help-49

[small warren]

yea XDDD

[civic lynx]

the x-values are already minutes, i think you mean you have to convert 1 tick into 3 minutes

but yeah you get the idea

so, if y=45 at 2 ticks in, how many mins is that

[small warren]

so should 0</=x</=6 be the solution for the interval?

[small warren]

6

[civic lynx]

yes and yes ^-^

[small warren]

Yayyy thanks! i also got another question wrong on my test but i immediately saw what was wrong, out of 20 questions i got 3 wrong for a total of 85% right!

[civic lynx]

this is unfortunately very accurate to trying to read scientific graphs in the field, they tend to be very crunchy pdfs with weird units. so you do always have to check 😔

[small warren]

that sucks because it lowers my gpa though :((

[civic lynx]

and now its 100% in your heart :)

[small warren]

Yea, i wasn't sure if the question was asking for the actual x units, or time, because both start at 0 minutes and 0 x

[small warren]

Mhm! ❤️ tysm

.close

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[small warren]

.i really appreciate the help!

[civic lynx]

. if it just says x, go with the unit (number of minutes). if it says "ticks on the graph," i would only count ticks then

[small warren]

.Okay ty! ❤️ noted

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[flat hollow]

Hello, can someone please assess whether my induction proof fulfills all the formalities and is logically sound

[flat hollow]

just want to check if my formatting is correct

[small jasper]

Looks good

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[last slate]

can i get help on 73

[last slate]

.

<@&286206848099549185>

[hushed mauve]

!15m, please

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[deep vine]

!xy

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[hushed mauve]

?

[deep vine]

...

[hushed mauve]

not sure what you expected by xy'ing a helper but okay

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[placid spoke]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[dawn vapor]

1. I don't know where to begin.

1

[placid spoke]

try the substitution u = a - x

[dawn vapor]

ok ill try it

thnaks

[robust quartz]

In 2 question use property 1 it will give $$\cos x^n$$ in numerator then add it with original I

[grand pond]

Not_light

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[tribal bramble]

is ts ass? https://soundcloud.com/akaeysiyas/icfms?utm_source=clipboard&utm_medium=text&utm_campaign=social_sharing&si=a66f3b8c3ee2493faa84bcf45689a2e8

[lethal path]

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[lethal path]

the help channels are only for maths questions, not for chatting

!occupied

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[dreamy hedge]

mb i just joined

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[shut canyon]

https://www.desmos.com/3d/lc7eep3sgz

seems to trace a trefoil projection on xy

I don't yet know how to draw this

x(t) = cos(t) · (3cos(t) + 2)
y(t) = 5cos(t)sin(t)
z(t) = sin (t) · (25cos²(t) − 1)

[midnight plank]

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[shut canyon]

maybe just sketch the functions and shows trefoil?

[shut canyon]

x = np.cos(t) * (3*np.cos(t) + 2)
y = 5 * np.cos(t) * np.sin(t)
z = np.sin(t) * (25*np.cos(t)**2 - 1)

#projection:
ax.plot(x, y, zs=min(z), zdir='z', color='lightcoral', lw=1, linestyle='--')  # XY plane (floor)

◇🖊️

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[delicate meadow]

I do not understand the note.

[toxic prism]

its abt adding directed angles

convention basically

[delicate meadow]

Ohok.

Thank yu.

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[shut canyon]

$$
A=\begin{pmatrix}1&2\2&2\end{pmatrix},;
B=\begin{pmatrix}1&3&1\2&1&2\end{pmatrix}
;\Rightarrow;AX\neq B
$$

[grand pond]

kytsu

[toxic prism]

what's the question

[latent wadi]

You can reduce B

[shut canyon]

… ⟹ ¬∃X ( AX=B )
maybe this means the same

Questions is AX=B has solution?

AX≠B eqm AX=B doesn't have solution eqm ¬∃X ( AX=B )?

[latent wadi]

Show what you've tried

[toxic prism]

That or you can say A is invertible

[shut canyon]

A invertible iff det(A)≠0

and square matrix

only square matrix have determinants

[toxic prism]

yeah

[shut canyon]

in B c₁=c₃ , I don't know if that matters

[toxic prism]

that's irrelevant really

A is invertible $\iff$ the linear map $T:\mathbb{R}^n\to\mathbb{R}^n$, T(X) = AX is bijective

[grand pond]

donkey

[toxic prism]

and the way bijectivity of f(x) gives that f(x)=b definitively has an unique solution as x = f$^{-1}$(b), same way

[grand pond]

donkey

[shut canyon]

Thank you @toxic prism and @latent wadi !
this exercise seems tricky, with monoids and left inverse, right solution, bijectivity of inverse

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[worldly pine]

do

[worldly pine]

No idea how to dp

[autumn canopy]

29?

[worldly pine]

not in the option

[autumn canopy]

I meant, are you asking about problem 29?

[worldly pine]

oh...yes

[autumn canopy]

Have you tried anything?

[worldly pine]

no

but reading it didn't make any sense

[autumn canopy]

Drawing these often helps

[worldly pine]

ok..let me try

[autumn canopy]

Ok so the start is something like this

After 6 minutes, what can you say?

Any info from the problem you can extract?

[worldly pine]

still don't know what to do here

[autumn canopy]

6 minutes translate to 0.1 hours, do you agree?

Since 1 hour = 60 minutes

[worldly pine]

yes

[autumn canopy]

Ok, so after 6 minutes, the car has driven what distance

0.1 * x kilometres, right?

[Because it's driving with x km/h

and s = v * t]

[worldly pine]

what is s

speed?

[autumn canopy]

I mean the displacement. displacement = velocity * time

so the displacement after 6 minutes (0.1 hours) is 0.1 * x kilometers, do you agree?

