you should use cantors to show that something is finite right
#help-33
still temple
restive sun
what do you mean?
still temple
to prove a set is finite
restive sun
I don't see how cantor's theorem would help
still temple
doesnt cantors thrm state that there is a "bijection" if you show two injections, like from f :X ->Y and G: Y-> X
and a bijection only exists between countable sets
restive sun
that's cantor-shroeder-bernstein theorem
and bijections can exist between uncountable sets
still temple
whats the difference/
restive sun
cantor's theorem is |P(X)| > |X| for any set X
still temple
ohhhh
that verymuch makes sense
when do you use that then
also there is cantors diagonalization argument
restive sun
well, you can show something is uncountable by showing it surjects onto P(N) I guess?
cantor's theorem is used in a bunch of random places for other results
diagonalisation shows there exist uncountable sets
still temple
cantor-shroeder-bernstein proves that two sets have an equal cardinality by showing 2 injections from the respective directions
so basically use cantor-shroeder for proving some things are the same cardinality?
still temple
and diagonalization is that if you show a surjection does not exist, it is uncountable right
wait i think that is wrong
restive sun
if you can show there is no surjection from the naturals onto your set, then your set is uncountable
but that's not really a consequence of diagonalisation
diagonalisation just shows that there exist sets that do not biject to the naturals
still temple
ahhhh idk why im so confused
trail flint
Do you think you could help me with this last step?
I don't understand the transformation that breaks it into _ + /n + /n^2
restive sun
this channel is occupied; you should open a new help channel
restive sun
but it looks like partial fractions anyhow
marsh citrusBOT
@still temple Has your question been resolved?
still temple
but it looks like partial fractions anyhow
how can you differentiate between countable vs countably infinite
restive sun
I mean
that's just finite vs infinite
it'll depend on your sets
but usually it should be fairly obvious which is the case
still temple
okay also, when you use diagonalization, how do you show NONE of the functions that exist provide a surjection
restive sun
you usually don't?
it can be hard to show something doesn't exist like that
and I don't really get what you mean by "using" diagonalisation
still temple
like to show something is not countable, like uncountable, you show that no possible surjection exists
restive sun
sure, I guess
still temple
how do you do that
restive sun
it'll really depend on the set
still temple
Is there any efficient way to sketch tangent graphs?
marsh citrusBOT
Is there any efficient way to sketch tangent graphs?
still temple
Like how would I go about sketching
Do i just
Transform every single point on the original tanx graph
compact crescent
Memorizing the features of the graph
hollow sparrow
basically just know the parent function, which is periodic so it's easy to repeat, then apply translations to the new graph.
ashen lichen
I need help understanding this question. How did the "student" know to use the identity shown in the question? and how is that identity used to solve for the exact value of that trig equation?
marsh citrusBOT
I need help understanding this question. How did the "student" know to use the i...
ashen lichen
usually I would just plug sin(pi/8) into my calculator and then plug the answers one by one until one of them matches, but I want to know how to actually solve this
mystic heath
,rotate
elfin berryBOT
restive sun
pi/4 is a pretty standard value to know (or derive from geometry)
so it's just using the double angle formula in reverse
ashen lichen
wdym in reverse?
restive sun
normally you use the double angle formula to split angles in half
it doesn't really matter, it's the same equation
it's just getting sin(pi/8) in terms of something with cos(pi/4) in it, which should be a known value
ashen lichen
whats the relationship between sin(pi/8) and cos(pi/4)?
restive sun
that's given by the formula in the question
cos(2x) = 1 - 2 sin^2(x)
where x = pi/8
ashen lichen
I know, but I'm confused about how I would get to that formula, or how I would know to use that identity if it wasn't given in the question
restive sun
the formula is non-trivial to derive
it's something you normally just memorise
along with the other double angle formulae
and you'd know to use it because pi/8 isn't a normally-known value for sin/cos, but pi/4 is
ashen lichen
.close
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celest silo
sorry but i think i have a brain lag
marsh citrusBOT
sorry but i think i have a brain lag
rotund pike
Multiply both sides by 9/2.
celest silo
but how r1 stays the same?
rotund pike
Why would r1 change?
celest silo
idk i thought r1, r2 and 6 should change
rotund pike
There is no r2.
elfin berryBOT
Chai T. Rex
rotund pike
r1 is a variable.
You simplify the left side by multiplying r1 times r1^2 to get r1^3.
You multiply both sides by 9/2.
