Mathematics - 2017-04-16 (page 5 of 6)

Space Otter

17:00

@DHMO please enlighten me oh great one

Ted Shifrin

What you did does not look correct, @SpaceOtter.

Lozansky

What's that integrand

DHMO

Hint:
$\dfrac{\mathrm d}{\mathrm dx} \cot x = -\csc^2 x$
$\dfrac{\mathrm d}{\mathrm dx} \csc x = -\csc x \cot x$

@Lozansky cosecant.

@SpaceOtter what the hell did you just call me in front of @TedShifrin and @AkivaWeinberger

and @AlessandroCodenotti

Akiva Weinberger

Haha

Space Otter

Lol

Simply Beautiful Art

17:01

@DHMO Haha..........

HAHAHA... wow, double laughs

Akiva Weinberger

@SpaceOtter Do you know the integral of $\sec$?

Lozansky

yeah but is it cos*sec or cos(sec)? because the first one is just one

Simply Beautiful Art

@DHMO You enjoy answering questions on MSE?

DHMO

@Lozansky cosec = cosecant.

Akiva Weinberger

Cosecant

DHMO

17:01

@SimplyBeautifulArt quite

Ted Shifrin

There is a cool way to derive these integrals using partial fractions. Do you know this, DogAteMy, @DHMO?

DHMO

@TedShifrin I did not

enlighten me, o great one

Lozansky

Lol okay

Akiva Weinberger

There's this really weird and hard-to-guess antiderivative of $\sec$. You can basically just put "co" in front of all the stuff I think (and maybe a few minus signs?)

@TedShifrin I do.

Ted Shifrin

Write $\dfrac1{\sin x} = \dfrac{\sin x}{\sin^2 x} = \dfrac{\sin x}{1-\cos^2 x} = \dfrac{\sin x}{(1-\cos x)(1+\cos x)}$.

4

Simply Beautiful Art

17:02

@DHMO Just don't be sweeping me off the leaderboard

Alessandro Codenotti

@Balarka did you work out whether there are subspaces of $\Bbb R^3$ with a finite fundamental group? I remember you were wondering about that a while back

Ted Shifrin

DogAteMy: $(\text{co}f)' = -\text{co}(f')$.

Akiva Weinberger

@TedShifrin "Why? Because it works, that's why"

Space Otter

@DHMO Intergrals of inverse trig functions Isn't in my nearest exam so I skipped that section.
So I should be able to do it without it right?

Anonymous

If f(x) is differentiable at a point but g(x) is not. Then is it compulsory for f(g(x)) and g(f(x)) and g(g(x)) to be non-differentiable at that point?

DHMO

17:04

@TedShifrin that is nice

Ted Shifrin

DogAteMy: I think we can justify that reasonably using the unit circle, considering $\int \dfrac{dx}y$.

Akiva Weinberger

@blue Take $f(x)=x^2$ and $g(x)=|x|$ as an example. $f$ is differentiable at $0$, $g$ is not.

Ted Shifrin

@SpaceOtter: No inverse trig involved here.

Alessandro Codenotti

@blue take $f=0$ for the first one and $f="a point at which $g$ is differentiable" for the second

Akiva Weinberger

And yet $f(g(x))=g(f(x))=x^2$ is differentiable everywhere.

DHMO

17:05

let's come up with pathological examples of $f \cdot f = id$ where $f : \Bbb R \to \Bbb R$

Simply Beautiful Art

@DHMO Need to fix our answers.

Space Otter

@DHMO Who are you talking to?

DHMO

@SpaceOtter anyone

Space Otter

How do you quote message from earlier in chat?

Anonymous

@AkivaWeinberger But g(g(x)) is non differentiable. Any idea about that?

Anonymous

17:06

Is it always so for all non differentiable g(x)?

Akiva Weinberger

@DHMO I feel like, for continuous functions, we just have $x$ and $-x$

BAYMAX

Hi all.

Akiva Weinberger

@blue Hm, let me think

DHMO

@AkivaWeinberger can you prove it?