[worldly pine]

ok yes

[autumn canopy]

Ok and how far has the pedestrian walked in those 6 minutes/0.1 hours?

[worldly pine]

2 × 0.1

[autumn canopy]

Yes

and that is 0.2 kilometers

Ok so after 6 minutes, the distance between them is 0.1 * x - 0.2 kilometers, right?

[worldly pine]

yes

[autumn canopy]

And this is when the pedestrian loses the car out of sight, right

The exercise says that happens after 6 minutes

and it was visible to him up to a distance of 0.6 kilometers

So this 0.1 * x - 0.2 has to be 0.6

right?

[worldly pine]

ok got it...8 km/hr is the answer after solving this

[autumn canopy]

yeah

[worldly pine]

👍

thank you kepe

[autumn canopy]

np

[worldly pine]

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[timid topaz]

Hello back again from earlier hoping someone can check 🙇

[timid topaz]

the second picture shows my solution regarding to the question but I'm not still sure about it 🙏

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@timid topaz Has your question been resolved?

[latent wadi]

2 is correct

For 1, I've told you the denominator should be n instead of n-1. But if that is not in your curriculum, then just use n-1 cause that's what your professor probably expects you to use

Number 3 seems wrong, there's too little working for me to tell

For number 4, the denominator should be n-2 (that's how it works for the standard deviation of residuals)

No 5 standard deviation is the square root of variance so -4

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[hybrid crow]

How might I continue?

[dreamy lichen]

might be easier to prove if ST = TS, then T = lambda I

T != lambda I seems like a difficult condition to utilize

Maybe it could work actually

try picking some vector from the basis, for which Tv != lambda v

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[dusty portal]

rip

.reopen

[midnight plank]

✅ Original question: #help-49 message

@hybrid crow Has your question been resolved?

[bitter bison]

Use the fact that the matrices associated with these linear operators will commute if they are simultaneously diagonalizable and then use this fact to show that if T commutes with everything then there are two operators that dont commute with the same eigenvalues.

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[analog oyster]

@worldly pine hello

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[shut canyon]

Objects are isomorphic to themselves?

a : 1a ⟶ 1a,
iso(a) if ∃ 1a ⟶ 1a ⟶ 1a,
construct it: 1a⟶1a ∘ 1a⟶1a via composition

is this valid?
intuitively a is iso if you can walk from its domain to its codomain and back, which we can build from the definition of a!

Oh I think this is only the left inverse

[bitter bison]

Take the identity morphism

[shut canyon]

b ∘ a = id

b ∘ a = ida. a ∘ b = idb

[bitter bison]

You dont have a sense of composition if objects in a category

That requires extra structure

By definition for each object exist a morphism Id in Hom(a,a) s.t Id(a)=a

[shut canyon]

What if the objects are identity arrows?

[bitter bison]

You would be working in a different category

There is a category of arrows tho and their morphisms are commuting diagrams

[shut canyon]

It is also possible to have arrows only theory, so I thought maybe use it interchangeably, but I guess I should mention it

Arrows only axioms, no objects, id arrows stand for objects

Oh here is the definition for isomorphic objects:

there is a 𝒞 arrow f : a ⟶ a that is iso in 𝒞, ie f : a ≅ a

[shut canyon]

I have this idea for a proof:

In the arrows-only definition identity arrows stand for objects, so we see if 1ₐ is iso in an arrows-only category:

1ₐ : 1ₐ ⟶ 1ₐ

∃ f : 1ₐ ⟶ 1ₐ iso(f):
construct 1ₐ ⟶ 1ₐ ∘ 1ₐ ⟶ 1ₐ and its inverse:
1ₐ ⟶ 1ₐ ⟶ 1ₐ
∴ iso(1ₐ)

Writing object a for identity morphism in the objects-and-arrows definition for 1ₐ.
1ₐ ⟶ 1ₐ in the arrows-only definition is a ⟶ a, which is then an isomorphic arrow.

oh I didn't show the iso f

[bitter bison]

I think the identity being iso just falls out of it being the identity but it isnt hard to show that it is one

[bitter bison]

Also remember $1_{A}: A \to A$

[grand pond]

Calisto

[shut canyon]

in arrows-only A is an arrow, so even identity has arrows as domain and codomain, I think

[bitter bison]

Oh okay cool

[shut canyon]

More intuitive for me

[shut canyon]

I still find it tricky, but once I understand these proofs it seems simple, so it's a good exercise

Thank you for your help @bitter bison

[bitter bison]

Ofc :3

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[graceful ferry]

Can someone check my work?

[fallow scarab]

How did you get your L and D

[graceful ferry]

From previous questions, one sec

[graceful ferry]

In this part they requested alpha=5

I just subtituted

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[graceful ferry]

<@&286206848099549185>

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[graceful ferry]

<@&286206848099549185>

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@graceful ferry Has your question been resolved?

[sharp coral]

assuming your matrix factorization is correct then it's correct

[graceful ferry]

.solved

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[sharp fiber]

D - F ok?

[midnight plank]

@sharp fiber Has your question been resolved?

[sharp fiber]

I’m sort of not sure of D3 and E 1-3

<@&286206848099549185>

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[night axle]

Part D seems fine

E 1-3 too

F 1-2 seem right too

[timid spade]

D3 is wrong

[night axle]

What’s wrong in d3

[timid spade]

1/0 is not the reason

[night axle]

It’s also cooked since out of domain true

[timid spade]

the reason is that lnx gives values outside of the domain, if x<e

[timid spade]

yes

[night axle]

Close enough

Still things not being defined

[timid spade]

not 1/0

[night axle]

Yes

[timid spade]

which means the answer is wrong

[night axle]

Yes

I am illiterate

[timid spade]

fair enough lmfao

[timid spade]

np with E1-3

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[near geyser]

Should I use a comparison test?