Then, you do the multiplication on the right.
celest silo
I understood now, thank you so much
rotund pike
You're welcome.
pallid mortar
dusty jolt
marsh citrusBOT
marsh citrusBOT
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dusty jolt
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wicked notch
marsh citrusBOT
wicked notch
I can't figure out how to solve the ysinx part of the equation
I used the product rule and got ysinx + cosx but I don't know which part to mulitply by (dy/dx)
willow owl
$\frac{d(fg)}{dx}=g\frac{df}{dx}+f\frac{dg}{dx}$ where f and g are 2 functions of x
elfin berryBOT
calculus is fun
worn plank
try using this notation
willow owl
so now let f(x)=y and and g(x)=sinx
willow owl
**calculus is fun**
and use this to get the derivative of ysinx
marsh citrusBOT
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tepid hollow
dk how to to start this
marsh citrusBOT
dk how to to start this
wild fiber
Put them into a matrix
left merlin
and then do the algorithm
tepid hollow
Wat is the algorithm
marsh citrusBOT
@tepid hollow Has your question been resolved?
undone shuttle
Can someone show me how to solve ex no 1?
flat bear
Please show me the steps how
$\frac{d}{dx}\sqrt{xy}$
turns in to
$\frac{y}{2\sqrt{x}}$
elfin berryBOT
Sinatio
gilded cape
shouldn't that be sqrt(y)?
still temple
surely our mean $\ds\pdv x \s{xy}$ btw
elfin berryBOT
still temple
not like it matters computationally
still temple
surely our mean $\ds\pdv x \s{xy}$ btw
you*
flat bear
this is what i rewrote its from a lecture in microeconomics, im having trouble getting to the expressions marked in red by derivating
gilded cape
looks like numerators are missing square roots
flat bear
this is the next step, are they correct? is it some shortcut theyre hiding or?
marsh citrusBOT
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tired peak
marsh citrusBOT
tired peak
Why does a^2 being divisible by 7 mean a is also divisible by 7
tired peak
Ok let me think about that
tired peak
7 is prime
Sorry, I am kind of dumb, why does that work?
main idol
suppose a isn't divisible by 7
rapid ledge
its an equality a^2 has to be divisible by 7
its equal to 7*(something)
if that helps at all
main idol
then 7 doesn't doesn't appear in the prime factorization of a^2
tired peak
its equal to 7*(something)
7*x = a^2
tired peak
then 7 doesn't doesn't appear in the prime factorization of a^2
I can kind of see it
I could just write it down but I don't like doing that
rapid ledge
Oh I read your question wrong
tired peak
a*a is divisible by 7
and because 7 is prime
then a is a multiple of 7?
otherwise a^2 can never be divisible by 7
@main idol Sorry for ping, can you confirm this?
rapid ledge
If I remember it has to do with Euclids lemma
if you have a prime that divides ab then it divides a or b
if its aa then obviously
divides a
rapid ledge
thats how i learned it at least
tired peak
If a prime p divides the product ab of two integers a and b, then p must divide at least one of those integers a or b.
Euclid's lemma
ok
rotund pike
its an equality a^2 has to be divisible by 7
They're asking why a is divisible by 7.
Oh, never mind.
rapid ledge
yeah i read it wrong
rotund pike
Didn't scroll down.
tired peak
They're asking why a is divisible by 7.
I am asking that
wait
I am confusing myself I think
tired peak
They're asking why a is divisible by 7.
If a^2 is divisible by 7, how does this lead to a being divisible by 7.
rapid ledge
If a prime p divides the product ab of two integers a and b, then p must divide at least one of those integers a or b.
rotund pike
If a isn't a multiple of 7, then a^2 can't be a multiple of 7. Take the contrapositive.
That's the basic argument.
rapid ledge
it divides the product (aa) so it must divide (a) or (a)
tired peak
If a prime p divides the product ab of two integers a and b, then p must divide ...
If a prime (7) divides the product aa of integer a, then 7 must divide a.
rapid ledge
Yeah
tired peak
a*a is divisible by 7
ok but does this logic also work? Look at the sentences below
rotund pike
ok but does this logic also work? Look at the sentences below
Do you see how a not being a multiple of 7 means a^2 can't be a multiple of 7?
Like if you factor a into primes, and it doesn't have 7 as a prime, a^2 can't somehow have 7 as a prime.
The only way for a^2 to have 7 as a prime is if a had it.
There are no other primes that combine together to give you 7.
So, you can't duplicate all the primes in the factorization of a, none of which are 7, by squaring a, and combine two of them to get 7.
So, a^2 can't have 7 as a prime unless a does.
tired peak
Like if you factor a into primes, and it doesn't have 7 as a prime, a^2 can't so...
