help-49

[surreal moon]

,rccw

[glacial fiber]

It is correct?

@surreal moon

[surreal moon]

!noping

[midnight plank]

Please do not ping individual helpers unprompted.

[rose trout]

One quick way to check if your solution is correct is to plug back into your original equations.
In this case, it doesn't look like your solution satisfies both, so something went wrong.

I'm not sure why you considered the equation 2x+3y=2, as it appears in neither of your equations.

Your solution isn't particularly graphic either, I think you should rather plot the lines directly on the plane and get the intersection from that.

[midnight plank]

@glacial fiber Has your question been resolved?

[midnight plank]

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[glacial fiber]

Not slove i want a easy way to slove

[glacial fiber]

.reopen

[midnight plank]

@glacial fiber Has your question been resolved?

[rose trout]

I explained what was wrong with your solution, and what you should do.

Where are you stuck?

[midnight plank]

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[hollow thicket]

Two circles, whose radii are 12 inches and 16 inches, respectively intersect. The angle between the tangents at either of the points of intersection is 29°30'. Find the distance between the centers of the circles.

[hollow thicket]

hello im back

how and why? please explain

@last slate

I did the math and got 8.11 inches, which is the correct answer. But...there is another answer which is 27.1 inches. I don't know where that came from. I used Cosine Law.

a=12 in, b=16in, C=29°30'
c=distance between the centers of the circle

by Cosine Law
c²=a²+b²-2ab cos C
= 12² + 16² - 2(12)(16)cos(29°30')
= 144 + 256 - 384cos(29°30')
= 400 - 384cos(29°30')
c = sqrt(400-384cos29°30')
c = 8.11 inches

the problem is on number 12

[midnight plank]

@hollow thicket Has your question been resolved?

[hollow thicket]

<@&286206848099549185> wanna try???

[glossy ibex]

weird

[hollow thicket]

yes, it is

[glossy ibex]

29°30' is 29.5° right? (Just checking)

[hollow thicket]

yes

[glossy ibex]

yeah I have no clue, sorry for being unable to help

[midnight plank]

@hollow thicket Has your question been resolved?

[midnight plank]

@hollow thicket Has your question been resolved?

[zenith cradle]

Not an explanation, and probably even just a coincidence, but $\sqrt{400+\cos 29.5^{\circ}}\approx 27.1$

[grand pond]

Crystopher

[zenith cradle]

Actual reason, there are two ways to draw the figure, which means that the problem description gives room for interpretation, more specifically regarding how the tangents intersect, in the following images you can see the two possible versions of this problem, in both cases, the blue lines represent tangents, the red is the searched distance and the bold black are the radii of the circles.

[hollow thicket]

WOW! This is what's on my mind but I can't comprehend because I was focused on finding 150°30'.

Thank you so much for your effort @zenith cradle it did bring in a new perspective

I got it now. I now know where 150°30' came from.

Thank you so much @zenith cradle and @last slate 🥰🥰🥰

.solved

[midnight plank]

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[last slate]

mb

was sleeping

[hollow thicket]

.reopen

[midnight plank]

✅

[hollow thicket]

.solved

[midnight plank]

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[midnight plank]

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[last slate]

what’s wrong with the question? I’m not really asking for help, but i’m asking for how the question can be phrased better :)

[last slate]

for one i guess n != -1 should be specified clearly cuz f(x^n) kinda works like differentiation then it’s inverse should be integration

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[last slate]

idk how to solve thiss... uhh its abt quadratics.. sum and product of roots.. anyone please assist me in solving this
final answer is 8x^2-36x-27=0

[clear void]

Do you know how to find the roots of this equation

[last slate]

yes

[clear void]

Yea then

[last slate]

ill just do p= and q= ?

[clear void]

Yea

[last slate]

x1=p x2=q

(\frac{3\pm \sqrt{15}}{2})

[grand pond]

ffffffffffffffffffffffffffflashy

[clear void]

Yea

Now can you calculate

[last slate]

i just square and multiply ?

[clear void]

$p^2q$

[grand pond]

Giraffe Theory 🦒

[clear void]

Yea

[last slate]

idk if this is true or not but i got \ (\frac{-9+3\sqrt{15}}{4})

[grand pond]

ffffffffffffffffffffffffffflashy

[clear void]

,w roots 2x^2-6x-3=0

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by axiomofnochoice

[clear void]

,w (3+sqrt(15))(3-sqrt(15))^2/4

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by axiomofnochoice

[clear void]

Sry

[last slate]

take ur time

[clear void]

Just multiply pq first

[last slate]

-3/2

[clear void]

Then multiply it with p again

[last slate]

damn thats much more simple

[clear void]

That'll be easy

[clear void]

Yea

[last slate]

same answer

[clear void]

Now for pq^2

[clear void]

You mean pq^2

?

[last slate]

yes

idk if i calculated it incorrectly or sum

[clear void]

Yea

do pq(q)

[last slate]

same answer

[clear void]

Are p and q same

[last slate]

(\frac{-9+3\sqrt{15}}{4})

[grand pond]

ffffffffffffffffffffffffffflashy

[clear void]

?

[last slate]

the only thing that differ is the plus minus

[clear void]

Then there should be differences in the pq^2 and p^2q

(Of plus minus)

Did you spot the error?

[last slate]

wait

[clear void]

Yea

[last slate]

i cant

can u tell me where did i do it wrong

[clear void]

Yea both of the signs should be negative

Right?