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[solid iris]

what would you compare it with?

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.reopen

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[onyx tide]

1103 Say whether the following graphs are functions. State their domain and range if they are.

It’s b?

[sharp fiber]

a function requires there to be at most one output for every input

which of the graphs shown from that image do not meet that criteria?

[lethal path]

vertical line test ^

[sharp fiber]

yep

[onyx tide]

Exactly I did that

[lethal path]

b is a function, yes

there are also other functions

[sharp fiber]

so which ones from that image get ruled out

[onyx tide]

No wait

It’s c

I thought b was c

[lethal path]

c is a function, or is not a function?

[sharp fiber]

does b pass the vertical line test?

[onyx tide]

It’s not

[sharp fiber]

does c pass the vertical line test?

[onyx tide]

State their domain and range if they are.

?

What is that

[sharp fiber]

domain are all the possible 'x' values that can be mapped onto that function

range are all the possible 'y' values that can be mapped onto that function

start with a

what is the domain of a

using the definition I gave you

[onyx tide]

(4,3) or (4,-3)

[sharp fiber]

we see that 'a' is line segment that is excludes the values on either end

[onyx tide]

I mean the c

[sharp fiber]

start with the negative most x value of 'a'

what is the negative most x value?

[onyx tide]

4,4

[sharp fiber]

look at function 'a'

and pay attention to the x axis

[onyx tide]

Okay it’s 4

[sharp fiber]

is 4 the most negative x value for function 'a'?

[onyx tide]

Yes…

Oh negative

-2

Then

[sharp fiber]

yes

what about the positive most value of 'x'

[onyx tide]

4

[sharp fiber]

yep

is -2 inclusive or exclusive?

[onyx tide]

Wdym?

[sharp fiber]

if inclusive, you write with this '('

if exclusive, you write with '['

[onyx tide]

What is that

[sharp fiber]

inclusive means it includes -2

exclusive means it doesn't include -2 but includes all the numbers positive to it until you reach the other end point

normally in a graph, inclusive is marked as a filled out circle

exclusive is marked as a circle not filled out

[onyx tide]

Okay so domain is max and min of x or y axis?

@sharp fiber

[sharp fiber]

domain is min and max of x

don't forget inclusivity / exclusivity

[onyx tide]

Okay and range

[sharp fiber]

range is min and max of y

[onyx tide]

???

Okok

Ok now

Ik what excludivity means

[sharp fiber]

answer this, is min of x inclusive or exclusive?

[onyx tide]

To a?

a)

Question

[sharp fiber]

function a, yes

[onyx tide]

Inclusive

[sharp fiber]

why?

yes

correct

but why

[onyx tide]

Lower or -2

[sharp fiber]

it is because the circle is filled out

[onyx tide]

Yes

[sharp fiber]

you see the circle is black

[onyx tide]

But it can be -2,-3,-4

[sharp fiber]

which means it includes -2

[onyx tide]

Ok that

[sharp fiber]

so you know you start with '('

sorry

I mean '['

'[' = inclusive notation

[onyx tide]

Black circle is <
White circle is (

[sharp fiber]

black circle means inclusive, which means [

[onyx tide]

It isn’t [

It’s <

[sharp fiber]

empty circle means exclusive, which means (

[onyx tide]

Okay

[sharp fiber]

so tell me the domain of a

you already know half of it

deduce the other half

[onyx tide]

<-2, +4)

[sharp fiber]

the left half is correct, although we normally write inclusivity as [ or ]

[onyx tide]

I think in my country it’s >

[sharp fiber]

if your instructor said to use < or > for inclusivity, then use brackets

ok

[sharp fiber]

right half is basically right

[onyx tide]

Ok ok

I know now

[sharp fiber]

except you do not have to write the '+'

[-2, 4)

is sufficient

[-2, +4) is redundant

since if a value does not have a negative sign, it is considered positive

[sharp fiber]

this is different. when you express ordered pairs in terms of domain, range, you want to use [ or ] for inclusivity

what is the range for a)?

you got the domain right, now what is the range?

[onyx tide]

Range?

I said

[sharp fiber]

you stated that the domain is [-2,4) which is correct

what about the range?

[onyx tide]

Idk

Oh it’s y

Axis ?

@sharp fiber

Okay it’s

<-2, 4)

@sharp fiber

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[onyx tide]

.reopen

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✅ Original question: #help-49 message

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[twilit field]

I will use this method

here , X=(3V-U/2);Y=2V-U/2

$f(R,S)= \frac{6}{7} \frac{3V-U}{2};J =1/2$

[grand pond]

wai

[twilit field]

so $\frac{3}{14} 3V-U$ is the new joint density

[grand pond]

wai

[twilit field]

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[fervent burrow]

guys, lets say if this side was closed by another shape like in the picture, would the side still count towards the perimeter?

[latent wadi]

Yeah, if I understood your question correctly

[fervent burrow]

ok lets say u need the perimeter of this sector

do u understand that

[latent wadi]

The one you bolded?

[fervent burrow]

no the pizza slice u see

that one

we want the perimeter of that

not this black thing

[astral canyon]

The "r cm" side woul not

[lyric charm]

no, OB wouldn't count towards the perimeter of the entire big shape

only bits that separate inside from outside count

[fervent burrow]

Okay, and what if it was only the sector that we wanted the perimeter of

[lyric charm]

then you would count it still & ignore the other shit attached to it

[fervent burrow]

Oh alr thanks!

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[leaden seal]

I don't understand why the NB force has a angle of 30 degrees

[leaden seal]

I'm assuming it's got to do with trig

[lethal path]

nope, it's just similar triangles

[cerulean oyster]

its just clarifying that the force is being applied perpendicular to the bar

[midnight plank]

@leaden seal Has your question been resolved?

