help-39

[rose sentinel]

The question clearly indicates two enamels yet the answer only accounts for one

Agreed?

[vocal lark]

Yeah

[rose sentinel]

Alright well then that’s as far as I’ll bother going

Thanks for the help

.close

[pearl pond]

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[vocal lark]

.reopen

[pearl pond]

✅

[vocal lark]

@rose sentinel

[rose sentinel]

Yo

[vocal lark]

Half of 144pi, which is the correct answer (right?) Would be the amount needed of EACH enamel coating, the white and the blue

So yeah I agree with you

[rose sentinel]

Yes

Wait yeah

Yeah I see you just never multiplied it

Okay alright then so the book is wrong

5th time so far

Out of like 1.2 million so far

Thanks

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[vernal grove]

Is this correct?

[marsh tangle]

yeah its correct

also nice username lmao

[vernal grove]

bet. lol ty I hate this course

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[ocean steeple]

$\int t^2 \cos t dt \$
$u = t^2, v = \sin t, du = 2t dt, dv = \cos t dt \$
$ t^2 \cdot \sin t - \int \sin t dt \cdot 2t \$
$t^2 \sin t + 2 \cos t \frac{t^2}{2} + c \$

[ocean steeple]

what is wrong with the way I integrate this by parts?

<@&286206848099549185>

[sly dagger]

uh

what did u do with the second integral?

[jolly parrot]

Tomi

[sly dagger]

bro explain what u did with the second integral? it literally just looks like u integrated both seperately?

[ocean steeple]

im using the integration by parts formula

[sly dagger]

yeah obv

[ocean steeple]

then what did I do wrong?

[sly dagger]

u cant just integrate sint * 2t dt like that

there are 2 functions of t being multiplied

[ocean steeple]

can you stop spamming this shit?

?

then why are you annoying me?

this is my help thread, feel free to open a new one

[sly dagger]

start your own help channel if u have a question

#❓how-to-get-help @whole mulch

[ocean steeple]

so what would I use there, do the integration by parts again?

[sly dagger]

yes

[lone sedge]

Yes

Again tsint

[ocean steeple]

so do it over the new integral which is sin t dt \cdot 2t

[sly dagger]

yes

[trim thistle]

You could also the chart rule thing

Useful when u have to do Integration by parts more than once

And faster

[pearl pond]

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[native surge]

is this formula correct?

the 2 - x/

[viscid dew]

how can you have some $\int f(x)\text{ dy}$?

[jolly parrot]

nameless individual

[native surge]

i dont get it?

i changed the x to y

[viscid dew]

uhhhh oh i see

looks fine to me

[native surge]

is the r with 2 - x correct?

i feel like its wrong some how

?

[viscid dew]

.

[native surge]

is it always like the variable - X or can it be x - variable

im kinda confused on the r

[viscid dew]

it can be anything

[native surge]

how

i kinda guessed that x - 2

actualy shouldnt it not be x + 2 ?

@viscid dew😅

[brave schooner]

should be $x-2$

[jolly parrot]

caspar

[brave schooner]

if you are rotating around x = 2

[native surge]

can you explain how to like get that

wait isnt it x + 1

no

?

@brave schooner?

[brave schooner]

let me do the problem and ill explain after, need some refreshing

[native surge]

alr

ty

[brave schooner]

yes its x+1

didnt read the question properly first time sorry

[native surge]

ok

i see now how

thankyou

[pearl pond]

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[pulsar stump]

whats this equate to and why?

[pulsar stump]

this is generalized union for big U, and cartesian product for <a,b>

[pine jay]

Start with the Cartesian products

Usually I see index notation for unions like these but I guess it’s already naturally indexed the way it’s a set of two objects

[pearl pond]

@pulsar stump Has your question been resolved?

[pulsar stump]

i am, what is it

[pine jay]

Do you not have a definition from your notes or textbook?

[pulsar stump]

basic one, im asking to check what it is

so what is it

[pine jay]

So in general with your notation it’s defined as

<A,B> := { (a,b) : a in A and b in B }

You asked about the Cartesian product, the thing you sent now is for the union..

[pulsar stump]

ik what cartesian product is

[pulsar stump]

do you not see the union sign?

[pine jay]

Uh yeah, I told you to start with the Cartesian products

[pulsar stump]

theyre there and theres nothing to do with it

i just want an =... answer

[pine jay]

Is it not better if I guide you towards the answer?

[pine jay]

Also I’m guessing you missed curly brackets for 0 and 1 here?

[pulsar stump]

actually thats how you write the cartesian product (0,{2, 3})

first element is in Natural numbers for example and the second element is a subset of Natural numbers

i.e $<a \in N,b \in P(N)>$

[jolly parrot]

Ayanokoji (always ping me)

[pine jay]

Alright so why write <> ? Your question implied it was a calculation to be done.

That’s just an ordered pair and not a Cartesian product

Unless your notation for ordered pair is <>

[pulsar stump]

yes, 2 pairs from the cartesian product NxP(N)

it is a calculation, what is the set resulting

[pulsar stump]

from this

[pulsar stump]

it is, more obvious

() is a hell to use for everything

[pine jay]

Okay good, that was not obvious from your message. Let’s move on

[pulsar stump]

alright

[pine jay]

So at the moment there’s another problem

You have a set of ordered pairs

Your Cartesian product gives a set of pairs, so something is missing here in order for you to take the union

[pulsar stump]

what is?

[pine jay]

So please show the original question.