[last slate]

what

p squared q or pq squared

[clear void]

-9-3sqrt15

[clear void]

This

[last slate]

that aint waht it says on my calculator

[clear void]

What

[last slate]

whichh?

[clear void]

pq^2

Right?

[last slate]

pq^2

p^2 q

[clear void]

Yea we denoted p and q differently

[last slate]

the only difference i notice is where it put the negative sign

[clear void]

Ohh

[last slate]

but i dont think that really matters... or does it?...

[clear void]

It's calcs fault

Look where the negetive sign is placed

[last slate]

😭

is there any difference

[clear void]

Yes

Can you spot the difference in neg sings place in both

[last slate]

pq^2 negative symbol is next to 9
p^2q negative symbol is next to division

[clear void]

$\frac{-a+c}{b} and -(\frac{a+c}{b)}$

[last slate]

equal?

[clear void]

No

[last slate]

o

wait the second one does the negative applies for denominator too

?

[grand pond]

Giraffe Theory 🦒

[clear void]

On of the numerator or denominator

[last slate]

what does that supposed to mean

[clear void]

$-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}$

[grand pond]

Giraffe Theory 🦒

[clear void]

Hope this helps

[last slate]

oh my god

[clear void]

Just give negetive sign only to the numerator

[last slate]

ok

[clear void]

so p^2q is actually \ (\left(\frac{-(9+3\sqrt{15})}{4}\right))

[grand pond]

Giraffe Theory 🦒
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[last slate]

ok i get it its the calculator's display issue

[clear void]

Yeap

[last slate]

okok proceed what now

(\frac{-9-3\sqrt{15}}{4})

[grand pond]

ffffffffffffffffffffffffffflashy

[clear void]

Yea

[last slate]

it should be this btw right

ok

[clear void]

Yea right

Now we have p^2q and pq^2

To find quad eq. With these roots

Just calculate (x-p^2q)(x-pq^2)

[last slate]

is it x^2 - (sum of roots) + (product of roots) = 0

[clear void]

That is also true

[last slate]

let me try using this first

[clear void]

Yea

[last slate]

sum = -9/2

[clear void]

,w (\frac{-9-3\sqrt{15}}{4}) +(\frac{-9+3\sqrt{15}}{4})

[grand pond]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by axiomofnochoice

[clear void]

Ywa but

[last slate]

product = -27/8

srry im kinda slow the internet connection is bad rn

so now do i use the formula uhm

[clear void]

Yea

It's alright

Dw

[last slate]

alpha + beta = -b/a

alpha * beta = c/a

[last slate]

-b/a = -9/2
c/a = -27/8

[clear void]

Does your answer match?

[last slate]

uhh wait

[clear void]

Yea

[last slate]

(x^2-\left(-\frac{9}{2}\right)x+\left( - \frac{27}{8}\right))=0

i knew it

[grand pond]

ffffffffffffffffffffffffffflashy

[last slate]

now idk if im doing this correctly or not

😭

[clear void]

Yea it's alright

[last slate]

how is the a=2 and a=8

[clear void]

No

It's not like that

[last slate]

oh

[clear void]

b is not 9

[last slate]

what is it then

[clear void]

Okay

Just multiply 8 both the sides

[clear void]

-36

[last slate]

oh

ok first we multiply sum of roots with 4 on numerator and denominator to get the denominator as 8 and then we multiply both sides by 8 to get rid of the denominator and since the RHS is just 0 everything multiplied by 0 is just 0.

omggg

hold on

are my statements true

[clear void]

Yeaa

[last slate]

ohhhhhhh

yess

[clear void]

Absolutely true

[last slate]

i see

and

negative and negative is positve

so

final answer is

[clear void]

YeA

[last slate]

8x^2+36x-27=0

omggggg

gg gg ggg gg g g

tysmmm

[clear void]

Yw

Although your give answer seems diff.

[last slate]

bro has been teaching for almost an hour on just 1 question 😭

[last slate]

idk

[clear void]

,w roots 8x^2+36x-27=0

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by axiomofnochoice

[last slate]

ohh i gave it negative

[clear void]

Yep it's alright

Yea

[last slate]

the answer is correct

tysmm

[clear void]

Calculators fault

Yw

Also I'm sorry if I was slow

But I like yhe way in wich

[last slate]

nonooo i appreciate ur patience lmao 😭

[clear void]

Person themselves do all things

Thank you

[last slate]

now may i close the channel

[clear void]

Cheers

Yea

[last slate]

.close

[midnight plank]

Channel closed

Closed by @pine saffron

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Be polite and have a nice day!

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[twilit field]

is this answer valid

[twilit field]

or do I prove the commutativity of addition of sets?

[visual tiger]

it seems like you didn't go well into why commutativity of addition over the whole space implies commutativity of addition for subsets

so maybe you should go and prove it rigorously

[twilit field]

ok, let me think of something

[shy rampart]

how rigorously is the answer required?

[twilit field]

this is axler, so very I guess?