Alessandro Codenotti

@DHMO is that pointwise product or composition?

DHMO

17:07

@AlessandroCodenotti composition

oh god what did i write

let's come up with pathological examples of $f \circ f = id$ where $f : \Bbb R \to \Bbb R$

I'm sorry for the confusion

Akiva Weinberger

@DHMO Wait, no, I mean $x$ and $c-x$

BAYMAX

If $a =b.c$ and $p$ is a prime s.t. $p^2$ doesnot divide $a$ then is this true $p$ doesnot divide both $b$ and $c$ ?

DHMO

@AkivaWeinberger that is nice

Anonymous

@AkivaWeinberger Sure. If we differentiate it we get g'(g(x))g'(x)...here g'(x) doesn't exist and I don't know about g'(g(x))

Ted Shifrin

@BAYMAX: Take the contrapositive of what you just asked.

DHMO

17:08

@BAYMAX take a=10, b=2, c=5, and p=2.

Akiva Weinberger

@DHMO No, clearly there's more actually…

DHMO

@AkivaWeinberger how?

Ted Shifrin

@DHMO: You're not right.

DHMO

@TedShifrin how?

Alessandro Codenotti

There's plenty of super ugly involutions if you don't require continuity

Ted Shifrin

17:09

Read it carefully.

Space Otter

Anyway...
$$\int(-1+u^2)^{-1} du$$

Akiva Weinberger

Let $\phi:[0,\infty)\to(-\infty,0]$ be bijective, with $\phi(0)=0$.

DHMO

@TedShifrin oh god

Ted Shifrin

@SpaceOtter: That is not a correct integral. How did you get that?

Akiva Weinberger

Then $f(x)=\begin{cases}\phi^{-1}(x),&x<0\\0,&x=0\\\phi(x),&x>0\end{cases}$ works.

Alessandro Codenotti

17:10

And plenty of continuous one too

DHMO

the letter "B" is a swear word in Mandarin

Ted Shifrin

@Alessandro: Did you ever prove that continuity of pointwise limit of sequence of continuous functions thing?

DHMO

what is $\phi$?

BAYMAX

Ok,If $p$ divides $b$ and $c$ then $p^2$ divides $a$ is the contrapositive.@TedShifrin

Ted Shifrin

No, @BAYMAX, if $p$ divides $b$ and $p$ divides $c$ ...

Akiva Weinberger

17:12

@DHMO http://chat.stackexchange.com/transcript/message/36746041#36746041

Anonymous

Can there be any such function where $g(x)$ is not derivable at a point but $g(g(x))$ is derivable at that point?

DHMO

@blue yes.

consider an involution

Anonymous

@DHMO Like?

Anonymous

What's that "involution"?

BAYMAX

yes@TedShifrin

Akiva Weinberger

17:13

Ah, I see. So:

DHMO

$g(x) = \begin{cases}x&x \in \Bbb Q\\-x&x \notin \Bbb Q\end{cases}$

Ted Shifrin

So now can you prove it, @BAYMAX?

DHMO

@blue $g$ is nowhere differentiable but $g \circ g$ is everywhere differentiable

BAYMAX

yes contrapostive is true so yes the statement is true

Akiva Weinberger

$g(x):=\begin{cases}-2x,&x<0\\-x/2,&x\ge0\end{cases}$ @blue

Alessandro Codenotti

17:14

@TedShifrin nope, I forgot to be honest. I tried to construct the indicator function of $\Bbb Q$ as a double pointwise limit of continuous function and it even seemed to work, until I noticed my construction would contradict the Baire cathegory theorem if it were right...

BAYMAX

I sometimes get into this trap that is while negating I made or , any strong point in this one @TedShifrin

Anonymous

@DHMO Ah, right. But can there be any such continuous function? Your function isn't continuous...

