help-13

[celest seal]

that means that P must be false

that's the only option that allows P → FALSE to be true

(again, you can google "modus tollens" for more, that's the argument form i'm using here)

(there is a lot of philosophical writing on modus tollens — some philosophers DO reject it)

(but in an introductory logic course, they typically don't get into the philosophical weeds and just give you the simplest version lmao)

[glass moth]

Isn't math just branched out logic from a set of axioms anyway?

[celest seal]

well, thats a different discussion lmao

some philosophers would agree with you, but i think most would disagree nowadays

that perspective is known as "formalism" and it isn't totally dead

but a lot of philosophers and mathematicians think mathematics tries to capture something more "fundamental" to "the structure of reality" than just the consequences of axioms

i.e. our logical rules are made up by humans and kind of arbitrary. like, they seem to WORK, but we never had a divine entity come down to Earth and explain Modus Ponens and Modus Tollens to us

yet despite the rules being kind of... made up, we end up with a system that seems to correspond to real things in the real world

in other words, the logical rules and axioms of mathematics are just the abstractions humans come up with to try to capture what mathematics "really" is

the answer as to what mathematics "really" is is a more controversial one

sine philosophers believe that mathematical "forms" "exists" in some metaphysical sense

like 1, 2, 3 arent just random symbols, they are concepts that are somehow manifest in the makeup of reality

(this is known broadly as "mathematical platonism")

others argue that "mathematics" is just a social construct, and our classification of things into "mathematics" vs "not mathematics" is just an arbitrary one made by society and educational institutions

this doesn't mean it's a lie, just that its boundaries are determined by its societal circumstances

[glass moth]

thats related to materialism vs idealism

[celest seal]

these standpoints arent necessarily incompatible

[celest seal]

correct, yes

you can see that this is a whole rabbit whole you can go down

probably beyond the scope of this server though, i am not an expert on this either

https://plato.stanford.edu/entries/formalism-mathematics/ and https://plato.stanford.edu/entries/platonism-mathematics/ are probably good starting points if youre interested in reading more

formalism vs platonism was the dominant "split" of 20th century mathematical thought

(culminating in Russell-Whitehead's Principia Mathematica as an attempt at a comprehensive formalistic system for all of mathematics, and then Goedel's incompleteness theorems showing that this goal was an unachievable one)

nowadays most mathematicians have mellowed out on the issue

[glass moth]

Actually I still don't understand Goedel

[celest seal]

but it's still actively discussed in philosophy departments

goedel's first incompleteness theorem is basically saying, given a "sufficiently powerful and useful" axiom system for mathematical reasoning, there will be statements that are consistent with those axioms but are not proven by the axioms

"sufficiently powerful and useful" is doing a lot of heavy lifting here

i'm using it as shorthand for "consistent, first-order, recursively enumerable, and capable of expressing natural-number arithmetic with induction"

all of these assumptions are important — if you don't have any of them, then the statement is false

which is kind of why it's so tricky to wrap one's head around

by "consistent with those axioms but are not proven by the axioms", i mean that we could add that statement to the axioms and not "break" anything

we wouldn't introduce any contradictions

but that the axioms themselves can't prove it

[glass moth]

I understand its saying that some statements in math are true but are unproveable

[celest seal]

(so we could also add the negation of that statement)

i dont really like that wording since i think its kinda misleading

like, if you have an axiomatic system A and some consistent-but-unproveable statement S, we could just add the statement S to the axioms of A

and then the new system (which we write A+S) would be capable of "proving" S

its just that this seems kinda... arbitrary

and the new system would still have examples of consistent-but-unproveable statements

[glass moth]

Yeah whenever mathematicians can't prove a statement they just add it as an axiom

[celest seal]

i think it's a much easier statement to understand if you've seen any abstract algebra

like group theory

without that background i think it's easy to fall into a lot of "pitfalls" about what it means

[glass moth]

So what was Goedels basically saying in summary

[celest seal]

do you want the philosophical "point" or the mathematical meaning?

[celest seal]

i wrote the mathematical meaning here, but i recognize that it's hard to follow if you're not familiar with the concepts (unfortunately i am not confident enough in myself as an educator to explain it more simply)

philosophically, the context of goedel's incompleteness theorem was that a bunch of formalist mathematicians wanted to "realize Hilbert's program" of setting up a system of rules (first-order axioms) that could express "all of mathematics"

"all of mathematics" is sort of like, intentionally vague here, since people didnt really know how to define that

like, they knew mathematics needed to be able to do arithmetic and algebra and geometry and analysis and stuff

statistics, etc

there was a rough communal understanding of what "all of mathematics" was

but no concrete definition

and they wanted to write down an unambiguous definition using logical rules and axioms

Russell and Whitehead even went as far as to publish an entire multi-volume book on this called Principia Mathematica

but then Goedel's incompleteness theorem rolled around and showed:

1. these attempts do not model all of mathematics — there are statements that are mathematically independent of the axioms (i.e. can't be proven or disproven by them)
2. it is impossible to have a "useful" axiomatic system that models all of mathematics

point #2 is particularly poignant since it's basically saying "even if you tried to fix Principia Mathematica by adding more axioms to cover more statements, new statements that it can't prove would pop up"

basically, it showed that the original goal of the Hilbert program was impossible

this didn't kill mathematical formalism as a philosophy, but it forced the formalists to reconsider/recontextualize a lot of things

[glass moth]

I see

[cedar kiln]

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[frigid stump]

how do i do this, more specifically, what are the bounds?