[shy rampart]

so you define the addition of subsets

[twilit field]

let V and U be finite dimensional vector spaces

then we have $V={v_1,v_2 \dots v_n}$
and $U={u_1,u_2 \dots u_n}$.

so $V+U= {v_1+u_1,v_2+u_2 \dots v_n+u_n}$

similarly $U+V ={u_1+v_1,u_2+v_2 \dots u_n+v_n}$

but addition is commutative on $F$ so $V+U =U+V$

[visual tiger]

even so here the rigorous proof is kinda trivial

[grand pond]

ƒ(Why am. I here)=I don't Know

[visual tiger]

no........

finite dimensional != finite

which still doesn't equal all subspaces

which don't even account for all SUBSETS

[twilit field]

I don't follow

[visual tiger]

oh and I didn't even realize that you wrote THIS

really?

[twilit field]

I thought this is the definition of addition of subspaces

[visual tiger]

... even IF we're taking the finite case

[visual tiger]

{1,3} + {3,4} is NOT {1+3,3+4}

you're missing out on SO MANY combinations

[twilit field]

oh right

I forgot that , oops

[visual tiger]

this is a FINITE SUM of subspaces

this isn't a sum of FINITE SUBSPACES

what even is a finite subspace

like you said V = {v_1,...,v_n} as if it made sense that V was finite

ESPECIALLY if you're working with R -vector spaces

[twilit field]

I mean I've heard of non-finite spaces, so I just wanted to mention that the space(or in this case subspace), I'm dealing with has n dimensions, where n is finite

[visual tiger]

EVEN SO

finite dimension

is not the same

as finite

[twilit field]

oh, ok

[visual tiger]

like

finite means a finite amount of elements

finite dimension means a BASIS WITH finite amount of elements

so it's not the subspace that has finite amount of elements

[twilit field]

I see

[visual tiger]

in any case, it's no use trying it with V or W being finite or finite dimension or whatever

since in any case

we're dealing with subsets in general

[twilit field]

hmm

ok

[visual tiger]

so

just a hint

prove double inclusion

and we're done

[twilit field]

that $V+U \subseteq U+V$ and vice verssa?

[grand pond]

ƒ(Why am. I here)=I don't Know

[visual tiger]

yes

[shy rampart]

yep

[frank wolf]

Yeah this is usually how one proves set equality

$A = B \iff A \subseteq B \wedge B \subseteq A$

[grand pond]

Pseudonium

[twilit field]

so I use this defn, right?

hmm, could I have another hint, not sure of how I'd structure the proof

like when we add sets, we add a element in A to every element in B

and then repeat that for all elements in A

right

[visual tiger]

sure

but you can use the set definition to the letter

and get exactly the good results

[frank wolf]

You can also think of A + B as the image of a function

[visual tiger]

like how do you prove "V + U subset of U+V"

[frank wolf]

There’s a function $A \times B \to V$

[grand pond]

Pseudonium

[frank wolf]

Sending (a, b) to a + b

Then $A + B$ is the image

[grand pond]

Pseudonium

[twilit field]

no, I get that but for instance {1}+{1,2}={2,3}

right?

[frank wolf]

Yeah

[twilit field]

ok, I take an element say $v_1=(x_1,x_2 \cdots v_n)$ $v_1\in V$ and $u_1=(x_1,x_2 \dots x_n)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

to begin, right?

[frank wolf]

Wdym by $v_1 = (x_1, \dots, x_n) \in V$

[grand pond]

Pseudonium

[twilit field]

the vector v_1 is in a n dimensional vector space

[frank wolf]

Right…

[twilit field]

so these are the components in each of these dimensions

[frank wolf]

So you’re choosing a basis?

[twilit field]

yeah

[frank wolf]

You can do so but it’s not necessary

[twilit field]

so I just assume $v_1 \in V$ and $u_1 \in U$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

where $ v_1$ and $u_1$ are arbitrary elements of V and U ?

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Sure…?

[frank wolf]

This is what you’ve gotta use

[twilit field]

I don't follow

[frank wolf]

So, how do you show $A \subseteq B$

[grand pond]

Pseudonium

[twilit field]

I take an arbitrary element belonging to A and show it always belongs to B

[visual tiger]

ok

So how do you show V+U subset of U+V

[frank wolf]

Mhm, so you show $x \in A \implies x \in B$

[grand pond]

Pseudonium

[frank wolf]

Yes that’s the next step

[twilit field]

$v_1 \in V+U \implies v_1 \in U+V$

[grand pond]

ƒ(Why am. I here)=I don't Know

[visual tiger]

ok

[frank wolf]

Sure

[visual tiger]

maybe don't take v_1 as a name

you could regret it

[frank wolf]

Yeah..

[visual tiger]

take x

[frank wolf]

Mhm

[twilit field]

oh

ok

[visual tiger]

so

[visual tiger]

the proof of this statement starts with

"Let x be in V+U..."

and it should end with

"thus x is in U+V"

[frank wolf]

Yes so $x \in V + U \implies x \in U + V$

[grand pond]

Pseudonium

[visual tiger]

how do we continue from here

[twilit field]

Let $x \in V+U $,which implies $x$ is made of two components say $x_1 \in V$ and $x_2\in U$

[frank wolf]

Wdym by made up of two components

[twilit field]

One vector from V and one from U

[frank wolf]

But how do they relate to x

(Also you’ve used the same variable for both)

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I thought $x_1+x_2=x$, but something tells me that;s wrong

[grand pond]

ƒ(Why am. I here)=I don't Know

[visual tiger]

no that's exactly it

you just needed to write it

[frank wolf]

So the way I would phrase this is

[visual tiger]

x = x1 + x2, where x1 in V and x2 in U

[frank wolf]

$x \in V + U \iff \exists v \in V, \exists u \in U, x = v + u$

[grand pond]

Pseudonium

[twilit field]

so $x_1+x_2=x$ and similarly $x_2+x_1=x$

[grand pond]

ƒ(Why am. I here)=I don't Know

[visual tiger]

it's not "similarly"

BECAUSE of commutativity

[twilit field]

ok

[visual tiger]

so if x_2 + x_1 = x

(recall x_2 in U and x_1 in V)

what can we say about x?