DHMO

@blue look at @AkivaWeinberger 's function

Space Otter

@TedShifrin u=cosx
$\frac {du}{dx} = -sinx$ so $dx=\frac{-1}{sinx}$
Then $$ \int cosecxdx=\int \frac {-1}{sin^2x} du$$
And $sin^2x=1-cos^2x$

DHMO

oh and my function is continuous at $0$ @blue

Ted Shifrin

17:15

Well, @BAYMAX, if you have NOT (....), then the negation is NOT NOT (....) :P

Akiva Weinberger

Then $g(g(x))=x$ (for example, $g(g(6))=g(-3)=6$), so $g(g(x))$ is differentiable everywhere but $g(x)$ has a corner at $0$. @blue

Space Otter

Latex takes me a while I just started

DHMO

@SpaceOtter and you still forget to do \sin x

Space Otter

What do you mean, I don't see it

Ted Shifrin

Oh, I see, @SpaceOtter. My apologies. This is exactly the method I suggested up there ^^^^ using substitution with partial fractions. It's identical.

DHMO

17:16

$sin x$ vs. $\sin x$

@SpaceOtter

BAYMAX

yes, I was saying this - negation of $p$ divides b and c is $p$ doesnot divide b and c ,while i made a mistake of writing this like I took negation of each that is $~$divide = not divide and [~and = or] , but it is not correct any point on this?@TedShifrin

Space Otter

DOH

DHMO

@AkivaWeinberger that's nice

Ted Shifrin

I can't follow that, @BAYMAX, sorry.

BAYMAX

yes, I was saying this - negation of $p$ divides b and c is $p$ doesnot divide b and c ,while i made a mistake of writing this

Space Otter

17:18

@TedShifrin How do you partial fraction that when the denominator isn't a product?

DHMO

@SpaceOtter make it a product

Ted Shifrin

But $u^2-1 = (u+1)(u-1)$.

Space Otter

DOH

Anonymous

@AkivaWeinberger Right. That's a great example!

Space Otter

Thanks Chaps, can't believe I didn't see that @TedShifrin

BAYMAX

17:19

Ok,why the negation of $p$ divides $a$ and $c$ is not this one :- $p$ doesnot divide $a$ or $c$?@TedShifrin

like why not take negation of all!

may be silly but still!

DHMO

(I know this is an old problem but bear with me)

Prove that $(-1) \times (-1) = 1$

Ted Shifrin

@BAYMAX: Your original sentence was "$p$ does not divide both $b$ and $c$."

The negation of NOT (....) is simply (....).

Simply Beautiful Art

@DHMO My English teacher said that double negatives turned things into positives, so yeah.

DHMO

@SimplyBeautifulArt bingo... not

Ted Shifrin

Oh crap.

Akiva Weinberger

17:21

@DHMO What axiom system

DHMO

@AkivaWeinberger ring

(you can try the peano extended to the integers if you like)

where $-x = [[(0,x)]]$

Ted Shifrin

@BAYMAX: There is often ambiguity with the language, but this should be interpreted as saying NOT ($p$ divides both $b$ and $c$).

BAYMAX

Yes negation is p doesnot divide b and c ?

or

p doesnot divide b or c

@TedShifrin

Ted Shifrin

You can rewrite it as $p\nmid b$ or $p\nmid c$, but why bother doing that when we want to NEGATE the NOT (....) statement?

Akiva Weinberger

@DHMO \begin{align}(-1)\times(-1)&=(-1)\times(-1)+0\\& =(-1)\times(-1)+(-1)+1\\&=(-1)\times(-1)+(-1)\times1+1\\ &=(-1)\times(-1+1)+1\\&= (-1)\times0+1\\&=1\end{align}

BAYMAX

17:27

oh I am little confusing I guess,you ask me a negtion of a statement and I will tell@TedShifrin ?

Lozansky

The scale factor from cartesian to spherical is $r^2 \sin \psi$ right?

DHMO

@AkivaWeinberger That's nice, I'll let you pass if you don't have time to prove the absorbance of $0$

Akiva Weinberger

@DHMO Isn't that a ring axiom?