[frigid stump]

i have the r bounds, just not sure what i should be using for theta, ive tried 0 to pi/4 and pi/4 to -3pi/4

[cedar kiln]

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[frigid stump]

<@&286206848099549185>

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[naive vigil]

hello

[naive vigil]

can someone help me fix this ______

[solid juniper]

honestly

wtf is that

[naive vigil]

fill the blank, to get bottom value

but the _ if we + to bottom, should get 100

[solid juniper]

quite an odd way to format asking that but ok

where is this from?

[naive vigil]

my friend gimme, and i can't fix it

just a little chalange

i should fill all the blank, and get the bottom value, and right value at right side

[solid juniper]

it might have more than one answer... let me see

[dusk goblet]

it should be more than one answer

[solid juniper]

yea it's like

6 unknowns and 5 equations

[dusk goblet]

is there some criteria for the numbers

[naive vigil]

if the pic upside not good, i draw better one

[naive vigil]

no rule, but you should done it with the right value at right and bottom

[solid juniper]

there are infinitely many solutions

[dusk goblet]

yea

because u can just say -1/2x+4y+6z=0 right

and there’s infinitely many different combinations

[naive vigil]

yes i only need 1 of it

[solid juniper]

formatting the blanks as
a b
c d
e f
you can choose any value of a and set the rest equal to these for a solution

[naive vigil]

can you fill it ?

[solid juniper]

this question is too silly for me to continue

[naive vigil]

haha sorry, i not really good at math and just try get some help here

😅

[cedar kiln]

@naive vigil Has your question been resolved?

[tropic oxide]

@naive vigil say, where did this problem come from...?

[tight sun]

If I were the one writing this question, I'd say that an additional paramater is that we want these to be positive integers in some "reasonable" range

[naive vigil]

so you can fix it if i said it should positive integers?

[naive vigil]

my college friend gimme a challange

[tropic oxide]

can you show this "challenge" exactly as it was given to you

[naive vigil]

it said the blank answer can't more than 100 too

[tropic oxide]

!original

[cedar kiln]

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

[naive vigil]

the .... can't be negatif & the max value is 100
can't put decimal into it,

[solid juniper]

dang zorn you're so smart

[tropic oxide]

also

@naive vigil what grade are you in?

or at least tell us if you are over or under 13

[naive vigil]

first year architecture

[tropic oxide]

k

so then, the COMPLETE problem is this:

• you have those 3 multiplication problems
• the left blanks have to add up to 100
• the right blanks also have to add up to 100
• each blank holds an integer between 0 and 100 inclusive

did i get that right?

[naive vigil]

yeah

[naive vigil]

right right

[tropic oxide]

and nothing else is given, yes?

[naive vigil]

yes,

[tropic oxide]

ok

right so then

let's call the left blanks x, y and z

[naive vigil]

okay

[tropic oxide]

we have two equations:

x + y + z = 100
x/2 + 5y + 7z = 100

for which we want an integer solution with all unknowns between 0 and 100

[cedar kiln]

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[tropic oxide]

bruh

.reopen

[cedar kiln]

✅

[naive vigil]

after that?

[tropic oxide]

i'd probably note first and foremost that x must be even

do you understand why

[naive vigil]

dunno, can you tell me?

[tropic oxide]

do you know what an even number is?

[naive vigil]

2, 4 , 6 , 8
like that right?

[vast pike]

0

[tropic oxide]

those are the first few positive even numbers yes

but can you tell me the definition

[naive vigil]

the number that can be 0 after divided by 2 or by it self?

is that true?

[tropic oxide]

half-right...

[naive vigil]

then i dont know what's more 😄

[tropic oxide]

you overcomplicated it actually

an even number is a number that is divisible by 2.

that's it

[naive vigil]

ahhh ok

[tropic oxide]

here you have x/2 as the right blank in the first line

x/2 must be an integer

therefore x is even

we might replace x with 2k for simplicity (ie to get rid of fractions)

2k + y + z = 100
k + 5y + 7z = 100

next i would observe that from the second equation there's not a lot of things that z can be

[naive vigil]

after that?

[tropic oxide]

the fact you keep asking "after that?" as if you expect me to just hand you the entire solution is kinda offputting ngl

would really prefer if you at least tried to think for yourself here

can you tell me all the numbers that z could be, based on the fact that 7z lies between 0 and 100?

[naive vigil]

i think 7z is the even number that can make k to be even too?

[tropic oxide]

no,

you don't know that 7z is even too.

in fact this has nothing to do with the previous argument about x...

[naive vigil]

is that around number 2 until 8?

i found the value already
that k = 46
y = 1
z = 7?

[tropic oxide]

,calc 46+5 * 1+7 * 7

[wraith dagger]

Result:

100

[tropic oxide]

checks out...

yeah

[cedar kiln]

@naive vigil Has your question been resolved?

[naive vigil]

yes, thank you to @tropic oxide for much help

[cedar kiln]

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[oblique pilot]

I had a question on how to solve question 5 in this textbook because I’m always getting 111ish degrees while the textbook says it’s around 90 degrees more

[cedar kiln]

@oblique pilot Has your question been resolved?

[oblique pilot]

<@&286206848099549185>

[cedar kiln]

@oblique pilot Has your question been resolved?

[pastel dew]

I understand why you got 111, you are one step away but missing a piece.

1. The hypotenuse of the triangle is around 8.54km and you have this correct.

2. To find the angle in the triangle, you can use the fact that tangent of that angle will be equal to the ratio between the opposite and adjacent side of the triangle. From this you find angle to be around 20.56 degrees.

Let me know if this step was unclear, I don't really know how to use mathematical notation on discord, but I could write on a piece of paper.

3. Because taxi travels south first, you instantly add 180 degrees (bearing is measured using a "clock method", so to go south you have to turn 180 degrees). Then you add the angle of 20.56 degrees which represents the angle to the west, for a total of 200.56 degrees.