[twilit field]

$V+U \subseteq U+V$

[grand pond]

ƒ(Why am. I here)=I don't Know

[visual tiger]

there is no statement about x here

you jumped the gun a bit

[twilit field]

$x \in V+U$ and $x\in U+V$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hard umbra]

this is a true statement but you need to demonstrate the flow of logic required to make each conclusion

[twilit field]

how exactly ?

[hard umbra]

a bunch of stuff has been said so far

but do you know which statement exactly it is that lets you conclude x is an element of U + V?

[twilit field]

not really, no

[hard umbra]

that's the problem i guess

maybe try writing out what you've understood of the progress so far in your document and show us what you've got

[twilit field]

OK

[hard umbra]

it's kinda hard to follow the logic through a bunch of split up messages

[twilit field]

ok, I have to eat now, so I'll be back with my progress in say 20 min

[hard umbra]

[midnight plank]

@twilit field Has your question been resolved?

[visual tiger]

you have this all reversed

if you "let x_1 in V and x_2 in U and let x in V+U"

what is the link between x_1,x_2 and x?

[twilit field]

oh, right

[visual tiger]

since you apparently chose them at random

just "let x in V+U. We write x = ... where..."

[twilit field]

okay, sorry

[frank wolf]

Honestly proving this is like

It requires a lot of familiarity with dummy variables and stuff

[twilit field]

is this fine

[twilit field]

like the dummy variables in definite integration ?

[frank wolf]

For propositions

Like

[twilit field]

ah, I skipped logic , thought I'll do it once uni starts

as it's rather dull

[frank wolf]

$x \in V + U \iff \exists v \in V, \exists u \in U, x = v + u$

[grand pond]

Pseudonium

[frank wolf]

Is also the same as

$x \in V + U \iff \exists a \in V, \exists b \in U, x = a + b$

[grand pond]

Pseudonium

[twilit field]

ok, that makes sense

yeah

[frank wolf]

Even though these aren’t like literally the same latex string

They mean the same thing

[twilit field]

I realise that, yeah

[frank wolf]

And so what you want to do is like

Use $x \in V + U \iff \exists v \in V, \exists u \in U, x = v + u$

[grand pond]

Pseudonium

[frank wolf]

But then for the other done, it’s best to use different dummy variables

So $x \in U + V \iff \exists a \in U, b \in V, x = a + b$

[grand pond]

Pseudonium

[frank wolf]

And then the idea is to define a, b in terms of v, u

[twilit field]

so $a+b=v+u$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Indeed, that’s what you want

But you want to define a in terms of v and u

And b in terms of v and u

[twilit field]

a=v and b=u

I don't see why dummy vars would be useful though, isn't my proof sufficent

[frank wolf]

This fails

And also illustrates why I needed to introduce the dummy variables

Because if you’re making that mistake then I’m not convinced your proof works

[twilit field]

hmm, so I should read proof writing before continuing with LA?

[frank wolf]

Well im not sure

I’m not a proof expert

[twilit field]

I mean, I could probably get the hang of dummy variables with enough LA proofs

[frank wolf]

Start with this one then

[twilit field]

ok, what should I do

[frank wolf]

Do you understand why this fails

[twilit field]

because a could be u and b could be v?

[frank wolf]

Not quite

We know that v is in V

[jagged saffron]

(I think picking it up as you go along like you've been doing is perfectly fine, you are getting better)

[frank wolf]

And u is in U

Such that x = v + u

[twilit field]

yes

[frank wolf]

From this, we want to produce the following

An a in U

And b in V

Such that x = a + b

[twilit field]

so a=u and b=v ?

[frank wolf]

That’s a better start

If you said a = v, there’s no guarantee that a is actually in U

That’s why your suggestion failed

[twilit field]

yeah, makes sense

[frank wolf]

Anyway

You still need to show x = a + b

[twilit field]

we know that x=u+v

and that a=u

[frank wolf]

Careful

[twilit field]

and b=v

x=a+b

[frank wolf]

This is what we’re starting from

[twilit field]

yes

[frank wolf]

So we don’t know this

[twilit field]

I substitute the relevant dummy variables into that

[frank wolf]

No im saying you can’t start from x = u + v

Because you haven’t told me why that’s true

We can start from x = v + u

[twilit field]

I'm defining x to be v+u

[frank wolf]

No

We don’t get to define x

x is given to us at the start

[twilit field]

ok

[frank wolf]

We start from:

x = v + u

a = u

b = v

And we must show

x = a + b

[twilit field]

yes

[frank wolf]

Lmk if you’re having trouble

[twilit field]

I don't know what to do now

[frank wolf]

So

You can do the following

$x = a + b \iff x = u + b$

[grand pond]

Pseudonium

[frank wolf]

do you agree with this?

