help-27

[supple narwhal]

Bruh

What you think

[steel sage]

Doesn’t look correct

(An element x of order m, order of x^i, say k. You will notice that m|ki

[supple narwhal]

What

[steel sage]

You don’t know that? if x^n=e then m|n

(Euclidean division of n by m)

[supple narwhal]

What

[steel sage]

What what?

[supple narwhal]

m|n

[steel sage]

| means division

[supple narwhal]

Ik

[steel sage]

m|n means n is divided by m

[supple narwhal]

Ik

But why

[steel sage]

n=km+r, x^n=e=x^r. m is the order means r=0

[supple narwhal]

Hmmmm

n=km+r?

[steel sage]

Euclidean division

[supple narwhal]

Bruh

You just saying euclidean division

[steel sage]

Like Euclidean division of 15 by 7
15=2 times 7 +1

[supple narwhal]

Il

Ik

But why bruh

Is that property or something?

Which property?

[steel sage]

You do Euclidean division of n by m

[supple narwhal]

It have a proof or something else

Why tf m|n

In cycles

[steel sage]

If x^n=e, let n=km+r, r<m x^n=x^(km+r)=(x^m)^k time x^r=x^r

[supple narwhal]

I just want to know it

[steel sage]

Thus r=0

Thus if x^n=e-> m|n

[supple narwhal]

Okay then m|mi

[steel sage]

order of x^i is k then x^ik=e-> m|ki

[supple narwhal]

m not k bro

[steel sage]

What?

You let k be the order of x^i

[supple narwhal]

Look at the question

[steel sage]

x^ki=(x^i)^k=e

[steel sage]

By this m|ki

[supple narwhal]

If |x^i| = m then gcd(m,i)=1 wdym by saying ki

[steel sage]

You didn’t prove that

I am explaining to you how it can be proved

[supple narwhal]

Yea but why you put k instead of m

[steel sage]

Okay I see the confusion

We ignore order of x^i is m for the time being

[supple narwhal]

Okay

[steel sage]

Then m|ki

[supple narwhal]

m|k or m|i?

Let k=m

[steel sage]

We don’t have that

[supple narwhal]

If m|i hmm

[steel sage]

But m’=m/gcd(m,i) , i’=i/gcd(m,i)
you have m’ | i’ k

[supple narwhal]

Im truly confused

[supple narwhal]

Okay im done with that

That’s fucking cycles notations im done with that

[steel sage]

minimal k such that m’ | i’ k is when k=m’, thus order of x^i is m’ =m/gcd(m,i)
order being m is the same as saying m/gcd(m,i)=m

[sudden flower]

(a+b)² =?

[supple narwhal]

Stop trolling

[supple narwhal]

Hmm bro

[steel sage]

Since m’, i’ are relatively prime to each other

[supple narwhal]

Can you explain what we conclude from m|ki

[sudden flower]

Just i wanted to confirm

[supple narwhal]

Again

[supple narwhal]

a^2 + b^2

[steel sage]

We divided by gcd(m,i) on both sides

[sudden flower]

Thanks

[steel sage]

of m|ki

[supple narwhal]

gcd(m,1) you mean right?

Since m=m*1

[steel sage]

gcd(m, i)

[supple narwhal]

Even basic algebra make me confused bruh in this shitty cycles

[steel sage]

order of x^i is k, we forget the condition that k actually is m
m| ki
divided by gcd(m,i) on both sides
m’ | k i’

[supple narwhal]

Oh

That’s right

[steel sage]

m’ and i’ are relatively prime. So it’s the same as m’ |k

[supple narwhal]

m’ and i’ are prime then

Relatively*

[steel sage]

Yeah

Minimal such k is exactly m’

So order of x^i is m’ =m/gcd(m,i)

[supple narwhal]

[steel sage]

(x^i)^k=e <-> m|ki <-> m’|k i’ <-> m’ |k
Minimal k such that (x^i)^k=e <-> minimal k such that m’ |k<-> k=m’

[supple narwhal]

Bro

Can we start from 0 please

[steel sage]

Okay

[supple narwhal]

We need show that
If sigma^i is ordered by m

Then gcd(m,i)=1

That’s what we want to prove?

[steel sage]

I use x for σ since I don’t want to switch keyboard. Let’s summarize my thought again:

x is an element of order m

[supple narwhal]

Okay

We suppose x^i is an element of order m too?

[steel sage]

No we don’t assume its order being m for now. Pretend we don’t know its order

[supple narwhal]

Then k its order?

[steel sage]

I will make it clearer

For any k
(x^i)^k=e <-> m|ki <-> m’|ki’<-> m’|k

<-> means being equivalent to

[supple narwhal]

You kidding me

[steel sage]

thus minimal k such that (x^i)^k=e (which by definition is the order of x^i) <-> minimal k such that m’|k

Thus order of x^i is m/gcd(m,i)

[supple narwhal]

How we conclude that

[steel sage]

I just showed you

For any k
(x^i)^k=e <-> m|ki <-> m’|ki’<-> m’|k

[supple narwhal]

Bro i want to say that im logician

It mean that what you think

[steel sage]

Those three <-> above, you still have any question regarding any one of them?

[supple narwhal]

m/gcd(m,i)

[steel sage]

I mean these three <->

[supple narwhal]

Okay and after?

m’|k what does mean

[steel sage]

(x^i)^k=e <-> m’|k
{positive k such that (x^i)^k=e}={positive k such that m’|k}
Take minimal both sides

order of x^i = m’

=m/gcd(m,i)

[supple narwhal]

Minimal is k=m’?

Okay that’s make sense

[steel sage]

Minimal positive k such that m’|k of course is m’

[supple narwhal]

Since the order of x^i is k then is the least possible positive integer

An least possible integer is m’

Since m’|k

[steel sage]

Yeah

From this point we can stop pretending we lost that condition
m/gcd(m,i)=m <-> gcd(m,i)=1

[supple narwhal]

wait

[steel sage]

All we have done are
Showing
(x^i)^k=e <-> m|ki <-> m’|ki’<-> m’|k
Choosing minimal positive such k both sides
Nothing else

[supple narwhal]

Okay

x^i is an element of order m

Then k=m

m/gcd(m,i) = m

gcd(m,i)=m

[steel sage]

Is equivalent to gcd(m,i)=1, yeah

[supple narwhal]

1*

[steel sage]

Yeah

[supple narwhal]

Good

Now we suppose gcd(m,i)=1

[steel sage]

So I understand your confusion. I should have made it clear we don’t use the condition of order being m only until the last step. I want to deduce the formula of order(x^i) in general first

[supple narwhal]

Yes i see

[supple narwhal]

We want to show that x^i is ordered of m

[steel sage]

So now you have a general result, order(x^i)=m/gcd(m,i) given order(x)=m
Better right

[supple narwhal]

Oh right

Didnt make
Order(x^i) = order(x) / gcd(order(x),i)
A general property of given i?