Oh ok, because if a is something * 7 then the 7 will also be squared
tired peak
ok thank you
rotund pike
No problem.
tired peak
That makes sense. I just didn't want to work through it and just write what the website says without actually understanding it
I understood the proof of root 2 being irrational but that was easier to follow
tired peak
No problem.
thanks for the help :)
rotund pike
You're welcome.
still temple
can someone help me with c please! :)
marsh citrusBOT
can someone help me with **c** please! :)
still temple
i know diff quotient is [f(a-h) - f(a)] / h
still temple
erm what
still temple
yea but you need to do difference quotient
tired peak
Oh ok
still temple
it's f(90° + h) and f(90°)
whole sleet
Alright so you plugged sin(x) into the diff quotient. What did you get?
still temple
can you use special triangles to evaluate 90°
whole sleet
Keep in mind sin(90+h) simplifies well
whole sleet
A nice fallback option here is to use the sum identity
sin(90 + h) = sin(90)cos(h) + cos(90)sin(h)
still temple
oh
whole sleet
Or I'd consider your angle h, and the other angle h+90 on the unit circle. There's a pretty nice way to link them
storm blaze
marsh citrusBOT
storm blaze
asking for 18
I am getting the image coordinates: (2,1), (2,-1) and (-2,-1)
but the book says:
marsh citrusBOT
@storm blaze Has your question been resolved?
narrow pecan
I am getting the image coordinates: (2,1), (2,-1) and (-2,-1)
Are you translating the center to the center and scaling the triangle?
slow wadi
trying to solve for x. answer key says x = -1 but I don't know how to get that. I keep getting -6/7
marsh citrusBOT
trying to solve for x. answer key says x = -1 but I don't know how to get that. ...
slow wadi
gleaming pecan
convert the given powers to a power with base 2, and also show your work
gleaming pecan
convert it into png
full orchid
your issue lies in the right hand side
note that you're multiplying $2^{4x+12}$ by an additional 2, so you have to add another 1 to the exponent
elfin berryBOT
accialto
full orchid
did you just rename the file extension to .png
full orchid
that doesn't work
slow wadi
woops
full orchid
**accialto**
here's the issue tho^
slow wadi
thank u im trying to understand it sorry
ig im confused how multiplying by 2 would add a 1
full orchid
exponentiation is just repeated multiplication, right?
so 2^5 is just 2*2*2*2*2
imagine you multiply 2^5 * 2, you get 2*2*2*2*2*2, or 2^6
you can do the same thing with variables
2^x is just 2 *2 *2 *... *2 x times
if you do 2^x * 2, it's 2 * 2 *2 *...*2 x times *2 again, which makes it x+1 times
slow wadi
tysm i just need a moment to ponder rq
gleaming pecan
if you do 2^x \* 2, it's 2 * 2 \*2 \*...\*2 x times *2 again, which makes it x+1...
$\text{look also at such an example:}\\\left( \frac{1}{9} \right)^{x+2}=9\cdot 27^{2x+5}\Leftrightarrow \left( 3^{-2} \right)^{x+2}=3^{2}\cdot \left( 3^{3} \right)^{2x+5}\Leftrightarrow \\3^{-2x-4}=3^{2 }\cdot 3^{6x+15}\Leftrightarrow 3^{-2x-4}=3^{2+6x+15}=3^{6x+17}\Leftrightarrow \\-2x-4=6x+17\\\text{etc}$
elfin berryBOT
Joanna Angel
gleaming pecan
exponentiation is just repeated multiplication, right?
oh not for you, but for liv 🙂
slow wadi
oh wait its cause you add exponents right
like theres the 2 to the power of one and you add it to the 12?
gleaming pecan
oh wait its cause you add exponents right
yes, in your case: 1 + 4x + 12
then you get 4x + 13
slow wadi
ok i get it now thank u that was my issue
gleaming pecan
yvw
slow wadi
.close
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wise oxide
marsh citrusBOT
round finch
y intercept is when x is 0
so his thing looks right
idk whats the issue
do u know the equation of the graph
so u can plug in 0 for x
wise oxide
I got it y was e-2
round finch
oh
i guess i didnt see the equation
for the graph
oh wait
u entered it
im dumb
i was looking at graph xd
silk sable
So the equation was e^(x+1) - 2
rotund pike
<@&268886789983436800> spam, probable scam, all other messages repeat the same thing
marsh citrusBOT
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marsh citrusBOT
main idol
<@&268886789983436800> spam
lone heart
<@&268886789983436800> spam
#help-14 message
I think it was from this btw
quaint hill
maybe i shouldve muted instead of banned
idk
opening a help channel to advertise a server with 1 member is just really sus behaviour
bleurgh ill lift the ban
it was overkill
main idol
that's been autoban for a while. keep it that way. my dos pesos
quaint hill
its autoban for users with 0 server history
prime nova
they were new
quaint hill
¯_(ツ)_/¯
main idol
¯_(ツ)_/¯
prime nova
kinda overkill
main idol
act like a bot, get treated like a bot
prime nova
Hello, How can I help you?
main idol
next step in evolution: avoid acting like artificial intelligence
marsh citrusBOT
Available help channel!
frozen python
why does ⁹log3 and 1-⁹log3 both equal 1/2
marsh citrusBOT
why does ⁹log3 and 1-⁹log3 both equal 1/2
iron meadow
why does ⁹log3 and 1-⁹log3 both equal 1/2
well because 1 - 1/2 = 1/2 : )
frozen python
well because 1 - 1/2 = 1/2 : )
well what is ⁹log3?
frozen python
well what is ⁹log3?