[twilit field]

yes

[frank wolf]

You may have heard before that

You’re not allowed to start with what you want to prove

[twilit field]

yes

[frank wolf]

In general for someone new to proofs, this is the correct advice to give

However, there is a way to do it legally

You just have to make sure you can use $\iff$

[grand pond]

Pseudonium

[frank wolf]

I.e. that you can reverse all your logic

So if I had just done $x = a + b \implies x = u + b$

[grand pond]

Pseudonium

[frank wolf]

Then I’m stuck

I won’t be able to prove x = a + b

[twilit field]

yeah

[frank wolf]

But we can reverse the logic

Because if we start from x = u + b

And a = u

Then we can deduce x = a + b

So $x = u + b \implies x = a + b$

[grand pond]

Pseudonium

[frank wolf]

And so overall $x = a + b \iff x = u + b$

[grand pond]

Pseudonium

[twilit field]

mhm

[frank wolf]

Can you continue? What’s the next step?

[twilit field]

so a=u=x

[frank wolf]

No, u is not (in general) equal to x

[twilit field]

then I'm not too sure

[frank wolf]

Go back to here

[twilit field]

one thing I don't follow is if x=v+u and a=u and b=v

why can't I use the fact that x=v+u

[frank wolf]

Wdym

Ofc you can

In fact, you have to use the fact that x = v + u

[twilit field]

no, but Iknow v=b

and u=a

[frank wolf]

Mhm

(Not sure how to help further without just giving it away…)

[jagged saffron]

I'm not convinced the introduction of these dummy variables is particularly illuminating

Feel like it's confused wai away from a two line proof

[frank wolf]

Hmm, how else would you do it

Like, we were trying it without dummy variables before

[twilit field]

this is my attempt at a proof

[jagged saffron]

Super close

Where and how have you concluded that x is in U + V?

[frank wolf]

Some typos and stuff

[frank wolf]

And yeah, this is the key step

[twilit field]

I'm not sure of how to do that

[frank wolf]

Hence why I introduced dummy variables..

[twilit field]

wait

[jagged saffron]

What's the relationship between x1 + x2 and x2 + x1

[twilit field]

they are equal

[frank wolf]

Because?

[twilit field]

commutativity of addition over F

[frank wolf]

Wdym by addition over F…?

But it is commutativity of something

[twilit field]

commutativity over a scalar field

[frank wolf]

That’s not really how you’d phrase it?

[jagged saffron]

We're not necessarily working in F

[frank wolf]

Indeed

[twilit field]

commutativity over the relevent field ?

[jagged saffron]

The field is F, but where do x1 and x2 live?

[frank wolf]

I mean strictly there doesn’t even need to be a field

Or at least, the original pinned question makes no mention of vector spaces

What are U and V subsets of?

[twilit field]

in subspaces V and U

[frank wolf]

Subspaces of what

[twilit field]

one minute

oops

[twilit field]

I missed a part of the question

sorry

This is the OG question

made a mistake while I was tping it

[jagged saffron]

Didn't make too much difference here but its rly important you either transcribe properly or send og question. The distinction between subspace and subset can be super important

[twilit field]

got it

sorry

[frank wolf]

Mhm

[jagged saffron]

Anyway just focus on one subset inclusion for a second maybe, you have x = x1 + x2 in V + U (x1 in V, x2 in U), then think about how and why you can rewrite that to look like something of the form (something in U) + (something in V)

And hence x be in U + V

[frank wolf]

Yep, that’s the idea

[twilit field]

as I said, commutativity

over the field on which V is defined

[jagged saffron]

Commutativity of what?

[twilit field]

of addition

right

[jagged saffron]

Addition in V which is a vector space

[frank wolf]

Mhm so

[jagged saffron]

You don't rly need to reference the field

[twilit field]

oh,ok

[frank wolf]

Addition is a map $+ : V \times V \to V$

[grand pond]

Pseudonium

[frank wolf]

It is this map which is commutative

Notice that F doesn’t appear anywhere

[twilit field]

yeah, it's commutative

[jagged saffron]

You are likely confusing this with proofs regarding elements of F^n that are done at the component level (so in F). Doesn't help that you called things x1 and x2 like you might call components of a vector

[twilit field]

ok, so what am I doing wrong

[jagged saffron]

Nothing right now

[twilit field]

like how would I do this

[jagged saffron]

Okay so how currently have you rewritten x = x1 + x2

[twilit field]

I've defined it to be that

[frank wolf]

Not exactly…

[jagged saffron]

Right but our goal was to rewrite how it looks

Using a propety you just cited

[twilit field]

commutativity ?

[jagged saffron]

Right commutativity in V (which x1 and x2 live in) let's us rewrite x1 + x2 as?

[twilit field]

x_2+x_1

[jagged saffron]

Now does this look like something that lives in U + V?

[twilit field]

yes

ooh

okay

[jagged saffron]

So starting from x in V + U you have concluded that x is in U + V

[frank wolf]

Mhm although note that when you write x = x_1 + x_2

This is not a definition of x

[jagged saffron]

Youll notice that your steps are reversible and you can run the argument backwards, which is what pseudo was trying to get you to see)

[frank wolf]

Yeah the symmetry here is again a little subtle if you’re not used to it

So, try to write the proof again?

[twilit field]

doing that rn

is this better?