[pulsar stump]

im trying to build up to it

f(0)={2,3}
f(1)={4,5}

if you see where i went with this

[pine jay]

So you left some pretty important information

The union is indexing through a

[pulsar stump]

but does it matter?

[pine jay]

Yeah since we can’t take the union otherwise

If you just have a union of a set of pairs

You can’t do anything

[pulsar stump]

teacher's definition

[pine jay]

Yup

You have a family of sets

[pulsar stump]

UF while F is what i drew

[pine jay]

And in our case we didn’t

[pulsar stump]

i see

[pine jay]

But now we can

[pulsar stump]

so what do i do here

[pine jay]

Okay so what is A?

[pulsar stump]

is the dom() A->P(B)?

[pulsar stump]

a set, undefined further

same for B

some sort of set, no givens

also not neccasarily H is a function

thats a follow up

[pine jay]

Are you sure this is the original question? If its undefined it’s best to leave the union as is

As you can’t really do anything

But maybe create a predicate of some kind with it

[pulsar stump]

yeah 100%

[pine jay]

My question is then, how did you get these numbers?

[pine jay]

Or these?

[pulsar stump]

i said A=N, B=N, and just took some numbers (small sample) to understand the union

[pine jay]

Ah I see

Sure let’s try

[pulsar stump]

purely to see what the result set will be

[pine jay]

So let’s maybe start slow

So let A = {0,1} for example?

[pulsar stump]

yes

P(B)={{2,3}, {4,5}}

[pine jay]

Is B the set of outputs for f?

[pulsar stump]

no, P(B) is

[pine jay]

And so B is arbitrary here too?

[pulsar stump]

as defined here

[pulsar stump]

yes

[pine jay]

Okay let’s try with this one

So since there’s only two elements from A, 0 and 1

The union which is indexing through A will just be a union of two sets

[pulsar stump]

yes keep going

but can you just write the =...

[pine jay]

Namely the union {(0,{2,3}),(0,{4,5})} U {(1,{2,3}),(1,{4,5})}

Do you see how?

[pulsar stump]

no

thats not true

[pine jay]

Hm did I miss something?

[pulsar stump]

,(0,{4,5}) is not defined as a singleton of a x f(a)

[pulsar stump]

only <0, {2,3}> is where f(a)=f(0)={2,3}

[pine jay]

Oh right, forgot f(1) was {4,3}

Ok let me rewrite it

Namely the union {<0,{2,3}>} U {<1,{4,5}>}

Do you agree?

[pulsar stump]

yes

not it's just the union of them

nvm

i think i got the answer

the range is A->B

and dom is A->P(B)

[pine jay]

Yeah and I think what we notice here is that this is just the same as the image f({0,1})

[pulsar stump]

well idk about that

[pine jay]

Oops

[pulsar stump]

ty mate

i have a lot of questions to do i really have to go

[pine jay]

I meant the graph set

Alright! Np

[pulsar stump]

oh we didnt learn that yet

.close

[pearl pond]

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[pine jay]

.reopen

[pearl pond]

✅

[pine jay]

@pulsar stump are you still there?

I missed a big error throughout this chat

Can you tell me what the Cartesian product {0}x{2,3} is?

[pulsar stump]

yeah {<0,2>,<0,3>}

[pine jay]

Yup, you wrote {<0,{2,3}>} earlier

And I somehow missed that

And didn’t correct it

So since you gave the correct answer i guess it was just a minor mistake right?

[pulsar stump]

yeah no worries

ty again mate

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[strange rose]

I need help with 5th iv)

[pearl pond]

@strange rose Has your question been resolved?

[strange rose]

<@&286206848099549185>

[ocean heart]

How far have you come so far?

[strange rose]

I took x=tan^-1 x and differentiated that after substituting

[timid isle]

I just want to know what course you're doing

obv maths/calculus but like what level (e.g a level, AP, degree etc.)

[abstract vine]

have you tried u sub?

[strange rose]

12th grade (highschool)

[strange rose]

yeah didnt work

[abstract vine]

can you show your previous results?

[strange rose]

pls bare with my hand writing

[timid isle]

dang your curriculums wild

[strange rose]

😭😭😭

isc board

[pearl pond]

@strange rose Has your question been resolved?

[civic crane]

Indian?

[strange rose]

yeH

[drowsy adder]

ye le

1/2 hona chahiye

you get the idea of keeping x as tan theta from 1+x^2 because of the identity

[strange rose]

yep got it thank you so much

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[strange rose]

hi

[strange rose]

I hv my maths test tomorrow on Differentiation and Application of derivates

AOD has sums where we have to find rate of change of some shapes and etc

does anyone have any tips on how to go about these sums

I always mess up here although my concepts of Differentiation are strong

any tips?

<@&286206848099549185>

[cosmic raptor]

you mean related rates?

[pearl pond]

@strange rose Has your question been resolved?

[strange rose]

yep

with reference to volumes and areas of shapes

<@&286206848099549185>

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[warm osprey]

Proof 0.999...998 = 1
x = 0.999...998
10x = 9.9.....98
9x = 9
x = 1
I randomly made up this proof. I just started Real Analysis and I learnt that two numbers are the same if you cannot find any number inbetween them. This would seemingly contradict the proof above because between 1 and 0.999...998 is 0.999...
What am I not understanding?