[steel sage]

Yeah, any i

[supple narwhal]

But again why my first prove is incorrect?

[steel sage]

I don’t think your solution even discussed i

You obtained x^mi=e

Then what follows doesn’t make sense

[supple narwhal]

You mean mi=1?

[steel sage]

(order(x^mi)=order(e)=1, doesn’t lead anywhere)

Yeah, mi =1 doesn’t make sense

[supple narwhal]

Order(x^mi) = mi

Order(e) =1

[steel sage]

Now you know which isn’t true

[supple narwhal]

mi=1

We have that property
Gcd*lcm=1

Gcd(m,i)*lcm(m,i)=mi

=1

It follows that gcd=1

[steel sage]

We just proved order(x^p)=m/gcd(m,p). Not order(x^p)=p

So you understand now mi=1 isn’t true

[supple narwhal]

Oh i think you’re right

[steel sage]

This doesn’t make sense. And the formula that does make sense was just proven

[supple narwhal]

mi=1 in general not correct

[steel sage]

Yeah

[supple narwhal]

Take m=8 and i=7

[steel sage]

Order is 8

[supple narwhal]

m*i != 1

[steel sage]

Anyway, any other question?

[supple narwhal]

Are tired?

Are you tired?

[steel sage]

No. Just want to make sure I made it clear

[supple narwhal]

Then can you help me to prove question 9

8*

[steel sage]

8 or 9

[supple narwhal]

8

[steel sage]

A finite group , order of an element | order of the group

I mean if |G|=n, then any x in G, x^n=e

If your group in 8 really is finite, you obtain a contradiction by this

[supple narwhal]

We suppose the group to be finite then?

[steel sage]

Yeah
and any m you can find a order m element in Ω

[supple narwhal]

The |Somega|=n

[steel sage]

Or simply

Ω has elements of order n+1

[supple narwhal]

m != n

[steel sage]

any positive m

Though m=n+1 will suffice

(1,2,…,m)

(f: N->N
restriction of f on {1,2,..,m} is m-cycle (1,2,…,m)
restriction of f on the complement is identity

[supple narwhal]

Why you saying that?

Mean cycle as function do you mean?

[steel sage]

f(1)=2,f(2)=3,…,f(m-1)=m, f(m)=1
f(k)=k for k>m

Order(f)=m

[supple narwhal]

And we can define an another cycle with order n

n != m

[steel sage]

I don’t see the point

[supple narwhal]

Since that we obtain there a infinite permutations

[steel sage]

Anyway, there exists an order n+1 element

Contradiction

order(f)=n+1 we can’t have f^n=e

n+1 doesn’t divide n

[supple narwhal]

Can we use upper bound concept

Suppose G is finite

Then there exist M ..

M such that forall elements in G

[steel sage]

Yeah

Order <=M

Contradiction

[supple narwhal]

Yea

We can take an element greater than M

That’s the contradic

Tion

[steel sage]

Yeah

Have to go running. Feel free to ask me anything you want DM.

[devout snow]

@supple narwhal Has your question been resolved?

[devout snow]

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[lost laurel]

So I'm trying to implement this code in python

[lost laurel]

import sympy as sp
import numpy as np
import matplotlib.pyplot as plt
import math as math

print("Please enter a  differentiable function whose roots you would like to find using fixed point iteration as a function of 'x'")
x = sp.symbols('x')
f=input("Enter your function\n")
f=sp.simplify(f)
print("The function you input has been registered as:", f)
print("Please enter the interval over which you'd like to find roots, along with the error as prompted ")
a=input("Enter the lower bound of the interval\n")
a=sp.simplify(a)
b=input("Enter the upper bound of the interval\n")
b=sp.simplify(b)
print("The lower bound has been registered as:",a,"\nThe upper bound has been registered as",b)
c=input("Please enter the desired  accuracy")
c=sp.simplify(c)
print("The accuracy has been registered as",c)
n=input("Please enter the desired number of iterations")
n=sp.simplify(n)
print("The desired number of iterations has been registred as ",n)
y=input("Please enter an initial value from which the search for roots will begin")
y=sp.symbols('y')
for i in range(n):
    z=f.subs('x',y)
    if z-y<=c:
        y=z
    else:
        print("Method failed")
print("The root is approximately",z)

I would like this checked ( also lmk if I should take such questions to coding servers in the future)

[rain summit]

oh my lord

[stone stump]

stay away from user input

it just bloats the program

for no benefit

[lost laurel]

The reason I included user input is to allow me to reuse this later on

[stone stump]

this is useless for later

you would have to rewrite it anyway

to put it into a function

[lost laurel]

Okay, noted

thanks

[light saffron]

are we good here?

[rain summit]

file reading

file reading

[light saffron]

How we doin fam?

[winter torrent]

wait you're not even using the interval (a and b), what's that for?

[ionic harness]

Holdover from the last method I assume?

[lost laurel]

I thought I would need it

I guess I'll remove it

[winter torrent]

your spec asked for four inputs so you should have the word input four times in your code

assuming you're supposed to do i/o and not just hardcode values into your program

[lost laurel]

yea

[devout snow]

@lost laurel Has your question been resolved?

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[devout snow]

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[lost laurel]

I'm sort of confused here, isn't it simply $\frac{1}{\theta^n}$?

[woven radish]

waimas

[lost laurel]

I feel like I'm missing something

[mint flume]

isn't it just theta?

3 theta - 2 theta is 1 theta so x cannot be more than 1 theta? right?

oh right

1/ theta^n yeah

[lost laurel]

huh?

[mint flume]

never mind carry on

[lost laurel]

we have to maximse L(\theta) here

but how would I do that here .....

[devout snow]

@lost laurel Has your question been resolved?

[lost laurel]

<@&286206848099549185>

.close

[devout snow]

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[rustic musk]

who mentioned me

[rustic musk]

oh sorry

i dont need help

who mention

[vagrant skiff]

.close the channel

[rustic musk]

.close

[dull parrot]

by mention you mean ping?