[oblique pilot]

so from south to west it is a addition of 20.56 degrees?

[pastel dew]

Correct

[oblique pilot]

So in essence the bearing is just looking for this?

Wait this makes so much sense now

The from A in the question is saying it’s the difference from true North on a to the point of c clockwise

[pastel dew]

That's correct :)

[oblique pilot]

Thank you so much I feel enlightened Lmaoo

.close

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[blissful saddle]

Why did he write b > M?

[blissful saddle]

And why is it strict?

What difference would there be in the proof if b > M wasn't stated?

<@&286206848099549185>

[cedar kiln]

@blissful saddle Has your question been resolved?

[blissful saddle]

<@&286206848099549185>

[cedar kiln]

@blissful saddle Has your question been resolved?

[blissful saddle]

<@&286206848099549185>

Hello?

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[sand cradle]

Can we find $x$ in $(x^2 + x)e^x = 5$ using the Lambert W function? It seems WA just gives an approximation, without using W.

[sand cradle]

.close

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[ornate girder]

blue was the correct solution but i got 2/5x

why do you have to add +1 when dividing 2 by 5x ?

[kindred storm]

With fractions, you can only cancel factors of the top and bottom, not terms.

So, (2 - y)/y isn't 2.

2 - y and y have no common factors, so you can't cancel any common factors.

You can do 2/y - y/y if you want.

[ornate girder]

so (2-y)/y

huh idk it just ask to get inverse of (2-x)/(5x)

wait chai can you draw it

[wraith dagger]

Chai T. Rex

[kindred storm]

Like that?

Oh, wait. You also shouldn't have a denominator there.

[ornate girder]

so i then i multiply 5yx with y?

[wraith dagger]

Chai T. Rex

[ornate girder]

yep but then i have to group y with 5xy

so i cant just add y's to both sides ? and cancel out 2-y right

[kindred storm]

You can add y to both sides.

I'm not sure what you mean by cancelling 2 - y.

[ornate girder]

you said this earlier i tohught thats what you meant by it

[kindred storm]

Oh, I misread what you did.

[wraith dagger]

Chai T. Rex

[kindred storm]

I think that's where the 1 disappeared.

Do you see why it would become y(5x + 1)?

[ornate girder]

no

thats what i was wondering

[wraith dagger]

Chai T. Rex

[kindred storm]

You divide all the terms by y and then move that y outside.

And you can check that it's correct by distributing the y.

Factoring divides the terms by the factor and puts the factor outside.

It's like the opposite of distributing.

[wraith dagger]

Chai T. Rex

[ornate girder]

ohh ok i see

[kindred storm]

So, if you continue with that, you'll have the + 1 they want.

[ornate girder]

ok thank you

[kindred storm]

You're welcome.

Just remember that y/y or anything over itself is 1.

[ornate girder]

.close

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[native lagoon]

Can do the Q, but is calculus always in radians?

[crimson sedge]

For angle measure, we do need to use radians. This is because the limits calculated for derivatives must be in radians to work as they do

If you consider a unit circle then the arc length is x radians

[mortal hemlock]

,rotate

[mortal hemlock]

radians is almost always superior to degrees

[upper abyss]

No need for any confusion in this case though:
The derivatives and integral formulas for trig functions are only correct in radians. They're different if you're working in degrees

[cedar kiln]

@native lagoon Has your question been resolved?

[native lagoon]

Interesting enough the answer was 4pi and I got 720, IF the values were angles they’d be equal, was interesting

Thank you both @mortal hemlock @upper abyss for your help!

[cedar kiln]

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[half jungle]

Can someone answer question 1 for me I’ve already solved it but I feel like I’ve done something wrong I don’t get the concept so idk I tried

[half jungle]

@ me when someone answers

[tulip stream]

first screenshot isnt clear, cant see value of f(x) at point x=-1

instead of taking screenshot from another device, take ss from that device directly

[half jungle]

It doesn’t let me

Here

Question 1 and 2

[tulip stream]

still cant tell if f(-1) = -1 or f(-1) = 0

but seems like f(-1) = -1

[half jungle]

Yeah it’s -1

[tulip stream]

since $$ f(-1) = -1 = \lim_{x \to -1^{-} } f(x)$$
we say f is left continious at x = -1

[wraith dagger]

Cyrenux

[tulip stream]

$x \to -1^{-}$ means we are approaching point x from left

[wraith dagger]

Cyrenux

[tulip stream]

and $x \to -1^{+}$ means we are approaching x from right side

[wraith dagger]

Cyrenux

[tulip stream]

but it seems like you grasped that already

we say f is continious at x = -1
IF limit exists on point x= -1 and equals to value of f(-1)

[half jungle]

Since the function isn’t right continuous at x=-1 then it jumps?

[tulip stream]

reminder that for a limit to exist at a point, left limit of that point and right limit of that point must be equal to each other

[tulip stream]

$$ f(-1) = -1 \neq 1 = \lim_{x \to -1^{+}} f(x)$$ so yeah

[wraith dagger]

Cyrenux

[tulip stream]

its NOT right continious

[half jungle]

Yeah

But the intervals part 🥲

[tulip stream]

which interval did you struggle on

for it to be continious on a interval, it must be continious on every point of that interval

oh also it seems like you confused part d with part e while solving @half jungle

[half jungle]

Yeah

[half jungle]

I mean like

I don’t really understand how to figure out when it’s left or right continuous

[tulip stream]

hmm, just to make sure, do you know definition of a limit

[half jungle]

if u mean how to solve by the definition of a limit yeah

But limit itself

No

Like in an understanding way idk no

[tulip stream]

definition of limit itself

[half jungle]

nope

[tulip stream]

hmm, do you know what left or right limit is?