[cursive wraith]

uhm

in which position is the "8"

if there's an infinite number of 9s

there is an ENDLESS supply of 9s

so there is no "end"

to put an 8 to

[craggy tendon]

0.99…998 < 0.99…999 <1

[warm osprey]

Not sure this is true

[craggy tendon]

idk either

[cursive wraith]

ok so

[warm osprey]

Makes sense I suppose

[cursive wraith]

if you put an 8

it means you have to put it after a FINITE amount of 9s

so

let's say 1000000 9s for example

and then an 8

that's possible

but you can't put an infinite amount of 9s

AND THEN put something else

because there is no "and then"

[craggy tendon]

i don’t thing this is true tbh, it makes sense intuitively but we are just going by definitions that 0.99..98 => there is a finite amount of nines, why?

[cursive wraith]

this is literally the definition of decimal expansion

a decimal expansion is a sequence of numbers

so the decimal expansion of pi looks like

first number : 1

second : 4

third : 1

then 5

then 9

etc...

you can put all those numbers in a sequence

so the REAL issue here is that 0.99...98 is not well defined when written like this

[craggy tendon]

Uhh i think you would first have to define 0.99..8 properly

[craggy tendon]

yws this

[cursive wraith]

so

as long as you have a well defined sequence of digits 0 to 9

you can define the number with that decimal expansion

[warm osprey]

👍 Thank you

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[grim fractal]

why does $\overline{c}+d = c+d$

[jolly parrot]

Ari

s

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[midnight haven]

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[visual canyon]

is this correct?

[visual canyon]

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[rich copper]

What do I do with the 3^-2/4^-2 and the 3^3/2^3 in this

[rich copper]

I got it now

Thanks

[jolly parrot]

faiyrose

[rich copper]

Oh yeah I noticed that

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[supple berry]

is there a way to simplify this first?

[supple berry]

to shorten the process

and avoid the messy roots and fractions lol

cuz, like, i evaluated it one-by-one lol

[warm current]

Sum to product, I presume

[supple berry]

OHHH YEAH

wait let me try

oh wait

[ocean hornet]

also kind of looks like sum of cot

[supple berry]

sum of cot?

[supple berry]

oh

[ocean hornet]

but sum to product is guaranteed to make progress i think, so probably worth doing that instead

[supple berry]

got it now

thanks

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[light helm]

where's pi/2 and 3pi/2 coming from

[tame apex]

Tanx = 0

It's wrong btw

I'm asking where I went wrong

I factored it

Got tanx = 0 and tanx = -1

Solved for that

[light helm]

you didn't solve tan(x) = 0 correctly

[tame apex]

Idk where I went wrong

[light helm]

tan(pi/2) isn't 0

tan(3pi/2) isn't 0

those are undefined

[tame apex]

Oh it's sin/cos

M

Oops

Jeez I am rusty

Sorry I haven't done trig in years

Ok more like half a year but so

Still

Ohhjh

2pi and pi then

Ok I got 0, pi, 3pi/4, and 7pi/4

@light helm is that correct

[pearl pond]

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[somber magnet]

My book asks me to decide if this is a vector space. Its definitely not given it doesn't pass the scalar multiplication closure rule

[somber magnet]

however i also dont think it passes the addition closure either.

here is the question:

decide whether or not this set is a vector space. The set of functions {f: R→ R | df/dx +2f =1}

[tacit mulch]

It also doesn't contain the 0 element

As long as you can show it breaks one of the conditions, it isn't a vector space. You don't necessarily need to worry about the others

[somber magnet]

howso, wouldnt 0*(1/2+Ce^-2x) = 0?

oh wait would it need to be another vector

are you refering to 0vector + V = V

or the 0*V=0

[tacit mulch]

I mean the zero vector, f(x)=0 does not satisfy that differential equation

[somber magnet]

oooh

gotcha

[tacit mulch]

But also what you said is true, it doesn't have closure under addition or scalar multiplication

[somber magnet]

wouldnt the zero vector be -1/2-Ce^-2x

gotcha

such that -1/2 +1/2 +-ce^-2x +ce^-2x = 0

no thats additive inverse

aah

[tacit mulch]

The zero vector in a vector space is the additive identity, like (0,0,0) in R^3

[somber magnet]

ah

and f(x) != 0

[tacit mulch]

For a set of functions like this, it's just the zero function f(x)=0, since adding that to another function g just gives g

[somber magnet]

right

[tacit mulch]

Yeah but you only need to show that it breaks one of the conditions for a vector space

So if you showed that it doesn't have closure with scalar multiplication, then that's enough to prove it's not a vector space

[somber magnet]

gotcha, thank you!

[tacit mulch]

Np 👍

[somber magnet]

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[sharp smelt]

so what all do I have to prove here?

[sharp smelt]

The union of two subspaces if a subspace of V if one is contained in another

and?

[vital estuary]

remember only if is a <=

[sharp smelt]

yes

oh

right

[marsh tangle]

doesn't even look like it should take more than a one-liner proof.

[sharp smelt]

so I just prove this

[vital estuary]

no

[sharp smelt]

I thought it was an iff statement

[vital estuary]

ou wait yes

english is dumb

[sharp smelt]

so will this suffice ?

[marsh tangle]

yes, works

[sharp smelt]

oh

thanks

my bad, it's supposed to be if and only if

not only if

[rough stream]

v1 ∈ V1 implies λv2 ∈ V2
... So what?

[sharp smelt]

V_1 is a subset of V_2

[west sapphire]

there's an even simpler proof here

if V_1 is contained in V_2 then their union is just V_2

which is a subspace, done

[sharp smelt]

oh right

but as I said, it's an if and only if statement

[west sapphire]

yea the other direction is the harder part of course

[sharp smelt]

what

sorry

what do I have to prove in the opp direction?