[leaden seal]

Thanks i see now

[lethal path]

no worries

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[lethal path]

'normal' meaning perpendicular is something you'll see more and more of

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[sturdy barn]

How do you work out (1.07)⁶?

[sturdy barn]

Is it 1.07x1.07x1.07x1.07x1.07x1.07?

[cerulean oyster]

yes

[livid python]

you could rewrite it using powers of 10
would make it like hella easier

[sturdy barn]

Oh

So the brackets

[sturdy barn]

Mean nothing here?

No difference?

[lyric charm]

i mean sure but that'll be long and painful

what do you actually need 1.07^6 for?

to how many decimal places, and in what context?

[sturdy barn]

I just need to find its value

[lyric charm]

!xy

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[lyric charm]

if it really does just say "Find 1.07^6" then STILL post an image of it so that we know that's what it says.

[sturdy barn]

1c

[lyric charm]

ok so you need to use a calculator

[sturdy barn]

That the one I need to do

[lyric charm]

Calculator revision

[cerulean oyster]

yeah, to do it in a calculator

[fierce osprey]

use calculator

[cerulean oyster]

press (
write 1.07
)
depending on your model, do the symbol ^ or xy
6

[sturdy barn]

Okay thanks

[cerulean oyster]

i cant believe im giving a calculator tutorial

[fierce osprey]

lmaooo

[sturdy barn]

Do I have to know what I am doing here?

Because I don't know I am just putting it in calculator

And getting the answer

[fierce osprey]

which question

[sturdy barn]

D e and f

1d 1e and 1f

[fierce osprey]

calculator...

[fathom night]

its just times it by itself

[fierce osprey]

just plug it into the calculator

[fathom night]

for example 1d its 1+13/100 x 1+13/100 x 1+13/100

the power shows how many times it is multiplied by itself

if that makes sense

[sturdy barn]

Ohhh

Okay thanks

[fathom night]

np

[fierce osprey]

do yk how to do question 2

[sturdy barn]

It's a new topic we did in class so I am not 100 percent sure but I will try it

[fierce osprey]

just know the equation I = P x R x N

[sturdy barn]

Exactly the method we did in class

[fierce osprey]

mhm

[sturdy barn]

Lowkey forgot it but I will check my notes

[sturdy barn]

Wait what does p.a. mean here?

[fathom night]

per annum

so like per year

[fierce osprey]

per year

yh

[sturdy barn]

If I am not wrong

2a is 105?

[fierce osprey]

yh

[fathom night]

yep

[sturdy barn]

Oh okayI think I understand it I try the others

Uh

About 2e

What do I do with the 18 months?

18 months would turn to 1.5 years?

[fierce osprey]

yh

[fathom night]

yup

[sturdy barn]

Okay thanks there goes my maths homework done

[fathom night]

yayy

[sturdy barn]

Not done yet😭 I will do chemistry homework last

I have so much to do because of chemistry

[fierce osprey]

goodluck

[sturdy barn]

Either way I will do my spanish homework now

[sturdy barn]

Thanks

Bye

.close

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[empty thistle]

hi, how would i find the absolute term (the summand (?) which doesnt have x in it) in the expression from the picture

[lyric charm]

$x \sqrt{x} = x^{3/2}$

[grand pond]

Ann

[lyric charm]

expand and simplify the powers using this, and then set the exponent on x to 0

and solve for n

[empty thistle]

what do i do with the combination number ?

[lyric charm]

leave it as-is

do what i said first

you'll find the value of n that makes the exponent on x zero

[empty thistle]

results say (11 choose 3) should be the answer

and 25 doesnt make sense anyways

mmmmmmmmm

ah (1/x^4)^n is not x^(n-4)

ok got it, thanks

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[main current]

Do you see how the one to the left of it is 60? Then, the two must add to make 90

If C is perpendicular to the beam, anyway

[leaden seal]

each I see

[midnight plank]

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[tacit jungle]

Hi, what's answer to this

[tacit jungle]

please I need answer I get on my 3 knees?

[neat oak]

HUHH

[graceful drum]

3 knees?

[tacit jungle]

3 knees

[graceful drum]

also we cant give answers

we can only guide you

[tacit jungle]

bro this is calculator

my calculator not calculating it

[neat oak]

why not

[graceful drum]

use google

[tacit jungle]

it's giving some negative decimal stuff i think broken i need it in fractions

[graceful drum]

show what you put in

[tacit jungle]

scroll up look at image that's what i put in

it give me some like this

[neat oak]

if you showed the question maybe we could see if you put in the values wrong

[tacit jungle]

what question

[neat oak]

cause what can we do looking at this

[tacit jungle]

ok fine

this is for 50k rectangles @neat oak

[graceful drum]

you need to give it in a/b form?

[tacit jungle]

I don't see what other form there is?

@graceful drum just put it in your ti and set it to frac mode for me if u have one of those expensive ti

[graceful drum]

10999790001/480000000

[tacit jungle]

that's precise answer?

[graceful drum]

yes

[tacit jungle]

tysm cherry

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[mortal falcon]

a terminal/final object x is such that for all other objects y in the category, there exists exactly 1 morphism y -> x

[shut canyon]

Thank you @mortal falcon

[mortal falcon]

try taking two generic terminal objects in a category, and examine the morphisms between them

[shut canyon]

what is generic?

[mortal falcon]

like arbitrary

i need to stop using that word everyone asks that 😭

just take 2 final objects in the category

[midnight plank]

@shut canyon Has your question been resolved?

[mortal falcon]

are you still stuck?

[shut canyon]

Trying to understand it, I don't yet, but it's fun!