[devout snow]

Channel closed

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Use .reopen if this was a mistake.

[vagrant skiff]

no one mentioned you

[rustic musk]

yes

[dull parrot]

you can check inbox to see who pinged you

[rustic musk]

ok

[devout snow]

Available help channel!

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[fossil haven]

can someone help

[tropic sinew]

do you have a specific question?

[fossil haven]

yes

[brittle inlet]

The guy was asking about trig functions

[tropic sinew]

ahh okay.. but that's still vague no?

[fossil haven]

can some one please explain what is sine, cosine, and tangent

[soft umbra]

Have you googled it in advance?

[fossil haven]

yea

[soft umbra]

Alright, could you share your understanding regarding them?

We can amend it for you if there is any flaw

[fossil haven]

all i know is that one of them is the hypotenuse and thats all i remember

.close

[devout snow]

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[somber dock]

guys can any one help me with numerical integration

[devout snow]

Please don't occupy multiple help channels.

[winter patrol]

Can you please stop spamming this in multiple channels.
Stick to #help-8.
and try to be more specific in your request.

.close

[devout snow]

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[loud monolith]

13. Define f(n) to be the sum of the positive integer factors of n. If f(n) = 360 and f(3n) = 1170,
    what is the sum of square of digits of the smallest possible value of f(2n) ?

[loud monolith]

Now Now N 's sum fo factors is 360

Upon multiplying with 3

how th ehell did it jump 1170?

shudn't it be like 363

[tidal mulch]

3 isn't the only new factor that is added upon multiplying by 3.

[loud monolith]

Give example

[rain summit]

hold the 💀 up

[loud monolith]

😅

[rain summit]

i immediately think of something

[tidal mulch]

2*3 has factors 1,2,3,6
Upon multiplying by 3,
2*3*3 has factors 1,2,3,6,9,18

[loud monolith]

Huh?

[rain summit]

prime factorization

[tidal mulch]

Sum of positive integer factors of n.

[rain summit]

assume n has a certain prime factorization

then 3n is just that prime factorization adding in a new 3 factor

[tidal mulch]

I don't know what's there to "huh" about it

[loud monolith]

Well........

Idk

So we multiply every thing with 3 ?

Like all factors

[rain summit]

f(n) is 360 right?

[loud monolith]

yup

[rain summit]

now do you understand this

[steel sage]

If power of 3 is a in n, f(3n)/f(n) by prime factorization, should be (3^(a+2)-1)/(3^(a+1)-1) and you have this fraction

[loud monolith]

ya ig

[rain summit]

if n is not divisible by 3

[loud monolith]

what......

[rain summit]

then the sum of all factors of n and 3 is 363

[celest frigate]

@loud monolith this is what the problem intends I believe

[rain summit]

so 1170-363 is the sum of your remaining factors

[loud monolith]

Mhm

[rain summit]

and if n is divided by 3, 1170-360 is the sum of your remaining factors

that's all i can think for now

hope anyone sees this :)

[steel sage]

I ended up minimal f(2n) is ||3024||, not sure whether my method has flaws. Checking?

[loud monolith]

Now how did you end up with that

[steel sage]

Anyway I found error anyway, fixing…

[devout snow]

@loud monolith Has your question been resolved?

[faint hearth]

||n=120,
f(n)=360, f(3n)=1170
f(2n)=744|| seems to be smaller

[steel sage]

Oh

[tidal mulch]

Let k be highest power of 3 dividing n.

f(3n)=1770
We also know
f(3n)=(3^{k+2}-1)/(3^{k+1}-1)*f(n)

So 1770=(3^{k+2}-1)/(3^{k+1}-1)*360
Try to proceed from here on your own, view this for hint.
||k=1 works
So n=3m where 3 doesn't divide m.
f(n)=f(3m)=(1+3)f(m) so f(m)=90
m=40 is smallest with f(m)=90.
So n=3m=120,2n=240, so f(2n)=744||

[steel sage]

I had a brain freeze, i somehow calculated f(3)=2. No wonder

[tidal mulch]

f(n)>=n+1 always.

[steel sage]

Got it. Yeah that one should be the minimal

[devout snow]

@loud monolith Has your question been resolved?

[devout snow]

@loud monolith Has your question been resolved?

[devout snow]

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[chrome falcon]

hello guys can anyobody help me with one

[chrome falcon]

maths qn

please

[pseudo basin]

image sending?

[chrome falcon]

yes

[pseudo basin]

a or b?

[chrome falcon]

both

can anyone solve and send it would be sucha great help

hello??
can anybody help

[pseudo basin]

https://tenor.com/view/we-dont-do-that-here-black-panther-tchalla-bruce-gif-16558003

!noans

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[chrome falcon]

can anyone help me out and tell me the concept

[pseudo basin]

do you know how to do proof by induction in general?

[chrome falcon]

yes kind of

[pseudo basin]

ok. can you write down your base case?

[chrome falcon]

i wwant the answer of B not a like the main theme.

is it as the areas are different?

[pseudo basin]

the areas are different, yes. but by how much? and how does this "missing" area actually show up?

[chrome falcon]

by one

64 and 65

[pseudo basin]

ok and can you spot where the extra 1 cm^2 of area is hiding in the 5×13 rectangle?

[chrome falcon]

no

[pseudo basin]

give it some thought. maybe try to make a really really accurate drawing on graph paper.

[devout snow]

@chrome falcon Has your question been resolved?

[chrome falcon]

no

[devout snow]

@chrome falcon Has your question been resolved?

[devout snow]

@chrome falcon Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

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[fair solar]

Why is there no 'help' channel?

[winter patrol]

wdym

[fair solar]

Got it

[winter patrol]

the server is set up so that everyone can get their own dedicated channel
and you've currently reserved one of them

[fair solar]

Hey Omega I'm 10th grader and i wanna get help in my mathematics.
Would you help me?