[half jungle]

no my teacher couldn’t explain the material clearly because we had to rush everything

So I just know the rules

[tulip stream]

few seconds, almost done

lets take a look at examples of this graph

lets take a look at graph 1 , f(x)

notice how this isnt a piecewise function

and f(a) is defined, and has a value

[half jungle]

Yes

[tulip stream]

now take a look at example 2
here g(a) is not defined, we draw a ball on part of the curve where its and we leave inside of this ball empty

although its not defined at that point of curve, it can still be defined at another point, look at example 2.5, see the red point above the point x = a

so for example 2 we have g(a) not defined
for example 2.5 we have g(a) defined and g(x) = c (for some real positive value c)
(c is positive because on the graph i gave its defined above the x axis)

also notice the orange and light blue coloring shade on the curve g(x) ,as x approaches point a

light blue represents approachment from the left
while orange represents approachment from the right
to point a

here they exist and are equal since they are approaching the same value as you can see

on both example 2 and 2.5

on example 3, right limit DNE ( Does Not Exist)
and on example 4, left limit DNE ( Does Not Exist)

when left limit = right limit at point x=a , we say limit of f(x) exists at point a

$$ \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) $$
$\implies \lim_{x \to a} f(x)$ exists and
$$ \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = \lim_{x \to a} f(x)$$

[wraith dagger]

Cyrenux

[tulip stream]

existence of left and right limit doesnt straight up mean that limit of that point exists though, a limit exists only if left and right limit are equal:

look at point x=0,
left limit is 2 while
right limit is 1
since they are not equal
\ $\lim_{x \to 0} f(x)=$ Does Not Exist

[wraith dagger]

Cyrenux

[cedar kiln]

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[cedar kiln]

Please don't occupy multiple help channels.

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[untold moat]

Prove that at least one number from the set of numbers $\sqrt{2}, \sqrt{3}, \ldots, \sqrt{1000}$ is irrational.

[wraith dagger]

π = ∞

[zenith nacelle]

huh √2 is irrational

[hollow minnow]

sqrt(2) is irrational

[zenith nacelle]

well known

[hollow minnow]

!xy

[cedar kiln]

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

[zenith nacelle]

in fact all are irrational except the perfect squares

[vast pike]

sqrt (prime) is irrational

$\sqrt{n}\in \mathbb Q \iff n=k^2,$ for some $k\in\mathbb N$

[wraith dagger]

everg

[heavy robin]

I think the issue is he has to prove it not know it lol

[zenith nacelle]

just prove √2 is irrational

[untold moat]

Bros This is the IMO 1959 question

[vast pike]

is it the six problem?

very difficult

[untold moat]

yes

[crimson delta]

well its not from IMO. so where is it from

[zenith nacelle]

wait lemme post rudins proof plz

[hollow minnow]

[minor crystal]

[untold moat]

Solution goes like this:
Assume, for the sake of contradiction, that all numbers $\sqrt{2}, \sqrt{3}, \ldots, \sqrt{1000}$ are rational. Then, consider the square of each of these numbers:

2
,
3
,
…
,
1000.
2,3,…,1000.

Now, consider the prime factorization of each of these numbers. For each number, the prime factorization is unique. Since these numbers range from 2 to 1000, they must include the prime numbers up to 31 (since
3
2
2

1024
32
2
=1024 is greater than 1000).

Now, consider the prime factorization of 2. It is
2
1
2
1
. Similarly, the prime factorization of 3 is
3
1
3
1
, and so on. Since each of the numbers in the list has a prime factorization consisting of prime numbers raised to the power of 1, when we take the square root, the resulting numbers will have prime factorizations with the exponents being multiples of
1
/
2
1/2.

However, the square roots of some of the numbers (like
2
,
3
,
…
,
31
2
​
,
3
​
,…,
31
​
) will have prime factorizations with exponents not equal to an integer. This means that at least one of these square roots is irrational, which contradicts our assumption.

Therefore, our assumption that all numbers are rational is false, and at least one of the numbers
2
,
3
,
…
,
1000
2
​
,
3
​
,…,
1000
​
must be irrational.

This problem illustrates the technique of proof by contradiction and relies on the unique prime factorization property to reach the conclusion. Keep in mind that while the solution may seem straightforward, the key is to recognize the appropriate approach and apply it effectively.

[wraith dagger]

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

[hollow minnow]

[crimson sedge]

very beautiful

[vast pike]

[minor crystal]

[zenith nacelle]

bruh

[untold moat]

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[distant fern]

how is it 7

[crimson sedge]

do you know Pythagorean theorem

[dusk goblet]

30 60 90 right triangle

[crimson sedge]

or recognise it is that ^

[distant fern]

can i send my working out

[crimson sedge]

sure

[distant fern]

@crimson sedge

[crimson sedge]

7^2 is NOT 14

[lucid nebula]

you also have to subtract not add

[distant fern]

why?

the shorter side always subtract no?

@lucid nebula

[lucid nebula]

yes

a^2 + b^2 = c^2

you are trying to find a

so c^2 - b^2 = a^2

[distant fern]

omg

i was doing so much mistakes..

i even noted them down

[lucid nebula]

its okay we all learned first

[distant fern]

and btw i have a question

[lucid nebula]

when i was learning to parallel park i hit a parked car

[fallow raven]

lol

I have a pretty small car so I kinda have just always been able to parallel park

[distant fern]

i did the first question correct and the second completely wrong cause i wasn’t concentrating

but the question

[lucid nebula]

i have a 94 camry its the size of a boat. luckily i dont drive this year

[distant fern]

im scared that cause im studying 2 days before the test i’ll forget everything i studied..

