, then what's the other root?
$n+11$
No.
$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1 r_2$
First of all $2n+18 -> \text{Integer}$
If one of the roots is $x=-n-11$, and the equation is of the form $x^2-(2n+18)x-(n+11)=0$
then what's the other root
no. that's overkill.
Polynomial long division
17:44
again, overkill.
What do the coefficients of your quadratic tell you about the roots?
sum of roots
product
Right. Here you've got the product of the roots as $-n-11$.
So if one of the roots is $-n-11$, what's the other one?
Right. So one of the options the rational root theorem allows here is that the polynomial is of the form $(x-1)(x+n+11)$.
What's the other option?
n+11 is a root
11 mins ago, by Abcd
If roots of equation $x^2 - (2n+18)x - n- 11= 0$, $n \in \mathbb{Z}$ are rational numbers for $n=a$ and $n=b$, then $|a-b|$ is: ?
And other root is -1
$(x+1)(x-n-11)$
17:47
yep
How would you prove that two matrices A and B such that AB = BA are diagonalizable with a same matrix of passage?
simplest point now is to compute the sum of the roots and compare with the given polynomial
I dont see why/how that works.
well, we know that the sum of the roots should be 2n+18 by examining the given polynomial
but we just said that the roots are either 1,-n-11 or -1,n+11
not a lot of options for the sum of the roots.
Oh. I just realized I left off a possibility.
What if -n-11=0?
@Semiclassical so?
17:53
If the last coefficient c=-n-11 vanishes, then the rational root theorem no longer requires that the roots be drawn from 1,-1,c,-c.
@Semiclassical I get n = -8 using 2nd pair
@Abcd yep
and $-28/3$ from first pair
@Abcd which isn't an integer, so you reject it
@Semiclassical We need one more value but
9 mins ago, by Abcd
11 mins ago, by Abcd
If roots of equation $x^2 - (2n+18)x - n- 11= 0$, $n \in \mathbb{Z}$ are rational numbers for $n=a$ and $n=b$, then $|a-b|$ is: ?
for n=b
17:56
Right.
And my point is that, in saying that the rational root theorem requires the roots be one of -1,1,-n-11,n+11
well, that statement is only true if the last coefficient is nonzero.
i mean, take $x^2-19x=0$.
the rational root theorem requires that the rational roots be of the form p/q where p divides 0 and q divides 1. but that means p could be anything.
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