⁹log3 isnt 1 is it?
frozen python
just written really weir
its wierd?
how do you usually write it?
yes its the base
iron meadow
well what is ⁹log3?
seems like a weird notation for log base 9 of 3 (i.e. 9 raised to what power is 3 ? ==> 1/2)
unreal shale
well what is ⁹log3?
1/2
elfin berryBOT
adam
unreal shale
yeah
unreal shale
where are you taking this math, it might be a different system
iron meadow
what country are you from @pei? im curious where that notation is used 😄
frozen python
where are you taking this math, it might be a different system
im indonesian
unreal shale
where are you from?
well i live in the UAE, but my school is doing the International Baccaluarate
unreal shale
im indonesian
very interesting
marsh citrusBOT
Available help channel!
edgy zinc
marsh citrusBOT
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
1
marsh citrusBOT
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blissful nova
Begin by Factorisation
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severe island
hi @stoic saddle (sorry for the ping), but for the exact same question i get the solution (thanks to you)
marsh citrusBOT
hi <@183668144404037632> (sorry for the ping), but for the exact same question i...
severe island
but why can’t i do 8C8 * 5C1 * 3C1
well i get pretty close with this to the actual answer so i wonder what i’m missing
severe island
but why can’t i do 8C8 * 5C1 * 3C1
this is basically choosing first vertex * choosing second vertex * choosing third vertex
severe island
i guess i meant one cuz i have the liberty to choose anything for the first vertex
8c8 is 1 too
stoic saddle
anyway the number of choices for the second vertex depends on the relative placement of the first two
could be 3 but could also be 2
severe island
anyway the number of choices for the second vertex depends on the relative place...
wait i’m confused, did you mean the third vertex?
severe island
oh okay okay, so basically it’s not deterministic?
stoic saddle
could say that
lean imp
marsh citrusBOT
lean imp
so my question here is the last step
b\ (a n b)
that is equal to b n (a n b) ^c correct?
proper zodiac
yes
lean imp
okay, so my question is can i seperate this when theres an intersection?
to get the mu(b) ?
lean imp
proper zodiac
hm weird? B and (A n B)^c arent disjoint
lean imp
well they would be if you think about it from a venn diagram standpoint
ah no shit so its just essentialy then the measure of b
proper zodiac
maybe im being dumb, but (A n B)^c would contain some of B
proper zodiac
ah no its a different result they're using
if X is a subset of Y then mu(Y\X) = mu(Y) - mu(X)
mu(Y) finite or whatever
you can prove that using just the definition
lean imp
so its mu(b) - mu(a n b)
ah no shit then the intersections cancel and your left with that term
proper zodiac
nimble parcel
If I wanted to prove that sin(x) was a total function from R to R then I would say that f⊆ R X R where f := { (x,y) ∣ y = sin(x) , x ∊ R } and then it would satisfy the 2 conditions needed for it to be a total function. My problem is the fact that I use the function sin(x) in my definition for the function sin(x) which kind of seems like circular logic is there anyway to avoid this or am I just misunderstanding something and this isn't circular logic or is there something wrong with the proof. Thanks.
marsh citrusBOT
If I wanted to prove that sin(x) was a total function from R to R then I would s...
proper zodiac
no its fine what you're really describing there is the graph G of the function f(x) = sin(x)
they are separate things
zealous tinsel
marsh citrusBOT
zealous tinsel
i was thinking the line would help but it just gives it as wrong
so i did 6 and 7 kg
got it as wrong
what do i do
i am stuck why this is not correct
<@&286206848099549185>
final skiff
so i did 6 and 7 kg
Howd u get 6 and 7 using the line
I think you might have read off the wrong number
zealous tinsel
just what number which was closest to the top of the line
please
tell me how you would do it
final skiff
Find the number (45) on the bottom axis, go straight up, once you hit the line, go straight left and see what value u land at
final skiff
Yeah its 5
zealous tinsel
does that mean the second answer would be 7 still
final skiff
yeah
marsh citrusBOT
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chilly patrol
Hello
chilly patrol
I need help with a question
It’s about trigonometric equations and plotting angles that fit the criteria
So I need to use the inverse of tan which I’m aware of to find the angle but I don’t know how to begin as there are two tans and they’re equal to 0 so what do I sub in?
Here’s the tan graph
<@&286206848099549185>
Where do I even begin with a question like this
marsh citrusBOT
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@chilly patrol Has your question been resolved?
marsh citrusBOT
@chilly patrol Has your question been resolved?