[mortal falcon]

do you want help?

[shut canyon]

∀Y ∃! !₁ :Y →T₁, !₂ :Y →T₂

[mortal falcon]

thats true, yup

[shut canyon]

I don't know how, I don't have question at the moment, I thank you for your help @mortal falcon!

[mortal falcon]

there's specific convenient objects we could put in for Y

this is sort of something where once you get the construction the problem is relatively simple

[shut canyon]

I am thinking maybe since I showed a≅a, it is a similar proof

also there seem to be symmetry in this:
∃g: y→x with
g∘f=idₓ & f∘g=idᵧ

[shut canyon]

like examples where this pattern applies?

[mortal falcon]

i mean moreso that you get specific useful information for certain Y objects

for instance, we could let Y = T1

[sand flame]

[mortal falcon]

that might be too big of a hint lol but see if you can finish the problem from there

[shut canyon]

everything has unique connection to terminal object, so terminal object has uniqe connection to itself?

[mortal falcon]

yup

that's one piece of information we have now

[shut canyon]

I don't understand why T1 and T2, is there not only one terminal object?

[mortal falcon]

if we have a morphism T1 to T1, then that morphism must be the identity morphism

because there's only 1 morphism from T1 to itself, and there's always an identity morphism between two objects

[mortal falcon]

how do you know that?

the point of the exercise is proving that all terminal objects are isomorphic

[shut canyon]

I thought there was only one

[mortal falcon]

there doesn't have to be only one

[shut canyon]

Okay

[mortal falcon]

for instance, Set has lots of terminal objects

but as you'll prove here, they're all isomorphic to one another

so in some sense we can think of there only being 1, because they're all "equivalent"

for now, assume T1 and T2 exist as terminal objects

they could be the same or different, we don't really know

regardless, we want to show they're isomorphic

[shut canyon]

this is why I thought, because.I read, in https://webhomes.maths.ed.ac.uk/~tl/ast/ast.pdf:

I think this is the same proof:

[mortal falcon]

yes thats the exact same thing as what youre trying to prove

[shut canyon]

you are saying before we know the proof we don't know if terminal things are just one or how many or what is going on really

[mortal falcon]

yes thats correct

but also, there being 1 object "up to isomorphism" isn't the same thing as there only being 1 at all

it's kind of a semantic difference, but e.g. we absolutely are allowed to have distinct objects in a category that are isomorphic to one another, that's the whole point of defining a notion of isomorphism

[shut canyon]

because you can replace things with different things, while structure holds

and see maybe how things are similar, and sort of the same

is this the pattern with this proof? I don't yet see how

[mortal falcon]

the proof doesn't care about any of this, but yes that is the core idea of an isomorphism

[shut canyon]

terminal objects are isomorphic so you can change the object and structure preserves

[mortal falcon]

all we need to do in this proof is show that for any two terminal objects T1 and T2, there are morphisms f1 from T1 to T2 and f2 from T2 to T1 such that f1 f2 is the identity morphism T1 to T1, and f2 f1 is the identity morphism T2 to T2

that's the definition of isomorphism

[shut canyon]

1≅{●}≅point

[mortal falcon]

dont focus on specific categories

the point of this exercise is that it doesnt matter if youre in Set or not

the axioms of categories are sufficient

the proof in that example is a great example of this

if you notice, they never actually used the fact that the category was Set to do that proof. the only Set-specific terminology was referring to morphisms as functions

if you instead just think about abstract morphisms in an abstract category, no step in that proof stops working

try walking through it, but specifically in the context of the arbitrary C categoru

[shut canyon]

holds everywhere in mathematics

Thank you @mortal falcon

[midnight plank]

@shut canyon Has your question been resolved?

[shut canyon]

I think I got it now more or less! experienced aha moments also

maybe it's a bit like applying a buff to a magic item in a video game,
apply terminal property to Y, so it has unique arrow from all objects:
Y=T,
any two items T₁,T₂ then have unique arrow from all objects, including from each other

T₁ ──u──▶ T₂
T₂ ──v──▶ T₁

v∘u=idₜ₁:
T₁ ──u──▶ T₂──v──▶ T₁ = T₁ ──idₜ₁──▶T₁,

u∘v=idₜ₂:
T₂ ──v──▶ T₁ ──u──▶ T₂ = T₂──idₜ₂──▶T₂

T₁≅T₂ 💫

ありがと @mortal falcon

[midnight plank]

@shut canyon Has your question been resolved?

[mortal falcon]

yes that looks good, and the important step is that any morphism from T1 to T1 must be the identity morphism since there's only 1 by uniqueness for terminal objects, and same for T2, so u and v must compose to the identity morphisms

and now youve shown that terminal objects are isomorphic in any category!

[midnight plank]

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[grand pond]

Renato

[tidal turret]

my question is what does this gcd condition mean

[visual tiger]

(this is supposed to give you a factor of f)

gcd means greatest common factor

[tidal turret]

i still get a cubic after long div

[visual tiger]

the other thing it tells you is what isn't a factor of f

[tidal turret]

do i need gauss lemma?

[visual tiger]

so for example, no use trying to factor by (x-5) or any other factor of x^6-1

[visual tiger]

I don't think so, no

[tidal turret]

there is a theorem

quotient root something

when the coefficients are in Z

[visual tiger]

you could use rational root theorem

[tidal turret]

that

[tidal turret]

because after polylongdiv i still get a cubic

any tips?

[visual tiger]

what's said cubic?

[tidal turret]

$\polylongdiv{5x^5+4x^4-21x^3-21x^2-20x+5}{x^2+x+1}$

[grand pond]

Renato

[tidal turret]

i get Q(x) = 5x^3 -x^2 -25x + 5

[tidal turret]

i still get a cubic

what now?