[winter patrol]

You'd need to post specific questions

[fair solar]

That's exactly what I got it

[fair solar]

Sure

[winter patrol]

that you're stuck on

[fair solar]

Sure

I'm having problems in number like
Continuity and infinity
And I don't know what does it mean
......... less than 0
Something like this

Teach me the concept of infinity since I am a 10th grader

[crystal dawn]

this sounds like something that would benefit from showing a question to show what you mean

[fair solar]

Sure I think I can't do rn as I have to go for pray

.close

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[zenith bronze]

My thought process is the green one and the answer given is in red

btw i just have problem listing the equation and i can solve the differential equation afterward myself

my bad, i messed up myself

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[analog fjord]

Can you guys help me with this. I did understood the comm.. and associ... what does this question even want like eg I tried the first want (a + b) + (c + d) to be a + (b + (c + d)) and from there what do I even do

[analog fjord]

I just want a lead

[uncut crow]

worst collection of problems i've seen in the past 10 minutes

[crystal dawn]

can you state the defn of associativity and commutativity?

[analog fjord]

commutativity = swap able and get the same answer
associativity = can the bracket be move and get the same answer

[crystal dawn]

eh... yeah, but I was looking to see the definition in algebraic terms

but one property some might forget about associativity is that you can apply it to remove brackets

[pseudo basin]

mmmm nope

not here

[crystal dawn]

oh oops

my bad

[pseudo basin]

here we have to engage in heavy duty bracketfucking

and we CAN'T write a+b+c in a setting like this. bc we have to have each and every application of assoc accounted-for

[crystal dawn]

I suppose OP was already partly correct with this move of
a + (b + (c + d))

then it remains to do the Kansas city shuffle to shift the inner pair around, then the outer pair, and once more the inner pair (of parentheses), if that makes any sense

[pseudo basin]

you want the b and d to be next to each other

notice assoc never messes with the order of the letters; only comm does

so the next good step is to swap the c and d around with comm

[devout snow]

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[fervent helm]

how to solve this?

[wanton thunder]

Hint: Fundamental theorem of calculus

[rare kernel]

differentiate both sides

[wanton thunder]

And use Leibnitz rule to differentiate the integral.

[fervent helm]

I'll have to apply the product rule right?

[rare kernel]

also put x=0 here for integration constant

[fervent helm]

ohh

[wanton thunder]

You will first need to justify that you can differentiate this expression using fundamental theorem of calculus

If you take (1+x^2) to the left hand side

[woven radish]

Prathmesh
Compile Error! Click the reaction for more information.
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[wanton thunder]

You can't do that because f(x) is not given to be differentiable.

[fervent helm]

$y = e^{2x} \cdot e^{\frac{\ arctanx}{2}} \cdot 2$

[woven radish]

Prathmesh

[wanton thunder]

Prathamesh you first need to observe that if you take the (1+x^2) on the right hand side

You have an expression that looks like the integrand of the integral on right hand side.

[fervent helm]

correct

[wanton thunder]

Now let $\int_{0}^{x}\frac{f(x)}{1+t^2}dt =g(x)$ then $g'(x)=\frac{f(x)}{1+x^2}$

[woven radish]

silicate

[wanton thunder]

So the expression reduces to a first order linear differential equation.

[fervent helm]

yea

[fervent helm]

I did that as soon as you said it

[wanton thunder]

You could not just differentiate because f isn't given to be differentiable

This equation is a consequence of Fundamental theorem.

Next step is solving the ODE.

I believe you already did that.

[fervent helm]

we get $\ln(1+g)=2x$

[woven radish]

Prathmesh

[fervent helm]

+c

now can we write this as $1+g = e^{2x}$

[woven radish]

Prathmesh

[wanton thunder]

Hmm

[fervent helm]

and differentiate both sides?

[wanton thunder]

I am not too sure but I think it should be a -2x instead of just 2x

[wanton thunder]

Yeah, you can do that, which will get rid of the constant in addition on our right hand side.

[fervent helm]

I don't think so

[fervent helm]

but why can we differentiate this and not the function in the question

[rare kernel]

i mean if u take 1+x^2 to the otherside and differentiate both sides it still works out

[wanton thunder]

You did not mention that when you hastily answered in the beginning.
Which is why Prathamesh asked if they need to apply product rule or not.

[wanton thunder]

Because this time there is no f(x) in our expression. The derivative of the integral exists from Fundamental theorem of calculus.

[rare kernel]

bro ofcourse u cant differentiate directly..the integral will still be there

[wanton thunder]

?

[rare kernel]

the integration term will still be there

[fervent helm]

I see

[wanton thunder]

The point I am trying to make is that instead of hastily dropping a message try to work with the person and provide hints instead.

[rare kernel]

"hastily"?

[wanton thunder]

Differentiate both sides

also put x=0 here for integration constant

[rare kernel]

i didnt give him some ground breaking insight here that will solve the whole problem i just told one of the main steps

[tropic sinew]

you can’t differentiate the original equation until you invoke FTC by defining the integral term properly

after that differentiating is valid

[wanton thunder]

@fervent helm still here?

[fervent helm]

$\ln(1+g)=2x+c \Rightarrow
1+g=e^{2x} \cdot e^c$

[woven radish]

Prathmesh

[wanton thunder]

Yes, now you can differentiate the expression. Use Leibnitz rule to differentiate the integral.

After that substitute 0 in the original expression and see what you get. That will be helpful later.

[fervent helm]

yea

[wanton thunder]

I don't know why but I feel like there is a negative sign that is missing.

Let me just solve the ode real quick.

,w y'-2y=2

[woven radish]

Results provided by WolframAlpha

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Requested by schizo_cat5

[fervent helm]

I got the answer

[wanton thunder]

Oh alright

[fervent helm]

10e^4

thanks

[wanton thunder]

Good luck

Oh I realised where I was messing up in my head when solving the ODE.

[fervent helm]

I had another question of vector algebra. Can I?

[fervent helm]

ahh no problem. Happens with me everyday

[wanton thunder]

Pick a new channel so the bot can pin your message.

[fervent helm]

alright

thanks

.close

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[devout snow]

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[fair solar]

I need help

[fair solar]

Anyone here?

[vagrant skiff]

You can just post your question

[fair solar]

Now lemme write question

I'm very confused.
GB Star exchanged some Nepalese rupees for American dollar at the rate of US $ 1= NRs. 110. After one week, Nepali currency devaluated by 10% in comparison to Us dollars.

If GB Star gained NRs. 33000 changing the US dollars again into Nepali rupees after one week, find the amount in dollars that he exchange.

Explain me

Anyone help me plsss
I want someone to explain me

[pseudo basin]

so first you exchange NPR into USD at 1 USD = 110 NPR

then the rupee devalues by 10% against the dollar

then you convert the same dollars back into NPR

and you end up with 33,000 NPR more than you started

did i understand you correctly

@fair solar

of course op goes offline

[devout snow]

@fair solar Has your question been resolved?