[frank wolf]

So typo, it should be $x_1 \in V$

[grand pond]

Pseudonium

[frank wolf]

Also this

You still don’t have this in your proof

Also it’s “Commutativity of addition in a vector space”, we aren’t considering vector fields here

[jagged saffron]

Yeah you are still not describing the relationship between x1 + x2 and x2 + x1 and how that allows you to draw a conclusion about where x additionally lives

[twilit field]

let me rewirite the entire thing

[jagged saffron]

Take your time

Reread the convo if you need

[twilit field]

oops

it shoudl be W+U

is this better>

[frank wolf]

There is a string of correct statements now but it doesn’t flow very nicely

Like, you don’t need commutativiry of addition to show that, starting from $x_1 + x_2 \in W + U$, that $x_2 + x_1 \in U + W$

[grand pond]

Pseudonium

[frank wolf]

So it’s unclear exactly where you’ve used Commutativity

[twilit field]

I need commutativity to show that $x_1+x_2 = x_2+x_1$ no?

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Yes

So then you should say that’s where you’re using commutativity

[frank wolf]

Not here

[twilit field]

ok

anythig else?

anything else wrong

[frank wolf]

You should rewrite it

[twilit field]

okay, will do

[subtle blaze]

The word is commutativity not commutivity

[twilit field]

got it

[frank wolf]

You don’t have U + W = x_2 + x_1? I’m not sure where this came from

Why would that even make sense…?

[twilit field]

U+W=x_1+x_2

right

[frank wolf]

No?

What does that even mean

[twilit field]

oh right

I meant it's an element

[frank wolf]

Ok also you’ve switched the indices for some reason

[frank wolf]

Being an element is different to equality

[twilit field]

I just realised that , sorry

[frank wolf]

Right…

[twilit field]

rather I got confused

oops

[frank wolf]

Maybe you see what I mean about making sure you have the definitions crystal clear

You won’t really be able to progress in maths otherwise

[twilit field]

yeah

[twilit field]

why is that wrong, isn't it supposed to be $x_2+x_1$

[frank wolf]

Yes but here you wrote x_1 + x_2 for some reason

Even though in your screenshot you had U + W = x_2 + x_1

(Or I guess technically you had U + W = x_2 + x_!)

[twilit field]

oh

ok

yeah

makes sense

[frank wolf]

(Which is still wrong)

[twilit field]

yeah, understood why

[twilit field]

other than that is the proof fine?

[frank wolf]

Well

You should probably rewrite it

Sorry I know that’s annoying but

[frank wolf]

Back here I actually thought you were done

But I wanted to check just in case

And then you wrote an incorrect proof

[twilit field]

Ok, will do

[jagged saffron]

you still have U + W = x1 + x2 which doesnt parse

[twilit field]

so U+W=x_2+x_1

?

[jagged saffron]

what does that mean?

[twilit field]

ah

[jagged saffron]

just emphasise the point at the end that x is in U + W and you're there

[twilit field]

okay

[jagged saffron]

afterall the whole point is to show that if x is in W + U then x is in U + W

so that conclude needs to be written to deduce the subset inclusion

and perhaps emphasise the point that x2 + x1 is an element of U + W (why)

[twilit field]

okay

thanks!

thanks @frank wolf and @jagged saffron .

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[zenith garden]

how did i do?

is my work right

[polar mortar]

looks fine

[zenith garden]

noice

i was a little worried it was kind of unorganized but if you can tell whats going on then its fine

thank you very much

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[slender walrus]

no real need to double check if y=7 is the solution

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[twilit field]

can I have this verified or refuted?

[twilit field]

just correct/not correct please

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

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[onyx moon]

Let A be a proposition that is not a tautology. Consider the proof system L+A, obtained from the regular proof system by adding A as an axiom. This means that in addition to the regular axioms, any statement derived from A by substituting any propositions in place of the atomic propositions becomes an axiom in the L+A system.

a. Prove that the system L+A is not consistent.

b. Prove that the system L+A is contradictory in the sense that there exists a statement such that both it and its negation can be proven in this system.

I need help with b mostly

[noble mural]

me after reading the question 😍 😊

[onyx moon]

?

[void pivot]

What is the "regular proof system" ?

[onyx moon]

Hilbert set of axioms,
A to (B to A)
(Not B to not A) to ((not B to A) to B)
(A to (B to C)) to ((A to B) to (A to C))

[void pivot]

I see

It's been a long while since I touched this, but I'll try to help

[onyx moon]

Appreciate

I just can’t find the relation between the the new axiom not being a tautology to prove some statement. What’s the relation between some model and proving

[onyx moon]

Maybe what I’m asking here would help us both?

[void pivot]

Hmm

[onyx moon]

And how that helps me? 😅
I mean I don’t even have to prove not A to A for that do I?

[void pivot]

yeah, it's not

[onyx moon]

There is some model where not A is true and A is false

[void pivot]

forget it.

[onyx moon]

Ok 🙂

[void pivot]

Yes, i'm very rusty

But maybe...

So, there is an asignment for which A is T, and another for which it is F

[onyx moon]

Yes indeed

[void pivot]

Now if you replace the assignment values T with tautologies and F with their negations (for the assignment in which A is F)

You can prove not A

and you also have A because it's an axiom

@onyx moon does this make any sense?

[midnight plank]

@onyx moon Has your question been resolved?

[midnight plank]

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[worthy wing]

x^2 is always positive

So it doesn’t matter if you approach to 0 from the left or from the right, it will be always a positive small number

[midnight plank]

@shadow osprey Has your question been resolved?

[worthy wing]

Could you give me an example?