DHMO

@Lozansky $\mathrm dx \ \mathrm dy = r \ \mathrm dr \ \mathrm d\theta$, if that's what you're looking for

Akiva Weinberger

That $a+0=a$?

DHMO

17:27

not this

Ted Shifrin

no, @BAYMAX, I'm done.

Akiva Weinberger

Oh, whoops

You mean $a\times0=0$.

Lozansky

@DHMO Looks polar to me

DHMO

@Lozansky oh, never mind

BAYMAX

ok ><

DHMO

17:28

@Lozansky you might like this page

Ted Shifrin

@DHMO: The usual thing you have to be careful about is proving that $(-1)\times a = -a$.

DHMO

@TedShifrin sure

Lozansky

@DHMO Yeah I couldnt figure out the name. Cheers

Akiva Weinberger

@TedShifrin I avoided that, I think

Ted Shifrin

You used it only for $1$, I think, DogAteMy :)

I usually write the argument a bit more succinctly, but I think it's essentially the same.

DHMO

17:30

T/F: $\forall A,B \in \Bbb R^{n \times n}: (A+B = I_n) \implies (A^2 + 2AB + B^2 = I_n)$

BAYMAX

Ok

The negation of "A or B" is the statement "Not A and Not B."

Sha Vuklia

In a game of bridge, the 52 cards of a conventional pack are distributed at random between the four players, in such a way that each player receives 13 cards. Show that the probability that each player receives one ace is
$$
\frac{24\cdot48!\cdot13^4}{52!}
$$
We're dealing with equiprobable outcomes, so I know that $\mathbb P(A)=\vert A\vert/\vert\Omega\vert$. I can see that $52!=\vert\Omega\vert$ and also that $48!$ stands for the number of ways we can distribute all cards except the aces amongst the players. Now I don't see exactly where $24$ and $13^4$ come from. Could someone explain th…

Akiva Weinberger

@TedShifrin That was the multiplicative identity property I used

$1\times a=a$

Ted Shifrin

I know, DogAteMy.

DHMO

looks like nobody is interested in my challenge

@ShaVuklia $24=4!$ is which ace each player receives

Meow Mix

17:32

OK this sounds like a stupid question but how do we generalize multiplication to all of the reals?

Akiva Weinberger

@DHMO $a\times0=a\times0+a\times0-a\times0=a\times(0+0)-a\times0$

Meow Mix

I mean, for the rationals it's pretty obvious

DHMO

@MeowMix using which definition of the reals?

Akiva Weinberger

$=a\times0-a\times0=0$

Semiclassical

I have a suspicion that this person isn't an English speaker: math.stackexchange.com/q/2237104/137524

DHMO

17:33

$13^4$ is where the ace is found in each player @ShaVuklia

@AkivaWeinberger nice

Sha Vuklia

@DHMO oh right, I didn't make that distinction

thank you!

DHMO

@TedShifrin are you interested or you know the answer already?

Meow Mix

I suppose using Cauchy sequences?

Daminark

Hey everybody, back!

Akiva Weinberger

@DHMO For what, this?

Ted Shifrin

17:34

Of course I know, @DHMO.

DHMO

@MeowMix then $x y = [[x_ny_n]]$

Ted Shifrin

Hi Demonark.

DHMO

@AkivaWeinberger yes

Meow Mix

oh pfft

Akiva Weinberger

Oh. Of course it's false :P

DHMO

17:34

@TedShifrin of course you're a professor

@AkivaWeinberger think thrice

Akiva Weinberger

Um…

I mean, unless $A+B=I$ implies $AB=BA$…

Ted Shifrin

Unless ...

Semiclassical

It does.

$A=I-B$ certainly commutes with $B$.

DHMO

lol I wanted @AkivaWeinberger to come up with the proof on his own

@TedShifrin I'm quite amazed by your speed

have you seen it before?

Ted Shifrin

No, @DHMO, but remember I've written a bunch of textbooks with linear algebra in them.