[devout snow]

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[fair solar]

.reopen

[devout snow]

✅ Original question: #help-27 message

[fair solar]

Yup it's exactly what I said.

[pseudo basin]

ok you're back

so tell me first things first:

after the devaluation, what's the new exchange rate?

[fair solar]

10% of old exchange rate

[pseudo basin]

incorrect and also not what i asked

first it says devalued BY 10% and not TO 10%. second the number of rupees per dollar goes UP

[fair solar]

You have to calculate from the old exchange rate and the percentage of devaluation

[pseudo basin]

ok well then YOU go calculate it, and tell me what it is

[fair solar]

I calculated it, it's $1 = NRs. 121 (with devaluated 10% against dollars)

[pseudo basin]

ok

so we exchange x rupees into dollars at 110 NPR/USD, then exchange that back at 121 NPR/USD, and end up with x+33000 NPR

can you find the value of x from here

[fair solar]

I am so confused to do this

In the textbook, the answer is $3000

[pseudo basin]

ok then let's take this step by step shall we

[fair solar]

But I can't calculate

[fair solar]

Yup

[pseudo basin]

when you exchange x rupees to dollars at 110

[fair solar]

Should I send you my rough?

[pseudo basin]

how many dollars do you get

if it's readable, sure

[fair solar]

So I have to rewrite in English

[pseudo basin]

what language is it in right now

[fair solar]

Nepali with Devanagri script

[pseudo basin]

yeah you can if you want. i don't speak nepali

or we could continue with my guidance instead.

[fair solar]

I'm a 10th grader

[pseudo basin]

do you want to translate your work or do you want my guidance independent of that

[fair solar]

Lemme convert it.
After observing my rough you may understand what I am unable to understand

[pseudo basin]

sure

[fair solar]

I'll be right back in 5-7 minutes

[fair solar]

You know what, somehow I calculated the answer correctly but I didn't know how.
Still check is it correct? @pseudo basin

[pseudo basin]

ok you took x as the dollar amount

yeah sure this works

so then, what remains in doubt?

[fair solar]

I somehow solved it but i couldn't understand it well,
How does it work and how do I understand these types of questions

[pseudo basin]

i mean

you took x as the amount of USD

you calculated how many NPR you get when exchanging it with each rate

you used the known value of the DIFFERENCE between these rupee amounts

there's nothing else to it

[fair solar]

Yup I got it already

[fair solar]

Ok

Let me come with another question

@pseudo basin

When a dice is rolled and a coin is tossed one after another,
What is the probability of getting 1 on dice and head on coin

[pseudo basin]

ok, what's your doubt about this one

[tender wharf]

hmm.. one after other sounds like bayes

[pseudo basin]

,rccw

[pseudo basin]

nothing of the sort

[fair solar]

Lol, i don't know how I solved it again.
I think I wrote every simple thing for a pic and I solved it carefully.

[tender wharf]

had a feeling

[pseudo basin]

your solution was methodical but also very long lol

[fair solar]

Yup, i tried in short and I wronged

Ok I think now I should close

Thanks for help

.close

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[light saffron]

hi what did I just doodle

[mystic scarab]

??

[fast garden]

this kinda triangular E you wrote is just sum notation; the subscript is where you're summing from and the superscript is what you're summing to

it substitutes every integer value between those into your expression on the right hand side

[light saffron]

I didnt get that

sorry

[fast garden]

that's okay

i assume you are just asking about notation, right?

[light saffron]

Well I actually conceptualized it

But what do you see

[fast garden]

conceptualized what exactly?

[light saffron]

I guess I wanted to describe gravity

[fast garden]

then, i am confused what you are asking about exactly ...

[light saffron]

What do you see from the formula

[tender wharf]

sum

[light saffron]

You just see a sum?

[fast garden]

i am sorry to tell you that it is nothing more than gibberish

[light saffron]

I disagree

respectfully

[fast garden]

okay

[light saffron]

First of all

Like oppenheimer said it is a sum

[fair storm]

Are you talking about the $\Sigma$

[woven radish]

Erebus

[light saffron]

no

I mean like what can you imagine from the formula

but I guess its nothing

[fair storm]

My opinion on it?

[light saffron]

I mean

Idk is this the right channel to ask this

[crystal dawn]

given that you don't seem to have a well-formed question I'm going to say this belongs more in a discussion channel

[fair storm]

It's nothing much

[light saffron]

.close

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[inland carbon]

how to find the limit when $x \rightarrow -1$ of \sum_{n=0}^{+\infty} x^{n^2}$

[woven radish]

bloubbloub
Compile Error! Click the reaction for more information.
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[inland carbon]

apparently it's 1/2 but idk why

[proud perch]

Uh this is the exact same as Σ (-1)^n. It's just an alternating sum of +-1 in pairs

So it doesn't really converge to 1/2 but the limit can just be treated as a geometric series

[proud perch]

Even and odd squares map exactly to even and odd numbers!

I'll leave that proof to you

[inland carbon]

ok but I'm interested in actual convergence

I should have specified $-1^{+}$

[woven radish]

bloubbloub

[inland carbon]

but I think I maybe got it thanks to you

[inland carbon]

how would you justify it in this case?

for x^n^2 I mean

[sand dove]

the series doesn't actually converge at x = -1

So no abel radial convergence th

Ok online I found a few things

[sand dove]

notably that this is the theta function applied to some specific values

[devout snow]

@inland carbon Has your question been resolved?

[inland carbon]

I mean I really want to say that the cesaro mean is uniformly convergent

since the cesaro value is 1/2 at -1 and we have pointwise convergence on (-1, 1), this would imply that the limit is 1/2

[devout snow]

@inland carbon Has your question been resolved?

[inland carbon]

moving to #real-complex-analysis so I don't have to keep this open

.close

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[gray oriole]

∀x∀y(x \neq y → ∃p∃q(p \neq q ∧ C(p,x,y) ∧ C(q,x,y)))

domain of x and y is the set of all points
domain of p and q is the set of all paths that connect two points
C(x,y,z): Path x connects two points y and z

[gray oriole]

Now I have to translate this into english

Any two distinct points are connected by at least two distinct paths?

$\forall x \forall y (x \neq y \rightarrow \exists p \exists q (p \neq q \wedge C(p,x,y) \wedge C(q,x,y)))$

[woven radish]

Sean [Ping On Reply Please!]