[lucid nebula]

dont be scared

you will do very well i am sure

[distant fern]

ok

i’ll update you’

anyways thanks for your help 💓

god bless you

,,afk

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[lucid nebula]

👍🏻

[cedar kiln]

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[distant fern]

wait

[distant fern]

@lucid nebula

x is obviously the longest side?

so why subtract

[tropic oxide]

no it isn't.

[vast pike]

hey Ann😉

[distant fern]

i’ve heard that you add when the right angle is opposite of the longest side

[tropic oxide]

!redir

[cedar kiln]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[crimson sedge]

[
{\t{hypotenuse}}^2 = {\t{side 1}}^2 + {\t{side 2}}^2
]

[tropic oxide]

you heard some kind of strange shortcut lol

[distant fern]

blame mathswatch 🤷‍♀️

[tropic oxide]

if there is a right angle, then the side across from it is ALWAYS the longest

[wraith dagger]

Qylo

[crimson sedge]

this is the general statement of Pythagorean theorem

[distant fern]

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[wraith dagger]

marulk

[idle tusk]

$\log_2$

[wraith dagger]

artemetra

[idle tusk]

both sides

well

yeah

but don't learn it as a rule

actually try and understand

$\log_2(2^y) = y$

[wraith dagger]

artemetra

[idle tusk]

this is how a logarithm is defined

essentially

log(3) is, well, log(3)

[idle tusk]

look at this

[wraith dagger]

marulk

[idle tusk]

this is all a logarithm is

yeah log_5 undoes it

[wraith dagger]

marulk

[idle tusk]

yep

precisely

[ornate girder]

ohh okay

gotcha

[idle tusk]

there are also a couple of useful log rules

[ornate girder]

like restrictions ?

[idle tusk]

no, more like properties

the order is weird on the picture

[idle tusk]

but rule 6 here is what i talked about

in any case

for your problem you just log_2 both sides

[ornate girder]

ok yea got it

thank you

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[wraith walrus]

Might just be me, but this question is worded a bit confusing. Any help on understanding what the question is even asking?

[cerulean sail]

“Differentiate r^3 with respect to t, and somewhere you want that to be 12 * [what you get when you differentiate r wrt t]”

[wraith walrus]

oh ok that makes more sense

thanks :3

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[uneven hound]

i have found

d|a^p+b^p

[lucid nebula]

im thinking about it

[opal basin]

Have you learnt modular arithmetic?

also, is this from MONT?

@uneven hound

[uneven hound]

yes for both quests

[opal basin]

okay, suppose some prime q divides a + b

then a = -b mod q

and you can expand $\frac{a^p + b^p}{a + b}$

[wraith dagger]

Jelle

[opal basin]

and look at it mod q

(modular arithmetic isn't strictly necessary, but it makes it a bit simpler I think)

[uneven hound]

this is equal to a^p-1 + a^p-2b .....

[opal basin]

should be a negative sign, right?

a^p-1 - a^p-2b .....

[uneven hound]

yeh

[opal basin]

(actually suppose q divides the gcd)

then this is 0 mod q

[uneven hound]

mhmm i'll try again bcs i already got into this point before

thnx

Can u help more pls

@opal basin

[opal basin]

You can substitute a = -b mod q

[uneven hound]

how

[opal basin]

a^p-1 - a^p-2b ..... = 0 mod q, if we assume q | gcd, right?

then you can substitute a = -b mod q, becuase q | a + b

[uneven hound]

ok i got it thnx

[lucid nebula]

i just took number theory, where did you get this expansion?

[opal basin]

You can factor a^n + b^n, if n is odd

google sum of nth powers factorization

[lucid nebula]

👍🏻

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@uneven hound Has your question been resolved?

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@uneven hound Has your question been resolved?

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[burnt crag]

hey

[cedar kiln]

Please don't occupy multiple help channels.

[burnt crag]

i need help

with a question

[slate lintel]

use your other channel

.close

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[mortal hemlock]

Damn you have the powers to just close a channel

Thats rad 😎

[cerulean sail]

Even helpfuls can believe the helpers role lets you too

[mortal hemlock]

Really? Ill try it

[cedar kiln]

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[thick obsidian]

this is just a basic general question to see if anyone has an answer, are there any sort of problem solving books youd recommend, like "how to solve it" by g. polya?

[royal loom]

#book-recommendations

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[floral pumice]

the answer would be 2,16 no?

[subtle harbor]

a good indication to check if a point is on f^-1 given a point on f(x) is to flip the coordinates of a point on f(x)

[floral pumice]

ok but i cant really tell if its 2,8 or 2,16

[subtle harbor]

are A and B your only choices?

[warm marsh]

Actually the question is wrong because you cant really define an entire function with just 2 points

[floral pumice]

thats what im saying

[warm marsh]

anything could have happened at f(8) or f(16)

[subtle harbor]

are those your only two choices?

[floral pumice]

no

[subtle harbor]

that doesn't help

[floral pumice]

YO ITS ONE OR THE OTHER

2,8 or 2,16

[subtle harbor]

yea

[subtle harbor]

I think its a log function?

[warm marsh]

none of the options can be said to be correct or wrong, the function needs to be more well defined, and you just cant say that it is a log function unless it is mentioned to be a log

[subtle harbor]

yea

[floral pumice]

welcome to AP bud

[subtle harbor]

but if I had to guess it would be a log function

[floral pumice]

so 2,16?