[visual tiger]

ok so apply rational root theorem on that cubic

you should get 6 possible roots from that

from those, you know 3 are impossible

so up to you to check the final 3

[tidal turret]

do you have the theorem def at hand or no

it was +-(leading term coefficient divisors / last term coefficient divisors)

[visual tiger]

if P(X) = aX^n + .... + b integer polynomial, then if x = p/q is a root of P (p and q coprime), it verifies q | a and p | b

[tidal turret]

that looks like a complicated way of saying q is a member of the divisors of a

iirc

[tidal turret]

is using gauss lemma only possibility?

[visual tiger]

you won't need to in the end

if you manage to find rational roots

[tidal turret]

by gauss lemma i mean rrt

[visual tiger]

not the same thing xd

rational root theorem is the easiest way

[tidal turret]

5 is prime

so the roots are either 5 or 1/5 or -5 or -1/5

because 1 and -1 doesn't work

[visual tiger]

forgot to mention 1 and -1

[visual tiger]

ok good

[visual tiger]

you're sure you can't eliminate one of those immediately?

[tidal turret]

the answer is 1/5

,calc 5(1/5)^3 -(1/5)^2-25(1/5)+5

[grand pond]

Result:

0

[tidal turret]

bingo

[visual tiger]

so you can factor by (x- 1/5) (I suggest factoring by (5x-1) if you can instead)

[tidal turret]

$\polylongdiv{5x^3-x^2-25x+5}{x-1/5}$

[grand pond]

Renato

[tidal turret]

P(x) =(x^2+x+1)(5x^2-25)(x-1/5)

in the field Q[X] isP(x) =(x^2+x+1)(5x^3-x^2-25x+5)

[visual tiger]

This is the best you can do in Q[X]?

[tidal turret]

well

[visual tiger]

didn't you just do better?

[tidal turret]

my bad

P(x) =(x^2+x+1)(5x^2-25)(x-1/5)

b^2-4ac = 1-4 = -3

[tidal turret]

this is in Q[X] field P(x) =(x^2+x+1)(5x^2-25)(x-1/5)

in R[X] we would need to expand 5x^2-25

in C[X] we would need to expand everything

[visual tiger]

factor*

[visual tiger]

and so factor x^2+x+1

[tidal turret]

i think i get the gist of it

is just remembering rrt and knowing how to polylongdiv

[visual tiger]

use all information given first to factor, then rrt, then you should be left with quadratics at most

that's if the exercise is well made

[tidal turret]

yeah i think this was doable

the gcd part complicated me for a bit

i appreciate the help

nothing really advanced, this kind of exercises makes me go back to my preuni classes

the previous one with roots of unity was completely different

completely new topic

anyways i appreciate it

.solved

[midnight plank]

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[fossil wagon]

these are positive vectors right?

[fossil wagon]

since arrow is pointing to the right

[west iron]

no such thing as a positive vector

the arrow pointing right (except for HG) means you could say the x-component is positive

[fossil wagon]

that means HG would be negative?

[west iron]

no

because there's no such thing as a negative vector either

[wise flicker]

Wrt others

[fossil wagon]

wait I thought there's things like equal vectors, negative vectors and zero vectors

[west iron]

i mean

you can take the negative of a vector

but no vectors are inherently positive or negative

every vector is the negative of another vector

[fossil wagon]

ok but if the arrows are heading to the right, the values would be positive right?

like the x and y component

[west iron]

the x-components should be positive if it's pointing right

the y-components are positive if it's pointing up

[fossil wagon]

oh

[west iron]

in optimization, you sometimes see v>0 to mean that v has all positive components

but that's probably not relevant

[fossil wagon]

if its like this that means the x is positve and y is negative?

[queen abyss]

Btw I'm thinking about studying for quants in finance.
I just don't know where to start and what should I start with.

[west iron]

Yes

[fossil wagon]

if its like this that means x is negative and y is positive?

[ripe fog]

yes (or more specifically the x component of the vector)
the main reason why you cant say that vectors are negative or positive or whatever is you can describe vectors in many ways. the most common is in x and y components, but you can also describe a vector as the sum of two other vectors, or two other perpendicular vectors (which you could call the a component or b component, depending on what you named the vectors)

[fossil wagon]

oh well this looks very familiar in physics under the resolution of forces

[ripe fog]

exactly

it doesnt make too much sense if you say something like "the velocity is negative" without defining what that means

[fossil wagon]

oh yeah cus velocity is a vector quantity

but what if the arrows are horizontal or vertical?

does it become 0?

[west iron]

horizontal means y-component is 0

vertical means x-component is 0

[ripe fog]

you'd still need to define what exactly you mean by 'zero'
also be careful, because there is this thing called the 'zero vector' where all components are just zero, so be careful with your wording as well

[fossil wagon]

alright then for this kind of qn, we need to have centre of origin right?

[ripe fog]

the points are defined by their coordinates, and so the x and y components of the vectors from the origin to those points are just those coordinates, so you can pretty much skip the origin step
you can introduce it if you want, and it can help in harder questions, but its not necessary

[fossil wagon]

oh so I just add the two matrices together?

[west iron]

the vector AC should encode the displacement of C, from A

think of it like telling you how to get from A to C

[ripe fog]

well yes, but be careful
this diagram is the question we have

adding vectors is like putting them head to tail and seeing where the end point is from the start
the problem here is that if we do that, the vectors will go too high and not enough right

[fossil wagon]

why cant it be drawn like this?

?

[west iron]

you can

that helps you see the components

its just usually vectors are drawn as a single straight line

[dusty glen]

---

If A = (1 - sin x) / (2 * cos x):

a) Show that 1 / A = 2 * (1 + sin x) / cos x

b) Find sin x and cos x in terms of A

---

[fossil wagon]

wait so how we add the two points then?

is it (-2 ,1) + (3 ,4)?