DHMO

17:37

@TedShifrin I know

Akiva Weinberger

@Semiclassical Oh!

Argh!

Ted Shifrin

DogAteMy: It's a trick trick question.

Alessandro Codenotti

@Ted so I want to prove (to begin with) that if $f$ is the pointwise limit of the continuous functions $\{f_n\}$ and $f$ is continuous at $x_0$ then $f$ is continuous in a nbhd of $x_0$. My problem is that regardless of how small I pick $\epsilon$ it's not true that there there is an $\tilde{n}$ such that $|f_n(x)-f(x)|<\epsilon$ for all $x\in(x-\epsilon,x+\epsilon)$ and $n>\tilde{n}$

BAYMAX

I think this is false

29 mins ago, by BAYMAX

If $a =b.c$ and $p$ is a prime s.t. $p^2$ doesnot divide $a$ then is this true $p$ doesnot divide both $b$ and $c$ ?

Alessandro Codenotti

Which is basically the same as saying that pointwise convergence isn't uniform... Derp

Semiclassical

17:38

If $p$ divides both $b$ and $c$ then $p^2$ certainly divides $bc$.

Alessandro Codenotti

Ok, nevermind, no hints yet, I think I need to change approach entirely

Akiva Weinberger

@MeowMix You'd want to prove that that sequence is still Cauchy and that it's independent of the choice of sequences for $x$ and $y$

Ted Shifrin

If all pointwise convergent sequences were uniformly convergent, life would be different, indeed, @Alessandro.

Daminark

That's very true @Baymax

Akiva Weinberger

(Existence and well-definedness)

Daminark

17:39

That'd be convenient for sure

Everything would just commute and life would be great

Ted Shifrin

We wouldn't need Lebesgue integration, for one thing.

Daminark

@Akiva Constructing $\mathbb{R}$?

DHMO

@Daminark multiplication of two numbers in $\Bbb R$

Daminark

Would that completely take care of the problems with Riemann integration @Ted?

BAYMAX

the above is of the form $P \implies Q$ right @Semiclassical@Daminark?

Mats Granvik

17:41

Are the Siegel zeros (conjectured/believed to be) the same as the Riemann zeta zeros?

Ted Shifrin

Well, the characteristic function of the rationals is the pointwise limit of a sequence of integrable functions, Demonark :P

BAYMAX

and while contrapositive we do negation $ Q \implies $ neagtion of $ P$

Daminark

@DHMO I appreciate that you're doing this by way of Cauchy classes and not Dedekind cuts

Semiclassical

$p|b$ and $p|c$ implies $p^2|bc$.

DHMO

@Daminark why? Dedekind cuts are much simpler

(I didn't choose Cauchy. @MeowMix did.)

Daminark

17:42

Verifying the field axioms are annoying

DHMO

That's true

Semiclassical

Verifying axioms in general tends to be tedious.

Ted Shifrin

That's probably an axiom.

7

Daminark

It's marginally better with Cauchy stuff

BAYMAX

and while contrapositive we do negation $ Q \implies $ neagtion of $ P$ , I s this true @Semiclassical

Daminark

17:43

Plus, like, Cauchy constructions generalize to metric spaces

I think we encounter those far more than Dedekind constructions

DHMO

but Dedekind cuts generalize to surreal numbers

Semiclassical

Sure. i.e. if $p^2$ doesn't divide $bc$ then either $p$ doesn't divide $b$ or $p$ doesn't divide $c$.

BAYMAX

ok so considering now negation of $Q$

Alessandro Codenotti

Dedekind completion generalizes to stuff in order theory and lattice theory (which is probably one of the reason set theorists prefer cuts over cauchy sequences)

Daminark

@DHMO So a friend of mine over the last summer did his REU paper on Conway-style game theory and surreal numbers

The definition of a game angered me so much

I'm just like "HOW IS THIS NOT CIRCULAR???!!!! >:("

DHMO

17:45

@SimplyBeautifulArt lolll

Daminark

And then I realized it's fine because vacuous truth of the empty set...