[subtle harbor]

yea

[floral pumice]

betttt

another question rq

[warm marsh]

log base 4 to be specific will be a good guess and taking the inverse will give 4^x and 4^2 is 16 so yeah (2, 16) is a good guess

[subtle harbor]

wait is this for ap exam

[subtle harbor]

[floral pumice]

yessir

nah but like midterm practice

[subtle harbor]

[floral pumice]

this is the 2 options i have

[subtle harbor]

yea

[floral pumice]

these other 2 is mid

[warm marsh]

for y = ab^x, ln(y) = ln(a) + xln(b), so the graph you should be looking for should look like y = ln(a) + xln(b), so find the line which touches the y axis at ln(a) and has slope ln(b)

[floral pumice]

ok so im a litte confused

is c or d right cuz your speakingmathanese

[warm marsh]

i made a mistake previously which i corrected so yeah check that, and show me how option d looks like so that i can see if it passes through origin. The function cannot pass through origin

unless a=1 when ln(a) = 0

[floral pumice]

im sorry chatters but this is all i have sketchy screenshots

take your finest of guesses my brother

[warm marsh]

for this option c would be the safest option

[floral pumice]

i thought similar

i am paranoid and i will keep having questions if your willing to stick with me

ill be as funny as possible

[warm marsh]

i wont really be sticking around so sorry about that

[floral pumice]

nah your good my bad og

im gonna cook up then

thanks

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[azure dock]

how do i do b)

[azure dock]

as this time theres a number with x^2

[rustic storm]

Thats cuadratic funcion

One moment

[azure dock]

so quadratic formula?

[rustic storm]

Bottom one is the formula

But im sorry i need to go to sleep

Gl tho

[azure dock]

oh ok no worries!

is this correct

<@&286206848099549185>

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@azure dock Has your question been resolved?

[azure dock]

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[tough spire]

This is a problem from a practice real analysis exam that I’m struggling with.

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[tough spire]

<@&286206848099549185>

[stoic gale]

Potentially: assume fnk is a uniformly convergent subsequence, use the definition of uniform convergence and see from there

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[rapid topaz]

8 b i don’t understand

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[azure dock]

how do i do this

[humble karma]

A newspaper is usually a rectangle. What is the formula for the area of a rectangle?

Once you have that, you can use the second bit of information for write the length in terms of the width, and since the area only involves those two values, you will be able to solve for one.

[static egret]

Make a drawing so it's easier to visualize

[azure dock]

could i also use the quadratic formula,

?*

[humble karma]

I mean you'll probably end up using it for solving at the end.

It's easy to say you can "use the quadratic formula", but you still have to set it up.

[azure dock]

oh

[static egret]

Yes

So basically

Draw a rectangle

The width is w and the length is w+8

[azure dock]

okayy

so 4 sides?

or js for 2

[static egret]

?

[azure dock]

this

like do i put it on all the sides?

[humble karma]

Use those values (w and w+8) to get an expression for the area.

[static egret]

4 sides

[azure dock]

alr

[static egret]

Ye

recall lw =area

[azure dock]

so 273 (w+8)(w)

= *

and then i do solving?

[humble karma]

Yes.

[proud jungle]

273=(w+8)(w)

[humble karma]

Distribute the right side and solve using the quadratic formula.

[azure dock]

okk

so after doing the quadratic formula, i got 13, and -21, -21 doesnt work so its wrong so 13?

or wait is it

13, 21

[proud jungle]

It's -21 and 13 so width cannot be negative then it's 13

Length is 13+8=21

[azure dock]

ohhh alr thanks

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[full harbor]

i am completely lost

[formal quest]

let's use some laws of logarithm about its bases

which one would be the best?

[full harbor]

Log b b = 1?

[formal quest]

well, that holds for every b, so it does not give useful information

do you know how to change the base of a log?

[full harbor]

I don’t think so?

[full harbor]

Oh wait yeah I do

Didn’t know it was called that

[formal quest]

you changed the base of a log from b to a ;D

this holds for any a(of course except negatives, 1)

now you may change the base and calculate freely with same bases

[full harbor]

Ohh ok I get it now ty

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[crimson sedge]

Is there an easy way to do this question that I’m not seeing

[cerulean sail]

,rccw

[cerulean sail]

Define “easy”

[crimson sedge]

Like

I don’t know where I can start

I have these formulas

[cerulean sail]

You have a regular pentagon - but also maybe think about the fact you have a couple of triangles in it…

[crimson sedge]

Should I use the second one

[cerulean sail]

Would be very useful to, yep!

What does that give you?

[crimson sedge]

108

So that tells me that

Each triangle has a sum of that?

[cerulean sail]

Not quite - but

[crimson sedge]

What does it tell me

That each angle of the pentagon is 108?

[cerulean sail]

You basically know all the purple’s are 108 - but it’s the orange you want to find, right?

[crimson sedge]

So 103 divided by 3

108

Or

I don’t know

[lusty grotto]

take triangle LMN

u know what angle LMN is right?

[crimson sedge]

Yeah

108?

[lusty grotto]

yes

[crimson sedge]

And it’s parallel so are the other angles the same

Or not parallel

The other word

[lusty grotto]

what kind of triangle is LMN

[crimson sedge]

Isosceles

[lusty grotto]

yes

what do u know about isosceles triangles

[crimson sedge]

They have equal sides

Well the 2

[lusty grotto]

what else

[crimson sedge]

I don’t know

Just 2 sides of equal length?

[lusty grotto]

yes there are two sides of equal length

[lusty grotto]

but there are also two other equal things

[crimson sedge]

The

Angles

?

[lusty grotto]

yes two angles in an isosceles triangle are equal

which ones exactly

[crimson sedge]

The two by the hypotenuse

I think I did something wrong

[lusty grotto]

hypotenuse is a term u use with right angled triangles

[lusty grotto]

can u tell me which two angles are equal in triangle LMN above?