[harsh sierra]

I would use the word points instead of endpoints

[gray oriole]

done

[gray oriole]

is everything else correct

[devout snow]

@gray oriole Has your question been resolved?

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[gray oriole]

.close

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[timber pebble]

physics

[pseudo basin]

do you have a question to ask?

[wide basin]

no

sorry

[pseudo basin]

ok then .close this

[brittle inlet]

What

[wide basin]

.close

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[gray oriole]

,rcw

[gray oriole]

how to prove that these graphs are not isomorphic?

I mean I can see that they are not, but how to write it

as a proof

[ionic harness]

Longest cycle?

Well no there are none

Sorry longest path*

[gray oriole]

yeah

nno cycle

is there an invariant for longest path

in the first one longest path is length 2

in 2nd one it is 3

whats that shrug for

😭.

[ionic harness]

Because I have no idea if there's an invariant for it

[gray oriole]

then how to disprove isomorphism

[ionic harness]

If directions count for chromatic number then their chromatic numbers are different

[gray oriole]

both are bipartite

[ionic harness]

So directions don't matter then

Damn

[gray oriole]

.close

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[short hare]

yay!

[devout snow]

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[gray oriole]

The second part of the question says at least 5

But I am getting ceil(800k / 200k) = 4

[ionic harness]

Yes but it says more than 800k
ceil(801k/200k) = 5

[gray oriole]

yeah got it

so i have to take 800k + 1?

[ionic harness]

Yep

[gray oriole]

cool

[glass canopy]

What are the pigeons and what are the holes?

[ionic harness]

That's also a nice one

[gray oriole]

100 million pigeons?

[glass canopy]

And how many holes - considering we're working in pennies

remember the "less than"

[gray oriole]

are there a hundred pennies in a dollar

is this the calculation

[glass canopy]

Yeah :)

[gray oriole]

why is it 100mil in the denominator

[glass canopy]

We're not working in the Carolingian monetary system lol

[glass canopy]

1 mil dollars is how many pennies?

[gray oriole]

oh ok

got it

[glass canopy]

Good job :)

[glass canopy]

i'd recommend a diagral

[gray oriole]

diagram?

[glass canopy]

yes^

[gray oriole]

arent there gonna be two buckets

[glass canopy]

Got to turn off my phone

[gray oriole]

one for friends one for enemies

[gray oriole]

.

[glass canopy]

two walls i guess

[gray oriole]

can this be solved using only the pigeonhole principl

.close

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[compact hawk]

Uhhh idk where to start

[pseudo basin]

join OB

the semidisk falls apart into 4 pieces:

• sector AOB
• isosceles triangle OBD
• the target piece
• minor segment CD

[compact hawk]

Okok then what?

[brittle coral]

use this information to express area of target piece in terms of the other areas which you can find

[devout snow]

@compact hawk Has your question been resolved?

[devout snow]

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[devout snow]

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[scenic eagle]

can anyone help me find a way to solve this with pure geometry and properties of tangents and parabolas? the coordinate geometry way includes some calculations and i wanna find a good method for it but i can't seem to find a way with just geometry

[trim merlin]

@scenic eagle What did u try ??

I dont think their is excessive calculation when using coordinate

[scenic eagle]

no yeah not a lot I've got that done but i wanted a geometry way

[trim merlin]

I dont think there is one with pure geometory u need to use the equations of tangents and all that

But I'll leave the helpers to it

[crude niche]

maybe this could explain it? i didnt watch the video though https://www.youtube.com/watch?v=M0v7krRq3mU

[scenic eagle]

i tried using the reflection property that the tangent is angle bisector to line joining P to directrix and focus and taking that angle as theta then making a rhombus by joining those points but idk it got too complicated

[scenic eagle]

this is just using coordinate geo ive done that alr

[crude niche]

alr

[scenic eagle]

<@&286206848099549185>

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[ionic harness]

This one seems fairly easy

[ionic harness]

\begin{proof}[Proof of \textbf{45}]
  Let ordered pairs be defined as $(a,b) := \{\{a\}, \{a,b\}\}$.

  ($\implies$) Suppose $(a,b) = (c,d)$. 
Then $\{\{a\}, \{a,b\}\} = \{\{c\},\{c,d\}\}$. 
Because these two sets are equal, we have $\{a\} \in \{\{c\},\{c,d\}\}$. 
But since $\{\{c\},\{c,d\}\}$ only has one singleton element, $\{a\} = \{c\}$, i.e., $a = c$.
Similarly, we must have $\{a,b\} \in \{\{c\},\{c,d\}\}$, which must be its doubleton element $\{c,d\}$.
Then $\{a,b\} = \{c,b\} = \{c,d\}$.
Therefore we either have $b = c$ or $b = d$.
If $b = d$, we are done; if $b = c$, then $\{c, b\} = \{c\} = \{c, d\}$ and thus $d = c = b$, completing this direction.

  ($\impliedby$) Suppose $a = c$ and $b = d$. Then $(a,b) = \{\{a\}, \{a,b\}\} = \{\{c\},\{c,d\}\} = (c,d)$.
\end{proof}

[woven radish]

Coolempire2026

[ionic harness]

\begin{proof}[Proof of \textbf{46}]
  By definition, for $S \in S$, we must have $S \notin S$, which is a contradiction.
  Therefore, suppose $S \notin S$. Then $S \cancel \notin S \implies S \in S$.
\end{proof}

[woven radish]

Coolempire2026

[ionic harness]

Looking for any comment/evaluation on these two proofs

[heavy current]

they seem alright to me

[ionic harness]

Let $|S| = n$ and write an $n$-bit binary word consisting of all 0's, where each bit corresponds to one element of $S$.

We will show that this correspondence is in bijection with the subsets of $S$, and therefore by iterating through all of the binary words with $n$ bits (e.g., through counting), we may obtain every subset of $S$.

[woven radish]

Coolempire2026

[uncut crow]

this is not combinatorics

[ionic harness]

Yeah I said I was skipping to the next chapter

We would come back to the combinatorics later

Although 47 is combinatorics

Slightly

[uncut crow]

[heavy current]

I think you've got the right idea

[vagrant skiff]

what if c=d

[uncut crow]

[ionic harness]

\begin{proof}[Proof of \textbf{47}]
  Let $\phi : \mathcal{P}(S) \to \{0,1\}^n$ be the correspondence described above.
First note that $\phi$ is well-defined, because each subset of $S$ may contain only elements of $S$, so it will be representable within $n$ bits.