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narrow breach
marsh citrusBOT
narrow breach
honestly dont know where to start
narrow breach
use polar form
in what way exactly...?
zenith python
try to express it in the form re^it
narrow breach
okay
zenith python
try to express it in the form re^it
(3+i)/(5+i) i mean
exponents are easy to deal with in polar/exponential form
marsh citrusBOT
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narrow breach
ngl im reading this but i still dont rlly know what to do
marsh citrusBOT
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still temple
can anyone help me or nah
marsh citrusBOT
can anyone help me or nah
still temple
hnelllo anyhone
trim quest
still temple
someone hlep please
marsh citrusBOT
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jagged chasm
marsh citrusBOT
jagged chasm
I need help on number 89 I can't seem to find a function that would make the function in 89 divergent.
restive sun
try replacing sin(x) with one of its extreme values
jagged chasm
try replacing sin(x) with one of its extreme values
so like 1?
restive sun
yeah
jagged chasm
so i can split the integral by bounds from 0 to infinity + the integral from 1 to infinity
but how would that prove the function to be divergent
restive sun
1/sqrt(2+sinx) is greater than 1/sqrt(3) for all x
because we've replaced sin(x) by its maximal value
so the integral of 1/sqrt(2+sinx) is greater than the integral of 1/sqrt(3)
but the latter is the integral of a constant over an infinite region, so it diverges
jagged chasm
oh really, a integral of a constat over a infinite region is always divergent?
thanks for the suggestion, but is there any way to utilize a p-integral or e^-kx
restive sun
think of it as an area
a constant always bounds an area underneath it
so if you integrate it to infinity, you have infinite area
jagged chasm
oh i see
so by direct comparison, because 1/sqrt(3) is divergent and is less than 1/sqrt(sinx+2), therefore 1/sqrt(sinx+2) is divergent
restive sun
not sure how to do it with those other techniques though
restive sun
so by direct comparison, because 1/sqrt(3) is divergent and is less than 1/sqrt(...
yeah
jagged chasm
ok i understand now thank you very much
however is it possible to use a p-integral?
or even the limit test
restive sun
the problem is 1/sqrt(2+sinx) is almost constant since it just wiggles up and down as sin oscillates
it'll be hard to compare it with a p-integral
marsh citrusBOT
restive sun
you too, take care
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sweet valley
How does having 3 equations and 3 unknows not give a unique solution?
marsh citrusBOT
@sweet valley Has your question been resolved?
sweet valley
<@&286206848099549185>
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queen echo
Can someone help with this?
queen echo
<@&286206848099549185>
marsh citrusBOT
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wind flame
Good afternoon, needing some help with a percentage break down for my module grading system- need 40% average to pass the module, 1st assignment graded 72/100 with a 30% weighting, what would I need on the other assignment to get a passing mark?
marsh citrusBOT
Good afternoon, needing some help with a percentage break down for my module gra...
wind flame
Also if you could explain the maths quickly behind it that would be a big help so I don’t have to bug you lot everytime I need to work it out hahhaha
<@&286206848099549185>
marsh citrusBOT
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winter notch
Please tell me how to solve this type of Function equation. I understand i need to put the 9 where the x is but i dont really know how to do it exactly because theres a fraction there. Please Explain.
marsh citrusBOT
Please tell me how to solve this type of Function equation. I understand i need ...
@winter notch Has your question been resolved?
blazing vortex
Please tell me how to solve this type of Function equation. I understand i need ...
substitude 9 for x. everywhere in f(x) that you see an x, write a 9
what will you get?
winter notch
it will have 1/3 9 + 4?
winter notch
no i dont know
blazing vortex
no worries. so it sounds like you've forgotten how to multiply fractions, but that's okay. we can work through it
winter notch
Im suppose to multiply the 1/3 by the 9 before adding 4?
blazing vortex
yes. you may have been taught "PEMDAS" or "GEMDAS" at some point
do you remember what those things mean?
winter notch
nope ive never been taught that as yet
ive been thought something else i think its call BODMAS
blazing vortex
they both essentially mean the same thing, but with different wording. i prefer using the "GEMDAS" wording
winter notch
Brackets Of Division Multiplication Addition Subtraction
blazing vortex
that's almost the same thing,but GEMDAS also considers exponents
winter notch
hmm okay
blazing vortex
Groupings, Exponents, Multiplication, Division, Addition, Subtraction
winter notch
i see
blazing vortex
so with your problem, are there any operations within groupings (parenthesis, etc)?
winter notch
Its called "Functions" thats in the CSEC Syllabus.
winter notch
so with your problem, are there any operations within groupings (parenthesis, et...
yes there are
blazing vortex
i'm afraid i'm not familiar with your CSEC syllabus. however, while there is a number in parenthesis, note that i asked if there is an operation in the parenthesis
blazing vortex
nope
operations would be multiplication, division, addition, and subtraction
note that (9) has a number in the parenthesis, but no operator
winter notch
GCSE i meant
winter notch
note that (9) has a number in the parenthesis, but no operator
okay thanks
blazing vortex
so, next, is there an exponent in the problem?
winter notch
so, next, is there an exponent in the problem?