[crimson sedge]

LMM and NLM

[lusty grotto]

LMM?

[crimson sedge]

Uhhh

LNM

[lusty grotto]

yes

LNM and NLM

u also know already that LMN = 108

[crimson sedge]

I think I got it

Hold on

[cerulean sail]

you even noticed the equal angles before!

[lusty grotto]

well then u have the answer

[crimson sedge]

Thank you

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[stoic orbit]

is it x^2+4

[dire geode]

No

Use factor theorem for roots of quadratics

Or fundamental theorem of algebra

[tropic oxide]

(8+2i)^2 + 4 isn't 0 so no.

[glacial dune]

the answer is x^2 - 16x + 68

[tropic oxide]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[glacial dune]

i was explaining

wait

[tropic oxide]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[tropic oxide]

you should not begin with the answer lmao

[glacial dune]

this is because if you use the quadratic formula, the part inside the root should evaluate to -4

(because we want it to be +-2i)

and we know that a is 1 (the question specifically wants the factor of x^2 to be 1)

so, the equation simplifies to $\frac{-b}{2} \pm \sqrt{(\frac{b}{2}) ^ 2 - c} = 8 \pm 2i$

[wraith dagger]

kahve

[glacial dune]

there, we can see that $\frac{-b}{2} = 8$

[wraith dagger]

kahve

[glacial dune]

so $b = -16$

[wraith dagger]

kahve

[tropic oxide]

honestly still kind of crap that you simply gave out the entire solution without so much as ensuring op is even listening

[stoic orbit]

wait its okay i got it 💀 seems that i forgot to multiply

thanks though 🔥

[cedar kiln]

@stoic orbit Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[fiery valley]

$$4^{x}* 8^{x + 1} = 16$$
$$2^{2x} * 2^{3(x + 1)} = 2^{4}$$

$$2x + 3x + 3 = 4$$
$$5x = 1$$
x = 1/5

[past oriole]

you wrote x+2 in the top line, it should be x+1

[fiery valley]

omfg

okok

thank you

[past oriole]

other steps look good though

[fiery valley]

holup

[past oriole]

you need to fix the whole thing

all the lines

[wraith dagger]

Maladroit

[fiery valley]

yea

there we go

[past oriole]

thats it

[fiery valley]

also, if i am to do this using logs, will the operation be multiplication or addition?

[past oriole]

if you log the whole equation?

[fiery valley]

in the exponential eq, its multiplcition

[fiery valley]

yea

[past oriole]

well, you did log it

[fiery valley]

$$(x \log 4) * (x + 1 \log 8) = \log 16$$?

[wraith dagger]

Maladroit

[past oriole]

you should use log base 2

[fiery valley]

since the expo eq. was a multiplication, is that how it is in log?

[fiery valley]

yea

[fiery valley]

but is this how it is going to look if i am to use logarithim

[fiery valley]

.

[past oriole]

(x+1) is in the brackets

[fiery valley]

yeah cause $$8^{x + 1}$$

[wraith dagger]

Maladroit

[past oriole]

yep

[fiery valley]

if i unbrack it, its gonna look messy

[hollow minnow]

[\log_2(4^x\cd8^{x+1})=\log_2(16)]

[wraith dagger]

Pure

[fiery valley]

why did we group 4 and 8?

because they have a common base? (2)

[past oriole]

no, because they are on the same side of the equation

[hollow minnow]

You can’t just apply log to the individual term of a product

,,\loglaws

[wraith dagger]

Pure

[fiery valley]

wdym

[past oriole]

you did it correctly, but formally you skipped a step

[fiery valley]

even if they have a different base, it still applied that way?

[hollow minnow]

[\log_2(4^x\cd8^{x+1})<>\log_2(4^x)\cd\log_2(8^{x+1})]

[wraith dagger]

Pure

[fiery valley]

is

[hollow minnow]

[\log_2(4^x\cd8^{x+1})=\log_2(4^x)+\log_2(8^{x+1})]

[wraith dagger]

Pure

[fiery valley]

ohhhh

so the rules for exponents are the same in logarithmic..

if product law you add exponents, just like in loga?

[fiery valley]

what law is the latter equation?

[hollow minnow]

Well log is the inverse of exponents basically

[hollow minnow]

First law

[fiery valley]

in exponent, you find the procuct?

product

in log you find how much the base needs to be raised to, to get the product ?

.close

[cedar kiln]

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[cedar kiln]

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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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[hollow storm]

Hello, can somebody help me understand this problem?

[hollow storm]

Let $f: X \to Y$ be a map between two sets. Let $A \subset Y$. Verify that $f(f^{-1}(A)) \supset A$, but the set equality $f(f^{-1}(A)) \supset A$ does not always hold.

[wraith dagger]

Tommy

[hollow storm]

Isn't $f^{-1}(A)$ by definition the elements in X that are mapped in A? And therefore $f(f^{-1}(A)) = A$

[wraith dagger]

Tommy

[hollow storm]

WHat am I getting wrong?

[crimson delta]

lets say f:R->R and f(x)=x^2. what about A=R

also you wrote the wrong subset signs in your statement

should be the other direction

[hollow storm]

yeah that's what i think as well

prof wrote it the other way i don't know why

[crimson delta]

profs also make mistakes

unlucky

[hollow storm]

but in that case the two sets are equal (?)

[crimson delta]

are they?

[hollow storm]

R = R

[crimson delta]

is for example -5 in the left set?

[cedar kiln]

@hollow storm Has your question been resolved?