  Now, suppose that $\phi(S_1) = \phi(S_2)$. Then $S_1$ and $S_2$ contain the same elements, and thus $S_1 = S_2$ by definition. Therefore, $\phi$ is one-to-one.

  Next, let $b$ be a binary word of length $n$. Let $T$ be the positions of the bits that are $1$ and let $S'$ be the set of elements corresponding (importantly, $b$ is of length $n$, so $S'$ is well-defined). Because $S'$ only consists of elements from $S$ (given that bits of $b$ correspond with inclusion/exclusion of elements of $S$), we must have $S' \subseteq S$, i.e., there exists $S'$ such that $\phi(S') = b$ for all binary words $b$ of length $n$.
\end{proof}

[woven radish]

Coolempire2026

[ionic harness]

Isn't that what's writen at the end

Ah are you saying then the doubleton element doesn't exist

[vagrant skiff]

yes

[ionic harness]

So cardinality isn't sufficient to mark inclusion

Mm yes I should have done the 'or' idea at each step

I would've ended with like 8 cases 🥲

But good shout

[vagrant skiff]

why eight cases

you get either {a} = {c} or {a} = {c,d}

the former is trivial

the latter immediately collapses back to {a} = {c}

just don't say it must be the singleton element

[ionic harness]

a in {c, cd} -> a = c or a = cd
ab in {c, cd} -> ab = c or ab = cd -> b = c or b = d

Okay 5 cases

It felt wrong writing it but now I know why 😆

[vagrant skiff]

what is that second membership step

[ionic harness]

The one I used in the proof?

[vagrant skiff]

{a,b} in {{c},{c,d}}

So the correct split is

{a,b} = {c} or {a,b} = {c,d}

[ionic harness]

Yes

and {a,b} = {c,d} gives b = c or b = d

This is what I did in the proof

[vagrant skiff]

so
{a,b} = {a} or {a,b} = {a,d}

because we know a = c

[ionic harness]

I did {c,b} = {c} or {c,b} = {c,d} but same difference

It's all up there 👀

[vagrant skiff]

right okay

yeah

should be fine

[vagrant skiff]

this also looks good to me

[ionic harness]

The answer is yes but the proof is abhorrent to write

Well I saw that, at least for students it is

Let me see if I can skip steps here

[vagrant skiff]

shouldn't be too bad

[ionic harness]

Only because I'm taking liberties

Ah wait

No this is the easy one

The hard one is distributivity

Yeah that's the abhorrent one

You're right

[vagrant skiff]

cooly is always right

[ionic harness]

\begin{proof}[Proof of \textbf{40}]
  Let $x in A \triangle (B \triangle C)$. Then either $x \in A$ or $x \in B \triangle C$.

\begin{case}
  If $x \in A$, then $x \notin B \triangle C$, so either $x \in B$ and $x \in C$, or $x \notin B$ and $x \notin C$.
  If $x \in B$ and $x \in C$, then $x \notin A \triangle B$ (because $x \in A$), and thus $x \in (A \triangle B) \triangle C$.
  If $x \notin B$ and $x \notin C$, then $x \in A \triangle B$ and thus $x \in (A \triangle B) \triangle C$.
\end{case}
\begin{case}
  If $x \in B \triangle C$, so that $x \notin A$, then either $x \in B$ or $x \in C$. WLOG this is just the first case again tbh. If $x \in B$, then $x \notin A$ and $x \notin C$, so $x \in (A \triangle B) \triangle C$. Otherwise $x \in C$, and thus $x \notin A$ and $x \notin B$, so $x \in (A \triangle B) \triangle C$.
\end{case}

  Therefore, in all cases, if $x \in A \triangle (B \triangle C)$, then $x \in (A \triangle B) \triangle C$. The other direction follows similarly.
\end{proof}

[woven radish]

Coolempire2026

[ionic harness]

That was too much work as is

[vagrant skiff]

wait

coolemplud

what does symmetric difference mean

because your definition isn't right

[ionic harness]

In one or the other but not both

[vagrant skiff]

yes

so why did you write x in A or x in B triangle C

[ionic harness]

either

x in A or x in B triangle C

XOR

[vagrant skiff]

either includes 3 possibilities

can be both

[ionic harness]

Either means it can't be both

[vagrant skiff]

[ionic harness]

,w either or

[woven radish]

Results provided by WolframAlpha

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

Requested by coolempire93

[ionic harness]

need something more specific wolfram

Maybe it 's a british/US difference

[vagrant skiff]

hm

[vagrant skiff]

okay maybe you're right

bruh it's messing with my brain

okay

seems correct

[ionic harness]

What type of english do you all speak that either-or isn't taught in grade school to mean one or the other 😂

They teach it to us with neither-nor

[vagrant skiff]

this blud is making fun of me now

[ionic harness]

No I'm trying to learn

This is very intriguing

[vagrant skiff]

I've worked with some native american logicians who say either or to mean OR and not XOR

so that's probably why I think it means OR

[ionic harness]

Do you mean native american as in native american native american

[vagrant skiff]

no I meant

[ionic harness]

Or like native to the US

[vagrant skiff]

uh american

honestly

just say but not both

avoid ambiguity of language

[ionic harness]

That's quite interesting, using either as inclusive or

[ionic harness]

I didn't even know it was ambiguous

(neither did cambridge and merriam-webster, it appears)

[uncut crow]

your cheekiness will be your downfall

[ionic harness]

But maybe I'll start including something extra

Just so it's clear

[vagrant skiff]

see https://math.stackexchange.com/questions/2130327/does-either-make-an-exclusive-or

[ionic harness]

Nasty

Who knew

Well keeping that in mind as we continue to 41

[violet wind]

wait

helpers have pin powers?

thats wild

[uncut crow]

i think they get cleared when the channel is reset anyway

[violet wind]

probably

[vagrant skiff]

lancey must have lied to me when they said i'm close to helpful

[violet wind]

anyways all of these problems are just like statements about logic

[uncut crow]

undeniable facts

[violet wind]

so like, set xor is associative

[vagrant skiff]

undenacts

[violet wind]

is just for all a, logic xor is associative w/ a in A, a in B, a in C

which is true because logic xor is associative

[ionic harness]

That is how proof exercises work yes

[violet wind]

so you can just like make a truth table

[ionic harness]

I had to try my hardest not to say "look, it forms a group ok"

\begin{proof}[Proof of \textbf{41}]
  Let $A$, $B$, and $C$ be such that $A \triangle C = B \triangle C$.