the answer would be 7
winter notch
i understand how to do it that way but theres another way i probably dont understand or im confused with
blazing vortex
i'm not sure what you mean
blazing vortex
did you write that?
winter notch
nope my teacher wrote it on the board and i took a pic of it before leaving the period
blazing vortex
i'm not quite sure what they mean by fg(-3). i think they mean f(g(-3))
winter notch
yes
blazing vortex
ok. remember how, for f(9), you just plugged in the number 9 everywhere there was an x?
winter notch
Yes
blazing vortex
so there's a couple of ways to understand problem b)
winter notch
Okay
blazing vortex
first, if you want to find f(g(x)), you do the same thing with g(x) as you did with 9 for f(9): you plug g(x) in everywhere you see an x in f(x)
winter notch
hmm okay can u show in an image like expressed above?
winter notch
okay i understand what u did there
blazing vortex
so if you want to find f(g(-3)), what do you think you need to do from here?
winter notch
i need to take the x below and cross out the x ontop, then i multiply it by the 1/3 outside that would give me 1 + 4 which is 5?
winter notch
i divided the 9 by 3 which gave me 3 to add to 4 to give me 7
blazing vortex
but first, you replaced all of the x's with 9's, yeah?
winter notch
yep
blazing vortex
so guess what you need to do with the -3 for f(g(-3))
winter notch
wait im confused sir, where did -3 come from
blazing vortex
blazing vortex
yep
winter notch
oh ok lemme work that out
blazing vortex
ping me when you did it, please. also helping others
winter notch
@blazing vortex sorry for the keep back, is the answer for the first one 1/-2?
blazing vortex
no worries. i did not get 1/-2
we can work through it together though
you got this far, yeah?
blazing vortex
we have to remember the order of operations again
what's the first thing we should do in this problem?
winter notch
what's the first thing we should do in this problem?
Im not sure sir
multiply the 1/3 by the stuff inside the brackets?
blazing vortex
remember GEMDAS?
G stands for Groupings. this means we want to solve everything in parenthesis first
winter notch
3/1?
blazing vortex
but we know that's not correct, yeah? because if we do it the long way:
winter notch
Ohhhh were so suppose to multiply the stuff with the stuff in the brackets
blazing vortex
the reason you can't cancel it out is because of that addition in the denominator. you can only cancel out like that if the entire thing is in both the numerator and denominator
winter notch
I understand
blazing vortex
so now you know that the part in parenthesis is supposed to be (-9)/(-2), yeah?
we can simplify that. do you remember how?
(now we can "cancel something out")
winter notch
since both is a minus number we can cancel out the 2 and the 9?
winter notch
Well i never learnt that actually that it can be like that. I always just thought of just -9 like a negative number. Never knew it could have a -1 there
blazing vortex
you can also remember, what is a negative number times a negative number?
winter notch
a positive number
blazing vortex
yep. and a negative number divided by a negative number?
winter notch
A positive number
blazing vortex
yep. so (-9)/(-2) = 9/2
winter notch
okay
winter notch
take the 1/3 multiply by the 9/2
blazing vortex
yep
winter notch
and we get 3/2?
winter notch
and we add 4 to the 3 which gives us 7/2 then we cancel out the 2 and the 7 and we end up with 3.5?
blazing vortex
i'm afraid not. it sounds like you may have forgotten how to add fractions
winter notch
oh
blazing vortex
remember, if you add or subtract fractions, you have to create a common denominator
winter notch
oh i remember
blazing vortex
a way to visualize that is, if you have half a pizza, and i give you two pizzas, do you now have 3/2 of a pizza?
winter notch
nope
blazing vortex
k, lemme know if when you solved it or if you need help
winter notch
k, lemme know if when you solved it or if you need help
is the answer 5/1/2?
blazing vortex
so we want to get a common denominator from 3/2 and 4, yeah?
what is the current denominator for 4?
winter notch
so we want to get a common denominator from 3/2 and 4, yeah?
yep the common denominator is 2
which makes the 4 into an 8
blazing vortex
nice
winter notch
and then add the 3 which makes 11
blazing vortex
ahhh, i see. i didn't understand that's what 5/1/2 meant
winter notch
thank you for helping me
winter notch
im confident in doing the rest by my self
winter notch
before you go, real quick
Yes sir
blazing vortex
it looks like the things to practice are:
-> substitution of values in for x (you seem to be getting much better at this now)
-> order of operations
-> knowing when you can or can't cancel in a fraction
-> adding/subtracting fractions
much respect for the effort you put in! not everyone has the determination that you do! i hope you have great success in the future 🙂
winter notch
Thanks sir
winter notch
it looks like the things to practice are:
-> substitution of values in for x (yo...