[west sapphire]

if V1 and V2 are subspaces and V1 U V2 is a subspace, then either V1 is contained in V2 or vice versa

[sharp smelt]

hmm, ok

thanks

[sharp smelt]

could I have another hint

[west sapphire]

yea, try a contradiction... it would go like this:

suppose V1 and V2 are subspaces and V1 U V2 is a subspace
suppose for a contradiction that neither is contained in the other

then there is some x in V1 but not in V2
and there is some y in V2 but not in V1

[sharp smelt]

other than the common 0 element, right?

[west sapphire]

see if you an find a contradiction

[west sapphire]

there might be other stuff in common too, but suppose neither is completely contained in the other

so each one has at least some elements not in the other

[sharp smelt]

okay

like this?

[west sapphire]

probably need more detail/justification for why x+y is not in V1 and not in V2

[sharp smelt]

as x \in V_1 but y \not in V_1, x+y isn't in V_1

similarly

I can argue for V_2

[rustic gate]

why isnt x + y in V_1?

why can you conclude that?

(also i think the proof for this without contradiction is a bit cleaner)

[sharp smelt]

y isn't in $V_1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

so $x+y$ isn't defined

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rustic gate]

why not?

V_1 and V_2 are both subspaces of V

so x and y are elements of V

[sharp smelt]

right

[rustic gate]

you can definitely do x + y

[sharp smelt]

can I have a hint in that case?

[rustic gate]

express y in terms of x and x+y

[sharp smelt]

what

how

[rustic gate]

y = some expression using x and x+y

[sharp smelt]

$y=x+y-x$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rustic gate]

yes

so if x + y was in V_1, then you'd have

,, y = \underbrace {x + y}{\in V_1} - \underbrace x{\in V_1}

[rustic gate]

the conclusion is?

[sharp smelt]

y is in $\V_1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rustic gate]

but we know that not to be the case

[sharp smelt]

yes

so we've arrived at a contradiction

[rustic gate]

so x + y \in V_1 is impossible

[sharp smelt]

So I basically repeat something similar for $V_2$ and I'm done, right?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rustic gate]

yes

[sharp smelt]

ok

thanks a lot!

.close

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[winged lance]

how do i solve for (b)

[regal herald]

if you mean 2

the odd terms can be written by a+(n-1)(2d)
n=1: a 1st term
n=2: a+2d 3rd term
n=3: a+4d 5th term

where n is the nth odd term

or do you mean 5b?

[winged lance]

yeah

5b

[pearl pond]

@winged lance Has your question been resolved?

[winged lance]

<@&286206848099549185>

[pearl pond]

@winged lance Has your question been resolved?

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[gusty flume]

A function of K^n->K^m with K being real or complex numbers is partially differenriable in all n coordinates at a given point. Is it necessarily always also continuous at that point?
Is the inverse statement also true? If not, what's a counterexample?

[gusty flume]

(I am assuming "yes" and "no" respectively, but that's mostly vague recollection that I think I remember hearing as a side note in some lecture)

[rustic gate]

id think that both are false

[digital cargo]

both are false

continuous does not imply differentiable obviously, and partial differential just means that the partial functions are continuous

[pearl pond]

@gusty flume Has your question been resolved?

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[grand sedge]

Can i please get some help with a)

[winter leaf]

Is this high level math?

[grand sedge]

this is stats

[winter leaf]

It is so hard

[grand sedge]

Its called statistics

[winter leaf]

ok

Do you know the answers?

[grand sedge]

<@&286206848099549185>

[grand sedge]

nope

[winter leaf]

help!

Are u still in the exam?

[grand sedge]

nope

@winter leaf this is called stats, its basically where you learn about probability

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[gilded hollow]

Could someone explain to me $p \rightarrow q$ the logic behind it?

[jolly parrot]

Merineth

[gilded hollow]

Afaik "if p then q" is the official meaning of it?

And it's only false if p = true nad q = false

Why is that the case? And is there a better way to remember it?

nel

https://tenor.com/view/cat-kiss-gif-21168766

[glass meadow]

I mean there are only 4 pairs of values to check

[vestal tapir]

One way is think of a real life example
In real life, if the kettle is boiling, then there is water in the kettle

That's the thing it's for

[gilded hollow]

Like for example if i were to rephase it as :

p: it rains
q: I bring an umbrella

[vestal tapir]

This is a worse example

[gilded hollow]

Hmm

We are given this on our formula sheet

ANd i can see that it's only false when p is true and q is false

I just mainly dont understand it?

especially the p = 0 and q = 1

If there is no water in the kettle, then it's boiling????

is true? xD

[glass meadow]

Taking @vestal tapir 's example

if the kettle is boiling, then there is water in the kettle

p = "the kettle is boiling"
q = "there is water inside"

The if-then statement is just a statement, it can be true or false

[gilded hollow]

Why did you swap p and q?

[glass meadow]

I didn't?

"if p then q"

[gilded hollow]

p = "there is water inside"
q = "the kettle is boiling"

should be like this?

[glass meadow]

No

[gilded hollow]

huh

[glass meadow]

"if there is water inside the kettle then it's boiling"

That's not a true statement in the real world

You can have water inside the kettle and have it not boiling

[gilded hollow]

If there is water in the kettle, then it's boiling
p q

[cursive wraith]

so almost, the "if" is not part of p technically

[gilded hollow]

Water in the kettle is the if and the statement that it's boiling is q?