Akiva Weinberger

There's a fine line between circular and inductive

BAYMAX

negation (P $\nmid$ b and P $\nmid $ c) is $P \mid b$ or $P \mid c$ right @Semiclassical

Simply Beautiful Art

@DHMO :P

DHMO

@SimplyBeautifulArt that censorship

Daminark

17:46

Honestly it felt like all of that subject followed from vacuous truth in the empty set

Ted Shifrin

well, most inductive proofs follow just from $1=1$ or some such, Demonark.

Simply Beautiful Art

@DHMO Well... if you want to <strikethrough>strikethrough</strikethrough> it...

Semiclassical

That's true, but I'm not sure why you're doing that version.

Simply Beautiful Art

Or however you do it

DHMO

~~strikethrough~~

Semiclassical

17:47

The relevant version has "and" and "or" swapped.

Daminark

Especially Conway induction... Like that stuff was dank and all, I want to check it out, but eek

It was aggressive

Simply Beautiful Art

Yeah

BAYMAX

yeah I did that t=right@Semiclassical

and , or swapped?

Koolman

0

Q: How many Digits in $3^{2020}$

I know n+1 = number of digits in $3^{2020}$ but how to find number of digits . How can we do it without using a calculator .

DHMO

@SimplyBeautifulArt ...

I fixed it for you

Simply Beautiful Art

17:49

...

Semiclassical

not(p|b and p|c) = not(p|b) or not(p|c)

Simply Beautiful Art

Oh...

xD

BAYMAX

yes, any mistake in my writing above@Semiclassical

Simply Beautiful Art

@DHMO That's why you should make your MathJax neat xD

Semiclassical

well, my point is that what I just wrote isn't the same as what you did.

DHMO

17:50

@SimplyBeautifulArt I didn't intend it to be edited

Ted Shifrin

@Semiclassic: You'll see if you scroll up a lot that I said at least twice that the negation of NOT (....) was merely (....).

Semiclassical

Ahh.

Simply Beautiful Art

@DHMO Nope, editing police coming through

BAYMAX

sorry but I am unable to find the difference.yes negation of $p \nmid b$ is $p \mid b$

and negation of and is or , so what is wrong in my statement?

Semiclassical

Like I said, it's not wrong. It's just not relevant to the problem you posed above.

BAYMAX

17:54

like can you point where I got deviated,

Like we need to do negation of $Q$

Semiclassical

Your question earlier could be stated as whether or not "If p^2 doesn't divide bc, then p doesn't divide b or doesn't divide c" is true

BAYMAX

and while doing that we need to negate $\nmid$ ,and right?

Semiclassical

The point, though, is that the implication is not "p doesn't divide b AND p doesn't divide c"

it's OR.

BAYMAX

actually this one -

19 mins ago, by BAYMAX

29 mins ago, by BAYMAX

If $a =b.c$ and $p$ is a prime s.t. $p^2$ doesnot divide $a$ then is this true $p$ doesnot divide both $b$ and $c$ ?

DHMO

T/F: The interval [-1,1] under the operation + forms a group

Zophikel

17:57

I'm having trouble discerning whether this is a valid approach to prove the following. For any points. $x$,$y$ in $\mathbb{R^{k}} let in $(1.)$ $$(1.)$$ $d_{1}(x,y) = max{|x_{j} - y_{j} | : j = 1,2,...,k}$ and $d_{2}(x,y)=\sum_{k}^{j=1}|x_{j}-y_{j}|$ show $d_{1}$ and $d_{2}$ are metrics in $\mathbb{R^{k}}

Semiclassical

That's exactly what I was paraphrasing.

"p does not divide b and c" is not the same as "p does not divide b and p does not divide c."

Alessandro Codenotti

@DHMO it's not closed wrt that operation

DHMO

@AlessandroCodenotti bingo

Semiclassical

"2 does not divide 4 and 5" is true. "2 does not divide 4 and 2 does not divide 5" is false.