[hollow storm]

for example for f(x) = x^2

we have {-2, -1, 0, 1, 2} --> {0, 1, 4}

in this sense the two sets are not equal?

[crimson delta]

"in this sense" ?

they are just straight up not equal

[hollow storm]

yes

[crimson delta]

on the left there are only numbers >= 0

on the right there are all real numbers

which are not the same sets

[hollow storm]

if A = {0, 1, 4}, f^-1(A) = {-2, -1, 0, 1, 2}

but f{-2, -1, 0, 1, 2} = f(f^-1(A)) = A

???

[hollow storm]

yeah

so how is f(f^-1(A) a subset of A?

[crimson delta]

well I said to pick A=R

what happens if you pick A={-1, 0, 1, 4}

then what is f^-1(A)

[hollow storm]

well -1 doesn't have an inverse

[crimson delta]

inverse is not quite the right word

it has no preimage

but that doesnt matter

intuitively, to calculate f^-1(A), you dont check every element in A and see which preimages you have

instead, you check every element x in the domain, calculate f(x) and check whether f(x) is in A

and if you do that, you get that f^-1(A) = {-2, -1, 0, 1, 2}

just like before

[hollow storm]

why include -1 ?

f(-1) is not in A (?)

[crimson delta]

f(-1)=1 is in A

[hollow storm]

ohh yeah now i see

so to prove that the equality of the sets is not always true i just need a counterexample, like f(x) = x^2

[crimson delta]

yes

to prove that something is false, always counterexample

[hollow storm]

okay!! thank you very much

all clear now

.close

[cedar kiln]

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• 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!

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[civic coral]

number like 1,2,3,5,7 which can not be factored no more

[olive crescent]

what about them?

[civic coral]

what they call

[olive crescent]

prime numbers?

[sturdy rose]

prime

[olive crescent]

*Numbers like
*cannot be factored anymore (or you can factor no more)
*what are they called?

[civic coral]

is it true that "perfect square is made up of two squared prime numbers"

[fickle valley]

💀

[civic coral]

thank you

[olive crescent]

with some extra conditions

https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem

oh sorry I misread you

[crimson delta]

!original

[cedar kiln]

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

[civic coral]

ayy, I'm trying to make some statement

mathematical statement

to solve questions more easily

that's it

[civic coral]

the claim(if it is true) will help me solving questions more easily

So it is true?

[olive crescent]

it is not true that every x^2 = a^2 + b^2 for some prime a, b

its prime decomposition contains no factor pk, where prime p ≡ 3 ( mod 4 ) p\equiv 3{\pmod {4}} and k is odd.

[civic coral]

ohhh

I should be more specific

[olive crescent]

this is the condition, which is kind of awkward to work with

[civic coral]

is it true that "x^2=a^2 * b^2"

[olive crescent]

not necessarily

[civic coral]

where x belongs to real numbers

and a and b are prime

[olive crescent]

...oh, we're at multiply

[civic coral]

a counter example?

[olive crescent]

not sum

uh, x=0.1?

0.01 is not the sum of 2 squares of primes

unless you mean x belongs to the natural numbers

[civic coral]

let constrict x to the realm of perfect squares

[olive crescent]

in which case x=4 gives x^2 = 16

your options are 3^2=9, 2^2 = 4 which is either 13 or 18

[tropic oxide]

what's the question again

[civic coral]

is it true that "perfect square is made up of the product of two squared prime numbers"

[olive crescent]

No

[civic coral]

a counter example?

[olive crescent]

...x=4

[civic coral]

emm

the claim has to be revised

[olive crescent]

x^2 = 16 and there are no a, b such that a^2 + b^2 = 16 and a and b are prime

oh product

[civic coral]

the result of the product of two squared prime numbers will always ends up a perfect square

[civic coral]

that is flawless

[olive crescent]

yes, that is true

[civic coral]

im so proud of myself

[olive crescent]

$p^2q^2 = (pq)^2$

[wraith dagger]

PrettyPrincessKitty FS

[olive crescent]

this is true for any p, q integers

[civic coral]

terffic

thank you so much!

my question solved now

[olive crescent]

is solved now

[civic coral]

thank you

.close

[cedar kiln]

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Remember:
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• 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.

[civic coral]

thats quite inappropriate

[outer topaz]

i get t - kmg = kma

6mg - t = 6ma

where do i go from here

[outer topaz]

how so

[civic coral]

nevermind, it is just me having a dirty mind

[outer topaz]

[lusty grotto]

there doesnt seem to be a question in the image u sent

[outer topaz]

oh my bad

[lusty grotto]

oh

so it says A descends with an acceleration g/3

[lusty grotto]

so a in these equations must equal g/3

that way u get a system of two equations with two unknowns

one unknown being t, the other being k

solve the system and u have the answer

[outer topaz]

ok thx

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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.

[oblique lynx]

can somebody please tell me how to solve this integral? Ive got it into this form
$$
\frac{1}{8}\int (1-\cos2x)^2(1-\cos2x)dx
$$
but dont know how to continue

[wraith dagger]

konxmok

[stone flame]

expand it

then separate into multiple integrals and calculate individually

[oblique lynx]

you mean into
$$
\frac{1}{8}\int (1-3\cos2x+3\cos^22x-\cos^32x)dx
$$

[wraith dagger]

konxmok

[subtle hinge]

yea now convert (cos2x)^2 in terms of cos4x and (cos2x)^3 in terms of cos6x

[oblique lynx]

how do i do those conversions? i really suck at trig

[subtle hinge]

wait i thot someone said u dont have to do that

idk

[stone flame]

no I misread what you said

[subtle hinge]

i only know the basics of integration

[subtle hinge]

oh ok

[subtle hinge]

and