[midnight plank]

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[viral dagger]

asking from pure curiosity, what will happen if 2 high sticks 1 away leaning ontop of eachother, i want to know:
-how the angles change on how far the8
sticks are leaning
-if the point that the stick touches follow a certain graph
-how the ratio of how far the sticks are apart and how long the sticks are changes how much the 2 questions abive change
although im not sure if this is way too high level or complex, if its so complex i cant understand it ill probably close it
the top part of the picture is for the infinite sticks thing, the bottom is for 3 that im trying to think about as a simpler part

[midnight plank]

@viral dagger Has your question been resolved?

[steel lantern]

So do you wanna know how the data changes as the wall is removed and the sticks start falling?

[viral dagger]

what?

[steel lantern]

The question is a bit long and confuses me

[viral dagger]

you get how the sticks are leaning right?

[steel lantern]

Yes

But we could try calculating the first 5 sticks and make functions out of that

[viral dagger]

for the first, i wanna know if theres a function for the angles

for the second, i wanna knkw if the blue points follow a curve or some sort

[steel lantern]

Give me a second ill try to find some paper and do that in a proper system. I dont think it should be too hard

[viral dagger]

alr thanks

[steel lantern]

Do you remember how to calculate the angles?

I would have to go and find my formulas if not

Nevermind. Found them

Now I got a vector for the first stick

I really dont like trigonometry so I will first establish the sticks and try the rest after that

Is there any time limit on this?

[viral dagger]

no this is just from pure curiosity

cause when i think about jt if everyone was leaning to eachother it would mean that a person would be laying flat assuming a person is a line

[viral dagger]

what does that mean

[steel lantern]

Do you know what vectors are?

For the next stick we will have to look at the angles actually

Damn

[viral dagger]

no not really

is it like an arrow

[steel lantern]

Yes!

Exactly

And s1 is the definition of where the arrow is and what direction it has

Now we calculate its angle

[viral dagger]

what does (7/0) mean

[steel lantern]

Those are ones

[viral dagger]

what

[steel lantern]

Can I call you

That would be much easier

[viral dagger]

im outside rn

sorry

[steel lantern]

Okay just look up some yt tutorials

[viral dagger]

huh

uh how about the (0<a<1) thing

ahat does that mean

[steel lantern]

It defines the start and end point of our arrow

[viral dagger]

ok

[steel lantern]

Okay now I am at a dead end

[viral dagger]

so the start would be (1/0) and the end would be (0,sqrt3)?

[steel lantern]

Basically yes

[viral dagger]

um ok

[steel lantern]

The undefined a means that rvery pint on that line is part of the vector

[viral dagger]

can i ask how you derive that?

[steel lantern]

Okay I dont know how you got this problem but I dont know how to solve it anymore

[steel lantern]

One second then yes

[viral dagger]

oh ok

[viral dagger]

wait using graphs i got

[steel lantern]

This is what I got

[viral dagger]

ok

[viral dagger]

this is tbe first stick btw

[steel lantern]

But sin-1 of the squareroot of 3 is undefined in my calculator

[viral dagger]

huh

[steel lantern]

And it should give us the angle you need

[viral dagger]

oh ok

thanks for your time and effort tho, i really appreciate it

[steel lantern]

Sec

If you wanna create a new ticket

Use this

It is the bit I found about this

[steel lantern]

Okay that helps A LOT i didnt see it

What point do the stick collide?

We can do that then

[steel lantern]

To create lines in a system (sry I wasnt taught in english so my vocabulary is bad for this) you need a base point

For stick one it is (1/0)

Then you need a directional vector

It defined where we are going to

It is multiplied with an unknown number which we can substitute later to get any point on a line

By limiting the possible values for this point we can limit its length

[viral dagger]

what

[viral dagger]

$(\frac{6+\sqrt{52}}{8},\frac{\sqrt{3}(6+\sqrt{52})}{8}$

[grand pond]

Skill_Issue

[steel lantern]

Okay that will be annoying to solve

But you can define thw other sticks with that

I told you how already, im gonna head off to my own maths now🫡

[viral dagger]

alr thanks, altjough i have no idea lol

alr thanks then, ill close this

.close

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[viral dagger]

cause this looks complicated

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[last slate]

Could someone please explain this

These are the steps before the verification

Do we usually verify after proving something ?

[plucky salmon]

So usually whenever we try proving limits, we first have to find whatever we want delta to be before actually proving the limit.

The first page you posted tries finding this delta

Then in the second page, we plug in that delta is equal to sqrt x which ultimately proves the limit under the epsilon-delta definition

[last slate]

This makes no sense to me

[plucky salmon]

What part of it doesn’t make sense?

[last slate]

I think i get the first part

But the second

" < (sqrt[epsilon])^2"

Where did that come from?

Are they squaring on both sides of the inequality 0 < |x|< (sqrt[epsilon]) ?

[plucky salmon]

So ultimately, by the epsilon-delta definition, we want to solve for delta in terms of epsilon such that for any epsilon we choose from the positive reals, |x-a| being less than delta would imply that |f(x)-L| is less than epsilon, where a is the value we’re approaching and L is the limit’s value.

We’ve taken epsilon to be some arbitrary positive real, and now we’re try to solve for delta in terms of epsilon

[last slate]

we already did that here

and i more or less understand it

Same case with the next example

What did they do?

[plucky salmon]

Not quite, in that step we were trying to formulate a satisfactory delta

[last slate]

Ahhh
I don't get it

[plucky salmon]

In the step you're confused about, we're substituting the absolute value of x for delta

[last slate]

Why did we substitute?