Noted
hardy trail
doing a practice calc test for grade 12, wanted to make sure this was my correct solution
marsh citrusBOT
doing a practice calc test for grade 12, wanted to make sure this was my correct...
hardy trail
<@&286206848099549185>
blazing vortex
doing a practice calc test for grade 12, wanted to make sure this was my correct...
i'm sorry, is that y = x * e^x or y = x - e^x?
hardy trail
y= x-e^-x
blazing vortex
when finding -e^(-ln(2))
it looks like you set that to be -2?
remember that -ln means that you're dividing
ah, i see, i didn't see the 2nd negative sign, cancelling it out
marsh citrusBOT
@hardy trail Has your question been resolved?
blazing vortex
<@&286206848099549185>
yep, i got the same thing
white brook
Super quick proof question:
If n<5, then n<10.
marsh citrusBOT
Super quick proof question:
If n<5, then n<10.
white brook
I was trying to see the best way to prove this, but it has been a while since last semester.
white brook
I don't have to do anything fancy? Just state the obvious?
proud ice
Assuming you've already established the total ordering of numbers
proud ice
5<10 should be a given, yea
tender sparrow
doubt
tender sparrow
DC is an Identity matrix here but i dont get how that makes B^n simplify
to CA^nD
sand fable
all of the DC in the middle cancel into the identity matrix
tender sparrow
can we do that
sand fable
so you have a C on the left, a bunch of As in the middle, and a D on the right
tender sparrow
in matrix mulitplication
we can just take the middle elements?
we dont have to multiply CAD as a whole?
sand fable
can we do that
yes because matrix multiplication is associative
sand fable
awesome
tender sparrow
thanks
sand fable
boreal rose
hi
boreal rose
I'm selecting n+2 integers from {1...3n} and n>=2 \in N
and like I have to prove there will always be an integer diff that is greater than n and less than 2n
I made the pigeon holes remainders when dividing by n+1
so there will be one hole with duplicate remainder
and sps they are integers with a,b
a = c(n+1) + b where c is 1 <= c <=2
now c = 1 automatically proves it
now for c = 2
we have a = 2(n+1) + b
and idk where to go from here
<@&286206848099549185>
raw hare
<@&286206848099549185>
hey cute penguin
boreal rose
hey cute penguin
what's up
boreal rose
Is this arithmetic progression 💀
nope 😭
still crow
if n = 3, then you have the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. You can select {1, 2, 3, 8, 9}
idk if i'm missing something
still crow
yeah no, this is a counterexample
boreal rose
if n = 3, then you have the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. You can select {1, ...
9 and 8?
wait
sorry
I mean
still crow
"greater than n and less than 2n"
boreal rose
wait
still crow
so idk if you've misrepresented the problem
boreal rose
"greater than n and less than 2n"
"Let n ≥ 2 be an integer. Suppose that we have selected n + 2 different integers from the set
{1, 2, 3, . . . , 3n}. Prove that there will always be two among the chosen numbers whose differ-
ence is more than n but less than 2n"
this is the problem in full
also 8 - 3 = 5
and 3 < 5 < 6
boreal rose
oh that's a different problem then
did I misrepresent
shoot my bad
oh nvm
my bad
still crow
yeah I read that as "you can always choose a number that is between n and 2n"
boreal rose
yeah I read that as "you can always choose a number that is between n and 2n"
mhm do you have any ideas
still crow
look at the remainders after division by n + 1 within the (n + 2) integers
boreal rose
look at the remainders after division by n + 1 within the (n + 2) integers
mhm that's what I did
for k = 1
with n+1
that's p easy
but with k = 2
with n+1
still crow
firstly, how many possible remainders can you have
boreal rose
you get a = b + 2(n+1)
boreal rose
firstly, how many possible remainders can you have
you can have n+1 remainders and you are choosing n+2 numbers
still crow
yup
still crow
ok there's a bit more work involved
still crow
but the rest isn't too bad
boreal rose
c = 1 is QED
boreal rose
but the rest isn't too bad
mhm
opengangs this is my first pset for combinatorics and graph theory
I can't mess this up
still crow
ok so let's take a bit at a time
boreal rose
ok so let's take a bit at a time
mhm
still crow
so two integers must share the same remainder among division by n + 1
do you understand up to here?
still crow
yeah so you're on the right track here
still crow
so with c = 2, you'd want to consider the remaining n possible integers that you could have chosen
boreal rose
so with c = 2, you'd want to consider the remaining n possible integers that you...