  Suppose, WLOG, that $x \in A$ but $x \notin B$.
  If $x \in C$, then $x \notin A \triangle C$ but $x \in B \triangle C$, a contradiction.
  Therefore $x \notin C$. But if $x \notin C$, then $x \in A \triangle C$ and $x \notin B \triangle C$, a contradiction.

  Therefore we must have $A \subseteq B$, and since the proof was without loss of generality, $B \subseteq A$.
  Thus if $A \triangle C = B \triangle C$, then $A = B$.
\end{proof}

[woven radish]

Coolempire2026

[violet wind]

here's another way

suppose A (+) C = B (+) C

uhh

im just gonna call it xor

suppose A xor C = B xor C

then (A xor C) xor C = (B xor C) xor C

applying associativity, A xor (C xor C) = B xor (C xor C)

A xor {} = B xor {}

A = B

[ionic harness]

XD not you applying the group action

I never actually proved that XOR preserves over =

As in applying it

[violet wind]

does it need to be proved?

surely that's just like a basic fact about logic

if you have any operation and you apply it to both sides of a true equality the equality stays true

[arctic field]

define f(A) = A xor C
A = B => f(A) = f(B)

[vagrant skiff]

Coriol Anus

[ionic harness]

Yes that does make sense

Now I have to figure out why that doesn't directly imply cancellation law

(which is what we proved here)

Like what operation can hopelessly mangle the objects so they're not equal anymore

Oh non-invertible one?

Yeah

Non-invertible one

Okay makes sense

[violet wind]

well that can turn non-equal objects into equal

but it can't turn equal into non-equal

thats impossible

[ionic harness]

Under which equal definition

😂 nonono

Yes I got it

[violet wind]

the one that satisfies the substitution property of equality

[ionic harness]

This is interesting

So cancellation law proves that the operation is one-to-one

Since it proves the reverse of this

And since XOR is well-defined we have the equality part dreyuk mentioned already

Interesting

[vagrant skiff]

the same proof works for any function symbol f

[ionic harness]

Yes

Only if it's well-defined though right

[vagrant skiff]

well a function symbol is automatically well defined

[ionic harness]

Lovely, very interesting this

[violet wind]

as long as it's a function

also you're proving the cancellation law, you can't use it

it's not a fundamental property of equality, while A=B => F(A)=f(B) is

[ionic harness]

Well yes

[ionic harness]

I meant that the cancellation law is what we proved here not the implication 😅

Because we definitely didn't prove that

But yeah

I should add that to my list of things

Prove that a function of the binary operation with a fixed element is one-to-one == prove that the operation has cancellation

I guess if you can prove it with inverted positions maybe that shows it's commutative

As well

Okay now the section on functions

I can learn what all these words that I'm using mean

I'll be honest I always called it an injection

Like bijection or surjection

What is injunction

[violet wind]

injection but worse ig

[ionic harness]

Jesus why is this section so long

I've just had basically everything there is to know about the basics of functions thrown at me all at once

Including asymptotic notation

And now partial functions

[uncut crow]

i come back here every once in a while and every time it’s still no combinatorics

[ionic harness]

Yes those will likely not be back till tomorrow

[iron sun]

are u asking for help for this or?

confusing

[ionic harness]

I am in the process of writing the proofs

It takes a moment for me to finish 😅

[iron sun]

yeah thats fine no hurry

was just wondering why u sent it

[ionic harness]

\begin{proof}[Proof of \textbf{34}]
  Let $f$ and $f \circ g$ be one-to-one functions and suppose that $g$ is not one-to-one.
  Then for some $x,y \in \mathrm{Dom}(g)$, $x \neq y$, we have $g(x) = g(y)$.
  But since $f$ is one-to-one, this means that $(f \circ g)(x) = f(g(x)) = f(g(y)) = (f \circ g)(y)$ and $f \circ g$ is not one-to-one, contradicting our original assumption.
  Therefore, if $f$ and $f \circ g$ are one-to-one, then $g$ is one-to-one.
\end{proof}
\begin{proof}[Disproof of \textbf{35}]
  I wish more was said about the codomains of these functions, but $g$ may be onto the domain of $f$, thus making $f \circ g$ onto, but $g$ but not onto its own codomain.

  In other words, I feel this question is malformed.
\end{proof}

[woven radish]

Coolempire2026

[violet wind]

likely it may be assumed that the codomain of g is the domain of f, but not necessarily the range of g

[arctic field]

one does not compose functions if codomain doesn't match domain

[ionic harness]

Oh I didn't know that 🤔

[violet wind]

well

[ionic harness]

Like sqrt(x) can be written as R -> R>=0
But I can compose that with f(x) = x +1 from R to R

Even though they don't 'match' one's a subset

[violet wind]

yea it's easy to expand the codomain of a function

or restrict the domain

[ionic harness]

Okay let me rewrite my answer

[vagrant skiff]

truly wiser words have been said

[ionic harness]

In the book all they said was range subset domain

So knowing to set them equal is helpful

otherwise I would have just like chosen f(x) = lnx and g(x) = sqrt(x) : R -> R

Ah they marked it in the definition

Sneaky

\begin{proof}[Proof of \textbf{35}]
  Let $f : B \to C$ and $g : A \to B$, where $f$ and $f \circ g$ are onto $C$, and suppose that $g$ is but not onto $B$.
  
  Suppose that there exists a subset $B' \subseteq B$ such that $f(B') = C$ (possible if $|B| > |C|$).
  Then if $g$ is onto $B'$, all the given conditions still hold, meaning that $f$ and $f \circ g$ onto does not yield that $g$ is onto.
\end{proof}

[woven radish]

Coolempire2026

[ionic harness]

And we have the simple example of
A = {1,2}
B = {1,2,3}
C = {1}

g(1) = 1
g(2) = 2

f(1) = 1
f(2) = 1
f(3) = 1

[sand dove]

If C is infinite then the comment about their cardinals is sufficient but not necessary

[ionic harness]

I didn't even think about the infinite case

I guess as a baby mathematician I should probably start thinking like that

[ionic harness]

How does this version look

[zenith spoke]

for the proof of 34, g(x)=g(y) implies (fog)(x)=(fog)(y) just because f is a well defined function, it doesnt have anything to do with the injectivity of f

[ionic harness]

Last one before I go