[plucky salmon]

Here, lemme add the missing step for you:

$|x^2|=|x|^2< \delta ^2=(\sqrt{\epsilon}$)^2

[grand pond]

Oreo
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[plucky salmon]

Since the assumption here is that $|x|<\delta$, we can substitute with the strict order

[grand pond]

Oreo

[plucky salmon]

Can you recall the epsilon-delta definition of a limit?

[last slate]

Ohhh
So we are just poving the converse ?

[last slate]

Yes

[plucky salmon]

Exactly

[last slate]

Ahhh

All of it makes sense now

One more

[plucky salmon]

So if we assume that |x-a| < delta, we ultimately prove that |f(x)-L|<epsilon with enough algebra

[last slate]

What do you mean by "strict order"

[plucky salmon]

<

[last slate]

Okay

Thanks a lot

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[onyx moon]

Let A be a proposition that is not a tautology. Consider the proof system L+A, obtained from the regular proof system by adding A as an axiom. This means that in addition to the regular axioms, any statement derived from A by substituting any propositions in place of the atomic propositions becomes an axiom in the L+A system.

a. Prove that the system L+A is not consistent.

b. Prove that the system L+A is contradictory in the sense that there exists a statement such that both it and its negation can be proven in this system.

I need help with b mostly, we use Hilbert set of axioms,
A to (B to A)
(Not B to not A) to ((not B to A) to B)
(A to (B to C)) to ((A to B) to (A to C))

[midnight plank]

@onyx moon Has your question been resolved?

[midnight plank]

@onyx moon Has your question been resolved?

[midnight plank]

@onyx moon Has your question been resolved?

[midnight plank]

@onyx moon Has your question been resolved?

[onyx moon]

.close

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[fallen heath]

does T1 + regular = T3 ?

(also T3 and regular isnt the same thing ?)

ok it seems T1 + regular = T3

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[last slate]

Let f : R -> R be defined by f(x) = sin(x^3). Then f is continuous but not uniformly continuous. [ True/ False ]

[last slate]

Composition of two uniformly continuous function is uniformly continuous.

x^3 is not uniformly continuous

Does that mean something ?

How do I begin ?

<@&286206848099549185>

[junior delta]

Ye

[last slate]

Help

[last slate]

The question

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sage helm]

I think I just need to use the monotone convergence theorem

[sage helm]

I know I can prove the sequence is decreasing with induction but I don't know how to word it properly

x_2 < x_1

then for some k I assume x_{k + 1} < x_k and with some algebraic manipulations I can show its decreasing

[strong lava]

xn+1 = f(xn)
You wanna prove f(x) <= x on the relevant segment

[sage helm]

uh

[strong lava]

then probably you could also prove that therefore, it converges to the nearest solution to f(x) = x (known as a fixed point) in that direction

[sage helm]

does my thing not work

[strong lava]

you didn't show anything

what's your thing

I'm just providing a framework for you to word it properly

[grand pond]

argon

[sage helm]

ignore the missing )

[sage helm]

with the verifiable base case

[strong lava]

yeah

[sage helm]

also all terms have to be positive therefore its bounded below

therefore this sequence converges

I just don't know how to write out the induction proof

[strong lava]

of convergence ?

[sage helm]

"Note that x_2 < x_1; now for any integer k consider..."

Does this work?

[sage helm]

what?

[strong lava]

prove by induction that P(n): "x{n+1} < xn"

[strong lava]

it's just a proof by induction

[sage helm]

Okay

I'll figure it out

thanks for the help :))

[sage helm]

I think this should work

.close

[midnight plank]

Channel closed

Closed by @sage helm

Use .reopen if this was a mistake.

[midnight plank]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[last slate]

can someone help me finding limit with function containing fractional part funciton?

[last slate]

help!

HELP

[fast igloo]

sure

post the question please

[shy rampart]

please open your own channel

[last slate]

oh ok'

$\lim_{x\to 2} \left[x+\frac{1}{2}\right]+{x^2}$

[grand pond]

pun pun

[last slate]

it has greatest integer function too but lets just focus on {x^2} part

[zealous schooner]

why does the x look like that

that's trippy

[last slate]

font

ig

it looks nice tho

[zealous schooner]

$\lim_{x\to 2}{x^2}$

[grand pond]

kheerii

[zealous schooner]

we want this

[last slate]

yea

[zealous schooner]

remember that $x=[x]+{x}$

[grand pond]

kheerii

[last slate]

yes

[zealous schooner]

so you can turn all fractional part limits to greatest integer limits

[last slate]

what about the square?

[shy rampart]

share plz 🙂

[zealous schooner]

$\lim_{x\to 2}\left(x^2-[x^2]\right)=4-\lim_{x\to 2}[x^2]$

[shy rampart]

*4 (right?)

[zealous schooner]

Can you solve it from here?

[grand pond]

kheerii

[zealous schooner]

Yeah my bad

[last slate]

\usepackage{ccfonts, eulervm}

hold on

[shy rampart]

thanks!

[last slate]

np

why is there 4?

[zealous schooner]

We substituted the 2 into the x^2 part of the limit, since that is a continuous function

[shy rampart]

$\lim_{x\rightarrow2}x^2=4$

[grand pond]

nameless individual

[zealous schooner]

The GIF function is what’s causing the (potential) discontinuity