[viscid dew]

so, if there is no water in the kettle and it's boiling, you can't disprove $p\implies q$

[glass meadow]

You're mixing everything up...

[jolly parrot]

nameless individual

[vestal tapir]

here's an OR statement "Josh went to the dentist or he's at home"
I don't know if A is true, I don't know if B is true, i know something worse, something less useful

that A + B > 0, arithmetically

like any OR statement it's equivalent to an implication
"if Josh isn;t home, he went to the dentist"

A → B means A <= B, "artihmetically"

[gilded hollow]

if p then q \
$p \rightarrow q$ \
if there is water in the kettle, then it's boiling

[jolly parrot]

Merineth

[gilded hollow]

How is this wrong?

[vestal tapir]

a kettle

[gilded hollow]

or is it reversed?

[vestal tapir]

this could be true for some situation

[gilded hollow]

if it is boiling, then there is water in the kettle

[vestal tapir]

but like, the other way around is more true

yes that one is always true

A being true is sufficient for B being true

B being true is necessary for A to be true

like, it's not an analogy

it's an actual implication, it's meant in the same way

so you're confused if you're trying to see how it applies to the topic, it's the topic

[gilded hollow]

Hmm i sincerely dont understand to be honest

i'll just copy from the formula sheet instead of tryingg to learn it

[vestal tapir]

the logic is that it's not just a binary operation, it's a statement that tells you something

that's what makes it confusing, that it's kinda both

b + c doesn't tell you anything, but A V B does

like when you see A V B in isolation it's usually to imply that A V B is true

@gilded hollowI actually want to figure this out, it's a common question and I never get what people mean, what are they missing

[gilded hollow]

Honestly it got to complicated for me to understand

[vestal tapir]

ok

[gilded hollow]

So i just decided that it's fine not to understand it

I do have another question however revolving tautology

[glass meadow]

I feel like you keep doing that and it's going to bite you later

[vestal tapir]

last attempt then
there are 16 different logic operations we can do on 2 variables, some are more useful, so you learn them
A → B is one of 16 operations, it's like saying A is less than or equal B, A <= B

[gilded hollow]

Perhpaps, but i also tend to get stuck on something and sit on it for hours until i finally figure it out but by that time i'm to exhausted to continue studying

[vestal tapir]

it's very useful, we have a thing for it in English

"if then", at least often, means A → B

just like we have and in English

[gilded hollow]

Show that the summary is a tautology: \
$p \rightarrow (q \rightarrow p \land q)$

[glass meadow]

\land?

[jolly parrot]

Merineth

[gilded hollow]

I can formulate in words why this is a tautology

[vestal tapir]

you probably just translate it to ORs

[glass meadow]

That's pretty easy to show by translating the "implies" into "not p or q"

[gilded hollow]

Ooh okay so i am supposed to transoform them?

[glass meadow]

I assume so

[gilded hollow]

Is the E_10 the one you meant Nel?

[glass meadow]

Yes

[gilded hollow]

$\neg p \lor \neg q \lor p \land q$

This is what i got when i transoformed it

[glass meadow]

Yeah, you can remove the parentheses

[jolly parrot]

Merineth

[glass meadow]

Now a little de Morgan's

Not quite

[gilded hollow]

$\neg(p \land q) \lor p \land q$

[jolly parrot]

Merineth

[glass meadow]

Right

Do you see the end?

No

[gilded hollow]

Hmm lemme think

I think it should be this?

[glass meadow]

Yes

[gilded hollow]

since we have "V"

it's always true

[glass meadow]

Yeah you have (not r) or r

[gilded hollow]

Yeah okay that makes a lot more sense

Seems way more plausible that i'm supposed to solve it by showing like that

instead of fomulating in words lol

[glass meadow]

Well yeah maths classes are going to ask you to use symbols, not words

[gilded hollow]

I def need more practice

Prove that it's a contradiction : \
$(r \land \neg q) \land (p \rightarrow \neg r) \land ( \neg p \rightarrow q)$

[jolly parrot]

Merineth

[glass meadow]

I have to leave but this is gonna be more of the same, get rid of the implications and use some de Morgan and whatever

[gilded hollow]

Instead of rewriting it, can i write it using a truth table?

We can easily conclude that the only way for the statement to be true is when all the statements between the $\land$ has to be true

[jolly parrot]

Merineth

[gilded hollow]

So we find every possible outcome for the statements to be true

[glass meadow]

That works too, it's just more laborious

[gilded hollow]

Yeah i realized that ;-;

very intensive thinking

I wouild like to do your method of simplifying it

but i'm having a hard time finding an appropriate formula

[glass meadow]

(r n ~q) n (p -> ~r) n (~p -> q)
r n ~q n (~p u ~r) n (p u ~q)
r n ~q n p n r n ~p n ~q
but p n ~p = 0

[gilded hollow]

Hmm i cant seem to understand the second step

you removed the parentheses

but what about the second part?

[glass meadow]

.

[gilded hollow]

Hmm but $p \rightarrow \neg q$ doesn't exist?