[wise flicker]

you can

express in terms of icap and jcap

[fossil wagon]

so it'll be (-1,4)?

[fossil wagon]

wat?

whats icap and jcap?

[west iron]

not important is what they are

What are the directions to get from A to C

[fossil wagon]

right

[west iron]

more specific

[fossil wagon]

from left to right?

[west iron]

how far right

[wise flicker]

unit vecotrs in specific direction ( x and y here)

[fossil wagon]

5 units

since -2 to 3 the difference is of 5 units

[wise flicker]

no, icap means direction along x axis and jcap for y axis

[dusty glen]

I need a help in a trig question

If A = (1 - sin x) / (2 * cos x):

a) Show that 1 / A = 2 * (1 + sin x) / cos x

b) Find sin x and cos x in terms of A

Any one can help me with this?

[wise flicker]

you cannot add values directly if they not in same direction

for eg, vector from origin to (-2,1) can be expressed as -2icap +1jcap

[fossil wagon]

I guess icap would be 5 units to right and jcap would be 3 units to up

[west iron]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[wise flicker]

how?

[west iron]

😭

Why are we doing this

[wise flicker]

crazy lol go on

[west iron]

You're not supposed to add the vectors, and if you were supposed to you would just do it componentwise

You're like forcing him to learn a bunch of random physics notation for no reason

[wise flicker]

hmm.. i thought theywould know since it's vectors

my bad

[fossil wagon]

cus from -2 to 3 its 5 units difference?

[wise flicker]

ignore me go with @west iron

[west iron]

Yeah so the "instructions" are 5 units right and 3 units up right

[fossil wagon]

yeah

[west iron]

Which you noticed we can subtract C-A to get these

These "instructions" form our new vector

What vector should represent 5 units right and 3 units up

[fossil wagon]

oh yeah

so if the direction is going towards the right, we subtract?

[west iron]

Well

The math will work out so that you're always subtracting

If AC was pointing left then C_x - A_x would be negative

Since C would be left of A

[fossil wagon]

well how would we know if AC is pointing to right or left?

?

[west iron]

Aside from drawing it?

If C is left of A, it's pointing left

Vice versa, pointing right

[fossil wagon]

I meant in a qn like this'

is it based off the arrow on top of AC?

?

[wise flicker]

order matters

AC != CA

[fossil wagon]

so if the arrow on top of AC is going towards the left, that means the direction is towards the left?

[wise flicker]

The arrow usually represents that AC is a vector

doesn't not say if it is towards right

[fossil wagon]

how then can we tell if the direction is towards the right or left?

is it based from the coordinates of the points?

[wise flicker]

the direction is from A -> C

[wise flicker]

yes to check if its right/left

!done

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

[fossil wagon]

.close

[midnight plank]

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[midnight plank]

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[obsidian glen]

I do not know how to begin

[wispy mortar]

this is the complete question?

[obsidian glen]

this is the complete question

[wispy mortar]

what do u have to find?

[obsidian glen]

prove this.

[midnight plank]

@obsidian glen Has your question been resolved?

[midnight plank]

@obsidian glen Has your question been resolved?

[bold peak]

What is this saying intuitively

[midnight plank]

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[tidal turret]

\textbf{1)} Find all the solutions ((x,y) \in \mathbb{Z}^2) of
[
12x + 19y = 7
\quad\text{and}\quad
x^{2} + 1 \equiv y^{2} \pmod{23}.
]

[grand pond]

Renato

[timid spade]

x, y are mod 23, ye?

[tidal turret]

excuse me?

[timid spade]

at the end, it says mod 23

[tidal turret]

so?

[timid spade]

does that apply to x, y or just the final eqn?

[tidal turret]

you are helping me or you are asking me?

[timid spade]

clarification on the question

[tidal turret]

i think the question is clear enough, and I dont see how you say that 23 | y and 23 | x anyways

[visual tiger]

x² + 1 is congruent to y², mod 23

[timid spade]

ah ic

[visual tiger]

Meaning (x²+1-y²) is divisible by 23

[visual tiger]

Do you know how to solve 12x+19y = 7 over the integers?

[tidal turret]

yes

we can start with that

gcd(12,19) = gcd(12,7) = gcd(5,7) = gcd(5,2) = gcd(1,2) = gcd(1,0) = 1

12x + 19y = 1

we can use extended euclidean or eyeball it

[visual tiger]

Btw, you don't really need to solve 12x+19y = 1 if you want to solve for ... = 7

The latter being easier to eyeball for a particular solution

[tidal turret]

we can take the expression modulo 7

and use modular arithmetic

[tidal turret]

i cant manage to eyeball it

[visual tiger]

Sorry I don't follow you

[tidal turret]

i was using bezouts, finding a particular solution and add the homogenous solution then multiplying by 7

[visual tiger]

You don't see an easy link between 12, 19 and 7?

[tidal turret]

oh

(-1,1)

[visual tiger]

That's not gonna give you all solutions at all

[tidal turret]

?

[visual tiger]

If you really intend to do it that way

[tidal turret]

is fine, I have my ways of finding the solutions of linear diophantines, but we eyeballed one particular solution

[visual tiger]

You need to FIRST multiply by 7 and then add homogeneous solutions

[tidal turret]

yes, find a particular solution to ...=1 then find the homogeneous solutions and multiply the particular solution by 7

i didnt meant to say that we will multiply the homogenous solution by 7

[visual tiger]

Yeah, only particular solution

[tidal turret]

(-1 -19k, 1 + 12k)

x^2 + 1 = y^2 (mod 23)

(-1-19k)^2 + 1 = (1+12k)^2 (mod 23)

this?