Zophikel

My intial idea was to use the definition of a metric space and for the beginning of the proof let $\mathbb{R^{k}}$ reside in a set $S$ and go from there

BAYMAX

18:11

yes@Semiclassical , I agree but I have not done that?

13 mins ago, by Semiclassical

"p does not divide b and c" is not the same as "p does not divide b and p does not divide c."

Ok ok

why aren't they same? My whole mistake was that!

sorry

why they are not same?

Semiclassical

Well, the example I just gave should suffice.

BAYMAX

Nice thank you very much @Semiclassical , I have to brush up that everyday.

sorry@TedShifrin

Semiclassical

It's the annoying fact that the not(A and B) is equivalent to not(A) OR not(B), rather than not(A) AND not(B).

Sha Vuklia

A family $\mathcal A=(A_i:i\in I)$ of events is called independent, if for all finite subsets $J$ of $I$,
$$
\mathbb P\left(\bigcap_{i\in J}A_i\right)=\prod_{i\in J}\mathbb P(A_i).
$$
I'm wondering why do we need $J$ to be a finite subset? Infinite products are defined after all, and we also have the following:
$$
\mathbb P\left(\bigcup_{i=1}^\infty A_i\right)=\sum_{i=1}^\infty \mathbb P(A_i),
$$
where $A_1,A_2,\dots$ are disjoint events in $\mathcal F$.

BAYMAX

so ,$a = bc$ and $p$ be a prime such that $p^2 \nmid a ,$$p \nmid b$ and $p \nmid c$ and now I think this is false!

I think this is the different version of my first one

Semiclassical

18:17

Here's a relevant example for your purposes. We have 24=4*6, and 2 divides both 4 and 6, and we have 4 divides 24.

BAYMAX

just did an edit!

Semiclassical

On the other hand: 4*5=20 is divisible by 4 despite the fact that 2 doesn't divide 5.

Whereas 2*5=10 isn't divisible by 4 and 2 doesn't divide 5.

So we can't conclude "4 doesn't divide bc" if we only know that 2 doesn't divide c.

BAYMAX

so is my above statement false!

Semiclassical

Perhaps. I'm a bit unsure which statement you're considering.

The original statement you provided is certainly true.

BAYMAX

3 mins ago, by BAYMAX

so ,$a = bc$ and $p$ be a prime such that $p^2 \nmid a ,$$p \nmid b$ and $p \nmid c$ and now I think this is false!

Semiclassical

18:19

Yeah, that's not terribly clear.

BAYMAX

so ,$a = bc$ and $p$ be a prime such that $p^2 \nmid a$ then $p \nmid b$ and $p \nmid c$ and now I think this is false!

Semiclassical

Sure. 4 doesn't divide 2*5, but 2 divides 2.

BAYMAX

as doing contrapositive $p \mid b$ or $p \mid c$ doesnot imply that $p^2 \mid a$

Nice one! thanks@Semiclassical

Semiclassical

np

BAYMAX

Now I need to brush up!

ok bye , have a nice day!.

Studentmath

18:37

I've got a question. Why does:
$\sum_{n=1}^N \{N/n\}<N$, where $\{N/n\}+\lfloor N/n \rfloor=N/n$? Or rather, is there a neat way to explain it?

Anonymous

@Studentmath $0 \leq \{\frac{N}{n}\} <1$

Studentmath

Yeah bu-

Jesus I am stupid. For some reason I thought of $\sum_{n=1}^N n\cdot \{N/n\}$

Thanks @Blue

Anonymous

np :)

Studentmath

I was like "wait but then I have to eliminate so many options!"

It's actually less than $N-2$, if $N>2$.

Anonymous

18:57

Can a certain point of intersection of $f(x)$ and $f^{-1}(x)$ not lie on $y=x$ ?

Anonymous

Can someone think of any example like that ?

Studentmath

@Blue well the may intersect at every point, or not intersect at all, too

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$Mathematics$ Mathematics