[jolly parrot]

Merineth

[glass meadow]

It's just p -> s where s = ~q

[gilded hollow]

Damn

[glass meadow]

You need to get used to forming variables for various things in your head

You can think of it as adding parentheses and considering "the whole thing in parentheses"

p -> (~q)

[gilded hollow]

yeah that makes sense

i see what you did now

$r \land \neg q \land \neg p \lor \neg r \land p \lor q$

[jolly parrot]

Merineth

[gilded hollow]

I got to this point

i see some de morgan now

$r \land \neg q \land \neg (p \land r) \land p \lor q$

[jolly parrot]

Merineth

[gilded hollow]

$r \land \neg q \land 0$

[jolly parrot]

Merineth

[gilded hollow]

And because of this rule where p and negative is 0 means the entire thing is negative

[glass meadow]

Careful, "and" typically has a higher precedence than "or", meaning you need parentheses around both "or" here

[gilded hollow]

Yeah i'm having it tought identifying when i can use paren or remove

[glass meadow]

Looking at that, it seems you pretty much can't remove parentheses ever, just move them around via associativity

But it's pretty restrictive

If you understand what you're doing, it's much more practical to remove parentheses in cases like p n (q n r) or (p u q) u r because that's what associativity gives you

But p n (q u r) is not the same as p n q u r

[gilded hollow]

Okay, i'll treat paren within logic the same as in math then

wont remove them :P

[glass meadow]

Look at the boolean algebra part, it's obvious that a + (b + c) = a + b + c and so on

It works the same

[gilded hollow]

Yeah but me removing only increases the risk of me doing wrong :I

I really despise when they include a excercise in the book but there is no solution at the end..

Assume that q is true (1) and that the statement is true. Determine p r and s

$(q \rightarrow (\neg p \lor r ) \land \neg s ) \land (\neg s\rightarrow \neg r \land q)$

[jolly parrot]

Merineth

[gilded hollow]

Before i start using the formula i can conclude that if i want to determine p r and s and that the statement is true.
Then both sides of the and symbol has to be true

$(q \rightarrow (\neg p \lor r ) \land \neg s )$

[jolly parrot]

Merineth

[gilded hollow]

$(\neg s\rightarrow \neg r \land q)$

[jolly parrot]

Merineth

[gilded hollow]

As in these two parts have to both be true

[glass meadow]

Yes though you don't really need to start by separating them

Doesn't really matter

[gilded hollow]

Can i keep at as is and just simplify?

the \right arrow parts?

[glass meadow]

Yeah

[gilded hollow]

Okay I’ll try that

[glass meadow]

The fact that they didn't put parentheses for the implications is pretty bad but I think we can assume that p -> q n r is p -> (q n r)

[gilded hollow]

$\neg q \lor ((\neg p \lor r ) \land \neg s ) \land (s \lor ( \neg r \land q ))$

[jolly parrot]

Merineth

[gilded hollow]

This is what i got after swapping the rightarrow with it's equivalent

[glass meadow]

You're missing a set of parentheses

[gilded hollow]

i am?

[glass meadow]

$\color{red}(\color{black}\neg q \lor \color{green}(\color{black}(\neg p \lor r ) \land \neg s \color{green})\color{red})\color{black} \land (s \lor \color{green}(\color{black} \neg r \land q \color{green})\color{black})$

[jolly parrot]

Nel

[glass meadow]

You correctly added the green ones but removed the red ones

[gilded hollow]

Ah i see

[glass meadow]

(odd that coloring in white makes it black and vice versa)

[gilded hollow]

I would assume i can use this?

[glass meadow]

Sure

[gilded hollow]

$( \neg q \lor (( \neg s \land \neg p ) \lor ( \neg s \land r ))) \land ((s \lor \neg r) \land (s \lor q ))$

[jolly parrot]

Merineth

[gilded hollow]

And now i can substitute q with it's equivalent value?

Since we assume q is true

[glass meadow]

I forgot the question

Ok so

$E = (q \rightarrow (\neg p \lor r ) \land \neg s ) \land (\neg s\rightarrow \neg r \land q)$

[jolly parrot]

Nel

[glass meadow]

E = 1, q = 1, find p,r,s

You can actually just start by substituting q

$E = (1 \rightarrow (\neg p \lor r ) \land \neg s ) \land (\neg s\rightarrow \neg r \land 1)$

[jolly parrot]

Nel

[glass meadow]

$E = ((\neg p \lor r ) \land \neg s ) \land (\neg s\rightarrow \neg r)$

[jolly parrot]

Nel

[glass meadow]

$E = ((\neg p \lor r ) \land \neg s ) \land (s \lor \neg r)$

[jolly parrot]

Nel

[glass meadow]

I'd say just apply de Morgan's here

[gilded hollow]

Oh I see.. I could’ve just substituted q with 1 immediately

[glass meadow]

Yeah

It doesn't matter too much though, you can very well use what you arrived at

$E = ( 0 \lor (( \neg s \land \neg p ) \lor ( \neg s \land r ))) \land ((s \lor \neg r) \land (s \lor 1))$

[jolly parrot]

Nel

[glass meadow]

$E = (( \neg s \land \neg p ) \lor ( \neg s \land r )) \land (s \lor \neg r)$

[jolly parrot]

Nel

[gilded hollow]

How could you remove the 1 at the start?

[glass meadow]

It's an implication

1 -> q just means q

I'm trying not to confuse you lol

[gilded hollow]

I see what you mean with it being 0

[glass meadow]

If you translate it into ~1 u q, it's 0 u q = q

[gilded hollow]

However

$0 \rightarrow ( \neg p \lor r) \land \neg s)$

[jolly parrot]

Merineth

[gilded hollow]

Can’t this be simplified to just

s has to be 0

[glass meadow]

No, why would it be?