hmm
we could have chosen something outside of the boundaries
liek b-1 numbers that are less than b (b numbers including b)
or something greater than a ig
still crow
if I take some other integer x, I want to look at |a - x| and |a - y|
boreal rose
that would be like 3n - (b + 2n+2 + 1) + 1
still crow
and maybe get some new range of values such that these differences do not lie between n and 2n
still crow
yup
boreal rose
n - 1
still crow
yup
boreal rose
WAIT
I done messed it up
we have n + 2 - (n-1) = 3 numbers left
and THESE NUMBERS MUST BE between a and b
still crow
would it be b + 3n - (b + 2n+2) + 1 =
wait this should be b + 3 + 3n - (b + 2n + 2) + 1
jks
we can take any of the terms below b + 3, or any between b + (2n - 3) + 1 and 3n
these will have difference outside of the range
boreal rose
these will have difference outside of the range
mhm
now we only have 3 terms 😦
what are we going to do with them
still crow
so this gives you n + 4 terms whose difference is outside of the range
but
we can't take all of them
boreal rose
so this gives you n + 4 terms whose difference is outside of the range
mhm
wait opengangs
we have 3 terms left right
boreal rose
where did the 3 come from
b + 3n - (b + 2n+2) + 1 = n-1
so I'm filling up the gaps
and now
everything has to between a and b
still crow
oh duh yep
still crow
ok i think i have the idea
there are only b terms <= b and so, we have (3n - (2n + 2 + b) + 1) = n - b - 1 terms that are above or equal to a
so in fact, there must be 3 terms between b and a
still crow
look at the smallest distance between these three terms and the two terms: a, b
smallest distance between any pair of these terms is (2n + 2) - 3 = 2n - 1
boreal rose
smallest distance between any pair of these terms is `(2n + 2) - 3 = 2n - 1`
wait explain this
everything else was good
still crow
think of these integers on a number line
still crow
We have that [b, b + 1, b + 2, b + 3, 2n + 2 + b]. The smallest possible distance would occur between any pair of adjacent terms
boreal rose
We have that `[b, b + 1, b + 2, b + 3, 2n + 2 + b]`. The smallest possible dista...
mhm ofc
still crow
so it must be at most (2n + 2 + b) - (b + 3) = 2n + 2 - 3 = 2n - 1
it could very well be smaller but it must at least be bounded above by the difference between b + 3 and 2n + 2 + b
boreal rose
We have that `[b, b + 1, b + 2, b + 3, 2n + 2 + b]`. The smallest possible dista...
smallest distance between what
still crow
smallest distance between any pair of elements in the interval
boreal rose
smallest distance between any pair of elements in the interval
mhm
yeah ofc
and the longest would be
(2n+2 + b) - (b+3)
cuz we have 3 terms (not including b and 2n + 2 + b)
still crow
yeah all of these distances is bounded by whatever the longest is
boreal rose
which is 2n-1
so we have the longest distance covered
but what do we have to do for > n
you might have to cook again ngl
still crow
not all of the distances between pairs of elements can be < n
i.e. there would have to exist some distance >= n
boreal rose
i.e. there would have to exist some distance >= n
why tho
oh wait nvm
I think I get it
but a formal line of explanation would be nice
still crow
pretty sure you can argue it by contradiction
like if all of the distances were < n, then (2n + 2 + b) - b < n which would imply that 2n + 2 < n which is impossible
so you can't have that all of the distances are < n
boreal rose
pretty sure you can argue it by contradiction
btw this argument is like at least 3
there can be more than 3
still crow
more than 3 what
boreal rose
more than 3 what
3 elements between a and b ofc
still crow
well, we showed there were 3 elements between b and a, no?
there are b terms that are <= b and (n - b - 1) terms >= a
boreal rose
well, we showed there were 3 elements between b and a, no?
at least 3
I feel like
yes?
nvm
I'm tripping
still crow
but anyways, if you believe that not all such distances are < n, then we are pretty much there
boreal rose
but anyways, if you believe that not all such distances are < n, then we are pre...
less than or equal to n
still crow
you can consider the intervals [b, b + n + 1] or [b + n + 1, a]
boreal rose
like if all of the distances were < n, then (2n + 2 + b) - b < n which would imp...
I think this is a better argument
@still crow lowkey this problem was a doozy
as a first time combinatorics question
not my fav
still crow
yea this is not an easy problem for someone starting out lol
i would prefer choosing n + 1 elements and showing there always exist two integers whose diff is divisible by n
but that's a bit easy
boreal rose
yea this is not an easy problem for someone starting out lol
mhm it's just umich math
they make it difficult for no reason
you seem like a smart person tho
one day I wanna do computational theory math
still crow
it's a fun problem but definitely does require a bit of bashing your head against the whiteboard
as with pretty much combinatorics in general
boreal rose
it's a fun problem but definitely does require a bit of bashing your head agains...
oh yeah
we are doing something tmr called ramses thereom
sounds egyptian fr
boreal rose
ramsey's theorem?
oh yeah this
boreal rose
good luck 🫡
may I ask you for help from here on
still crow
well, i might not be active but i do pop in every once in a while
boreal rose
well, i might not be active but i do pop in every once in a while
what is ur career goal
still crow
if you want to continue the convo, feel free to dm me
don't want to clog up this help channel
boreal rose
.close
agile kiln
Last question i have for tonight, im not really sure where i went wrong after 10 minutes of staring at my work so any help would be appreciated