[visual tiger]

Yep

[tidal turret]

now what

[visual tiger]

Well, what's the point of what you did?

What are we solving for

[tidal turret]

for (x,y)

[visual tiger]

So, since (x,y) can be expressed in terms of k

[tidal turret]

this whole last congruence, is a diophantine

[visual tiger]

What do we need to solve for

[tidal turret]

I pressume k

[visual tiger]

Yes

[tidal turret]

help

[visual tiger]

Try to expand and put everything on one side

[tidal turret]

(-1-19k)^2 - (1+12k)^2 + 1= 0 (mod 23)

[visual tiger]

Missed a 1

[tidal turret]

(-19k - 1)^2 - (12k + 1)^2 + 1 = 0 (mod 23)

,calc 19^2

[grand pond]

Result:

361

[visual tiger]

Hint: -19 is congruent to 4 mod 23

You can be efficient while computing congruences mod 23

[tidal turret]

(4k - 1)^2 - (12k + 1)^2 + 1 = 0 (mod 23)

(16k^2 -8k + 1) - (144k^2 + 24k + 1) + 1 = 0 (mod 23)

,calc 12^2

[grand pond]

Result:

144

[tidal turret]

-144 = ? (mod 23)

[visual tiger]

I'm not going to chew every bit of work for you, but yes that is helpful in simplifying the problem

[tidal turret]

-144 + 23 x 7 = 17

-144 = 17 (mod 23)

(16k^2 -8k + 1) - (144k^2 + 24k + 1) + 1 = 0 (mod 23)
(16k^2 -8k + 1) - 144k^2 - 24k - 1 + 1 = 0 (mod 23)
(16k^2 -8k + 1) + 17^2 - 1k = 0 (mod 23)
(16k^2 -8k + 1) + 17^2 + 22k = 0 (mod 23)
33k^2 + 14k + 1 = 0 (mod 23)
10k^2 + 14k + 1 = 0 (mod 23)

[visual tiger]

Alright great, that's the first big step completed

[tidal turret]

then what

b^2-4ac

[visual tiger]

Yes, it's either that immediately or rewriting the equation as k² + ... = 0 mod 23

[tidal turret]

196-4(10) = 196 - 40 = 156

[tidal turret]

care to elaborate on the latter

[visual tiger]

Well, you would have to multiply by the inverse of 10 mod 23 on both sides

Which you know exists by bezout (23 is prime)

[tidal turret]

yes

[visual tiger]

And that would give you k² + ...k + ... = 0 mod 23

[tidal turret]

gcd(10,23) = gcd(10,3) = gcd(1,3) = gcd(1,0) = 1

[visual tiger]

The idea would have been to find an easy factorization

[tidal turret]

let me think

23 46 69
7x10 = 1 (mod 23)

7 is the multiplicative inverse of 10 (mod 23)

[visual tiger]

Yep

So 7 × (10k² + ....) = 7 × 0 mod 23

[tidal turret]

(16k^2 -8k + 1) - (144k^2 + 24k + 1) + 1 = 0 (mod 23)
(16k^2 -8k + 1) - 144k^2 - 24k - 1 + 1 = 0 (mod 23)
(16k^2 -8k + 1) + 17^2 - 1k = 0 (mod 23)
(16k^2 -8k + 1) + 17^2 + 22k = 0 (mod 23)
33k^2 + 14k + 1 = 0 (mod 23)
10k^2 + 14k + 1 = 0 (mod 23)
k^2 + 98k + 7 = 0 (mod 23)
k^2 + 6k + 7 = 0 (mod 23)

[visual tiger]

98 is 6 mod 23

k² + 6k + 7 = 0 mod 23

In this form, it's pretty difficult to find the roots

So you would need quadratic formula again

[tidal turret]

b^2-4ac

36-4(7)

36-28=8

[visual tiger]

Yep

Now you need to find 8 = (...)² mod 23

To use quadratic formula mod 23

[tidal turret]

what

[visual tiger]

You know how quadratic formula for real numbers uses sqrt(b²-4ac)

Well you need to compute the sqrt(b²-4ac), but in the field of integers mod 23

[tidal turret]

sqrt{8} = sqrt{2x2x2} = 2sqrt{2}

( -6 ± 2sqrt{2} ) / 2

-3 ± sqrt{2}

k = -3 ± sqrt{2}
k1 = -3 + sqrt{2}
k2 = -3 - sqrt{2}

[tidal turret]

k needs to be an integer

so something is wrong

[tidal turret]

since k is guaranteed to be a number in between 0 and 22

we can try the k = 0, k = 1, k = 2, k = ... until k = 22 and we are guaranteed to find both roots

@visual tiger

,rotate

[tidal turret]

k^2 + 6k + 7 = 0 (mod 23)

64 + 48 + 7 = 119

[tidal turret]

is strange because the k that I get for before we took multiplicate inverse are different than the k for this

see the attached picture @visual tiger

strange, I might have made a mistake

70 = 23x3 + 1

[lusty python]

just ned to remind myself of the question

and i just made a table of remainders using my calculator

[tidal turret]

at first the problem looks hard but it was simple right?

Just don't mess up the arithmetic

.solved

[midnight plank]

Channel closed

Closed by @tidal turret

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[lusty python]

why squares in the congruence ngl

[midnight plank]

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[candid gust]

Hello

[candid gust]

Find the angle beta if a = 2√3, b = 2, and alpha = 2beta. Conclude that the triangle is equilateral.

This is the question i have. I missed every single lecture about this when my class had it so i do not know where and how to start

[surreal moon]

Not enough information

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[candid gust]

That is the whole task

[fallow scarab]

doubt