[gilded hollow]

Since no matter what p and r is

It’s always true

[glass meadow]

No, it matters what p and r are

You have an "and" here

[gilded hollow]

$0 \rightarrow (\neg p \lor r)$

[jolly parrot]

Merineth

[glass meadow]

s has to be 0, yes, but if r is 0 and p is 1, you get a (0 and 1)

[gilded hollow]

This is always true

[glass meadow]

Ok hold on

First of all it's not 0 -> ..., it's 1 -> ...

[gilded hollow]

oh right.............

man fuck my life

[glass meadow]

0 -> p is always true, but that's a different story

[gilded hollow]

It's so hard keeping track of everything

[glass meadow]

Hmm ok let's get back to this

Remember the kettle? "if the kettle is boiling, then there is water inside". Now you know that "the kettle is boiling" is true, so "there is water inside" is true as well

Does that make sense?

[gilded hollow]

Yeah i guess

it's just so unbelivably complicated to keep track of everything

I despsie when i have to "guess" and there isn't any method behind what i do

$(1 \rightarrow (\neg p \lor r ) \land \neg s ) \land (\neg s\rightarrow \neg r \land 1)$

[jolly parrot]

Merineth

[gilded hollow]

This is what i start with

[glass meadow]

That's basically intuition

[gilded hollow]

s HAS to be 0

i can see that just by looking at the first statement

[glass meadow]

Yes

[gilded hollow]

$(1 \rightarrow (\neg p \lor r ) \land 1 ) \land (1 s\rightarrow \neg r \land 1)$

[jolly parrot]

Merineth

[gilded hollow]

Can i write it as such now?

[glass meadow]

Just 1 instead of 1s, but yeah I guess

It's not rigorous

[gilded hollow]

I can also see that r cannot be 1

[glass meadow]

Yeah

[gilded hollow]

that also implies p has to be 0

[glass meadow]

Yes

[gilded hollow]

So they are all 0?

p r and s

[glass meadow]

No, you said s is 1

Wait no

You're correct

[gilded hollow]

Okay so it isn't a rigorous exercise it's more of a "see" question?

[glass meadow]

No the way you solved it is not rigorous

You just said "I can see that ..."

[gilded hollow]

I’m just worried that a similar question appears on the exam and I won’t be able to see it at all

[glass meadow]

Do you understand how to get to this point?

[gilded hollow]

Yea

You swap the q with it's appropriate value and then apply

[glass meadow]

Yes

Then you can rearrange the expression:

$E = \neg s \land ((\neg p \lor r ) \land (s \lor \neg r))$

[jolly parrot]

Nel

[glass meadow]

You have E = 1 so by that same rule of p n 1 = p you can deduce ~s = 1 hence s = 0

[gilded hollow]

Yeah, makes sense

And then we sub s = 0 into our expression

[glass meadow]

That's basically how you solved it, just a little more rigorous

[gilded hollow]

Okay nice

i'll be sure to remember that

[glass meadow]

Right so you get
$E = (\neg p \lor r ) \land (0 \lor \neg r)$

[jolly parrot]

Nel

[glass meadow]

0 u ~r = ~r so you can rearrange again

$E = \neg r \land (\neg p \lor r )$

[jolly parrot]

Nel

[glass meadow]

Same thing but for ~r which gives r = 0

[gilded hollow]

ooh neat

[glass meadow]

And you're left with E = ~p

So ~p = 1, p = 0

It might be more rigorous to say 1 <=> ... instead of E = ... everywhere

I'd say it's a matter of taste honestly

Like it may matter in higher order logic or something but not here

[gilded hollow]

Yeah i see what you mean

It isn't to hard

Luckily got some easy "prove" questions

$\neg (p \land ( \neg p \lor q )) \Longleftrightarrow \neg q \lor \neg p$

[jolly parrot]

Merineth

[gilded hollow]

$\neg p \lor \neg ( \neg p \lor q )) \Longleftrightarrow \neg q \lor \neg p$

[jolly parrot]

Merineth

[gilded hollow]

$\neg p \lor (p \land \neg q )) \Longleftrightarrow \neg q \lor \neg p$

[jolly parrot]

Merineth

[gilded hollow]

Well here comes the problem again when i'm unseure if i'm allowed to ignore parentheses

$\neg p \lor (p \land \neg q) = \neg p \lor p \land \neg q$?

[jolly parrot]

Merineth

[gilded hollow]

If that is true then i can solve it but not otherwise

[glass meadow]

How do you solve it if it's true?

[gilded hollow]

Hmm actually i take that back

$\neg p \lor p = 1$

[jolly parrot]

Merineth

[glass meadow]

Right so this is why precedence matters

$\neg p \lor (p \land \neg q) = \neg p \lor p \land \neg q \neq (\neg p \lor p) \land \neg q$

[jolly parrot]

Nel

[glass meadow]

The first two are exactly the same thing when you consider "and" to have higher precedence than "or"

(which is typically the case)

Your material doesn't deal with precedence though, so if you don't want to deal with it either, don't remove the parentheses

[gilded hollow]

$\neg p \lor (p \land\neg q)$

Am i right so far tho?

[glass meadow]

No the second one is a "and"

[gilded hollow]

mb

[jolly parrot]

Merineth

[gilded hollow]

This is what i meant

[glass meadow]

From this, yeah you're correct

[gilded hollow]

$\neg p \lor (p \land\neg q) \Longleftrightarrow\neg (q\land p)$

[jolly parrot]

Merineth

[gilded hollow]

So this is the point i get stuck at

[glass meadow]

Distribute

[gilded hollow]

aaaaaaah ofc

totally missed that one

$(\neg p \lor p) \land (\neg p \lor \neg q) \Longleftrightarrow\neg (q\land p)$

[jolly parrot]

Merineth

[glass meadow]

Ye

[gilded hollow]

and since (neg p V p) = 1

we have thus proven it

[glass meadow]

Well you're one de Morgan's away but ye

[gilded hollow]

Yea

nice

Completely forgot about distributive

When a question tells me to "negate"

what do they mean?

Negate and simplify : \
$p \land q \rightarrow r$

[jolly parrot]

Merineth

[glass meadow]

I guess just ~(p n q -> r)

[gilded hollow]

oh damn

How do you understand that ? D: I dont even know what they are asking for when i'm given a question like that lol

[glass meadow]

I mean it seems a little stupid, can you screenshot that question?

[gilded hollow]

Yeah

it's 15.20

It literally translates to "Negate the following statements and simplify the results"

$\neg r \land (p \land q)$

[jolly parrot]

Merineth

[gilded hollow]

This is what i got but idk if it's right

Oh nvm they had the answer in the book

and it's correct?

Kind of an ass question, no?

[glass meadow]

Yeah

[gilded hollow]

Shesh the next question asks me to form my own statement based on text....

does that seem normal?

[glass meadow]

I don't know I guess they felt like writing "negate and simplify E" instead of just "simplify ~E"

[glass meadow]

Understanding the language I suppose

[gilded hollow]

Is this a normal way to write=

I haven't encountered this in the book at all yet

[glass meadow]

Define normal

Yes these are usual mathematical symbols

[gilded hollow]

Why do they have different rows*?

Can this be written on a single row?

everything above t and then long left and right arrow t?

[glass meadow]

With "and" symbols to connect each row, yes

It's written as a list and a conclusion

If we have:

• a
• b
• c
  then:
• d

[gilded hollow]

Oooh okay

That's sooo odd tho, it isn't even mentioned in the book but he just started using it on the exams

Guess i'll just have to remember that

[glass meadow]

∴ means "therefore": https://en.wikipedia.org/wiki/Therefore_sign

[gilded hollow]

Yeah i assumed so but good to know

[glass meadow]

https://en.wikipedia.org/wiki/Logic
Look at the first image

It translates in english to:

• we know p is true
• and we know that "if p is true then q is true"
• therefore we can conclude q is true

[gilded hollow]

Ooh i see

Show that the following is correct : \
$p \land q \Rightarrow p \lor r$

[jolly parrot]

Merineth

[gilded hollow]

How do i prove this?

inferenslaws?

[glass meadow]

Is r supposed to be q?

[gilded hollow]

no

[glass meadow]

Oh never mind

[gilded hollow]

That's what got me hung up also

[glass meadow]

L5 followed by L4, yeah

[gilded hollow]

ooooooooh

i can combine

[glass meadow]

Of course

[gilded hollow]

hmm

$(p \lor q \rightarrow r) \land (p \land s) \Rightarrow r$

[jolly parrot]

Merineth

[gilded hollow]

$(p \rightarrow r) \land (p \land s) \Rightarrow r$

[jolly parrot]

Merineth

[gilded hollow]

Not sure how i handle the s tho

since s isn't listed on the formulas

I can transoform p into p V q

[glass meadow]

It's L5 again

[gilded hollow]

$(p \rightarrow r) \land ((p \land q) \land s) \Rightarrow r$

[jolly parrot]

Merineth

[gilded hollow]

Like so?

[glass meadow]

No

p n s => p

[gilded hollow]

oooh lol right

$(p \rightarrow r) \land p \Rightarrow r$

[jolly parrot]

Merineth

[gilded hollow]

$(p \rightarrow r) \land (p \lor q) \Rightarrow r$

[jolly parrot]

Merineth

[gilded hollow]

L_1?

[glass meadow]

L1 from here, yes

[glass meadow]

That doesn't help

[gilded hollow]

might be a weird question but

i cant go from right to left right?

Considering the right arrows?

$r \Rightarrow r$

[jolly parrot]

Merineth

[glass meadow]

You can't

That's why <=> exists

[gilded hollow]

Yeah i had a suspicion that was the case

Just had to verify

Well either way i've been at it for like 4.5 hours now and i can feel my concentration drifting

[glass meadow]

=> is "implies" while <=> is "if and only if"

[gilded hollow]

I'll continue from where i left of tomorrow + start with induction if you are good with that :P

[glass meadow]

(or "is equivalent to")

Lol you shouldn't expect me to be there

But good luck

[gilded hollow]

Well i am glad when you are ^^

[glass meadow]

[gilded hollow]

thanks for the help Nel

have a nice evening

.close

[pearl pond]

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[pearl pond]

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[quiet reef]

Help 5 b

[quiet reef]

,rotate

[quiet reef]

I have gotten the equation 2sin (3x) -1

but how do i find p when Y =0 ?

<@&286206848099549185>

[pearl pond]

@quiet reef Has your question been resolved?

[pearl pond]

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[pearl pond]

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>.
• Type the command .close to free the channel when you're done.
• 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.

[hazy radish]

I know what injection, surjection, and bijection is, but how do you prove a function meets some or all these properties?

[hazy radish]

For a small set of numbers in the domain and codomain is easy, just look for examples where it isnt true

but for larger sets it becomes less obvious

[west sapphire]

do you have an example?

[hazy radish]

not an example of proving a function but an example of coming up with a function