X=y=2 passes.
#elementary-number-theory
silent lantern
Ur other question
sacred junco
Wait which other question?
silent lantern
Then after n is a prime it doesn’t work, show that it doesn’t work for n= p to n=2p-1
Ur original one
Think about p!
As a common denominator
graceful notch
X cap N = N
sacred junco
Dude thats basically bertrand's postulate.
silent lantern
It’s not
graceful notch
the series 1/n is divergent
graceful notch
the series 1/n grows without bound, meaning it will eventually hit every natural number
swift shard
It will eventually go above every natural number. Whether or not it hits any of them isn't so simple
sacred junco
X=1^-1+2^-1+......+N^-1
graceful notch
oh right it might not hit them
sacred junco
We want such a number to be natural
So its pretty clear n=1
But does that prove it?
No
silent lantern
Pick a prime p.
1 + 1/2 + 1/3 + ... + 1/p is pretty clearly not an integer.Assume it is an integer, and we can make a common denominator as follows: (p! + p!/2 + p!/3 + ... p!/(p-1) + (p-1)! ) / p!
All terms of the numerator are divisible by p except for the last one which isn't (because p is prime, duh). Therefore the numerator isn't divisible by p and can't be divisible by p!.
Now, fixing p, we prove 1 + 1/2 + ... + 1/x is not an integer for p <= x <= 2p-1, by the same argument... the numerator of the common-denominator-expression contains exactly one term that's not divisible by p, therefore the sum of the numerator isn't divisible by p and the fraction isn't an integer.
from friend of mine
@sacred junco
sacred junco
Hmm interesting arguement
And also spot on.
But how would u conclude ur answer from that?
sacred junco
Im dumb af so yeah
sacred junco
Why do u do this for a prime?
Well thats what i would do.
And prove it wont be an integer for n+1
And then make the final case unless n=0
silent lantern
idk it was more intuitive for me
cuz i do olympiad so i think based on prime numbers
sacred junco
Ok so ill do it for naturals
Essentially the same arguement as yours.
I still havent figured out what u meant by. p/2p and p/2p-1
Idk the reference so......
graceful notch
1 is an integer.
suppose the sum to k-1, call it S, is an integer.
then, the sum to k is
S + 1/k.
suppose this is rational, then
S+1/k = a/b
where b>1 and gcd(a,b)=1
then
S(bk) = ak -1
k(a-Sb) = 1
so k|1, but k>1, a contrqdiction
silent lantern
$$\frac{p}{2p}, \frac{p}{2p-1}$$
graceful notch
does this work ?
stoic basinBOT
silent lantern
All terms of the numerator are divisible by p except for the last one which isn't (because p is prime, duh). Therefore the numerator isn't divisible by p and can't be divisible by p!.
replicate thta
for p to 2p-1
sacred junco
@graceful notch yeah i think so..thats what i also said
@silent lantern what good will that do? Sorry if im being intolerable but i serioisly dk
silent lantern
well i dont wanna spoil teh question
sacred junco
Dm
silent lantern
for u
sacred junco
Ive already solved it now with the nartural no.arguement.
silent lantern
lemme see if what @graceful notch posted works
sacred junco
Plus yaya's soln is basically the same as mine.
graceful notch
i dont think it's riggt actually lol
sacred junco
Yeah if that doesnt work then im wrong too.
graceful notch
uep whoops
sacred junco
Its equal to b
sacred junco
b=k(a-sb)
silent lantern
for what
graceful notch
for the questiob
graceful notch
k|b not b|k
sacred junco
Nopes.
silent lantern
k|b means k is a factor of b
sacred junco
Ok hold up u guys.
graceful notch
idk lel
silent lantern
not the other way round
sacred junco
S+1/k=a/b
silent lantern
where S is an integer
graceful notch
the proof cannot be right
1 + 1/2 is not an integer
silent lantern
he assumed its an integer to get a contradictino
sacred junco
Ok.
silent lantern
imao @graceful notch we're trying to prove it doesnt become an integer
sacred junco
The hell are we doing.
Ok see
1/x+1/y can never be an integer for x,y>2
Easily provable by induction
silent lantern
ohh i thought u were doing the sum of reciprocals
sacred junco
Nah not by induction.
Wait wait hold up.
Let me write this stuff.
If i cant do it
The. Please tell me ur complete method
a8
silent lantern
k
sacred junco
Read 2nd then 1st.
@silent lantern
The last bit isnt that rigorous but i had no clue how to prove thag
I mean the stuff on 2nd page.
silent lantern
@sacred junco NOTATION PROBLEMS
sacred junco
Wdym?
silent lantern
oops brain freeze
sacred junco
Because its wrong right?
silent lantern
so what u did is u assumed if n is an integer and sum is not integer
silent lantern
brb get off train
sacred junco
Dont bother its wrong.
silent lantern
yh it had a logic problem
sacred junco
No.
silent lantern
imao u assumed it was not an integer
sacred junco
A case exists which contradicts my proof
silent lantern
then proved it wasnt an integer
silent lantern
n is defined to be a natural number
ok nvm ur proof works
needs some tweaking
supupose sum from n=1 to n=k is not an integer
proof that for n=k+1 its not an integer
n ure done by strong induction
brave steeple
anyone know how to solve (5000001 * 1468162901)^11 ≡ x mod 33?
wide shuttle
well, since it's a lot of big numbers, i'll just assume that the size of the number doesn't matter, so we can generalize the equation
(ab)^c ≡ x mod 33
brave steeple
have to find the smallest nonnegative integer x
silent lantern
It’s Chinese remainder theorem @brave steeple
U know the number is 0 mod 3
And it’s also -5 mod 11
Which means it’s 6 mod 33
6 mod 33 to the power of 1468162901
Can be done by same method
Reduce the power mod 33 by the same method through considering mod 3 and mod 11. Compute the final simplified power, and do that mod 3
sacred junco
So ive solved 10,11,12,14
But 13 is sets which i have no knowledge of.
And 9th i can figure out
Need a little boost of idea.
@silent lantern
silent lantern
@sacred junco in other words, their intersection will be the numbers which can be represented simultaneously either as a sum of 7 , 8 or 9 consecutive integers
So if u let a + a+ 1 ... + a+6= b+b+1 ... +b+7= c +c+1 ... +c+8
Where a b c are integers, what do u get
sacred junco
7a=8b+7=9c+8
But they didnt use different variables.
In all the sets its the same n.
@silent lantern
sacred junco
Oh ok.
The sum is equal
Yeah i get it.
Currently trying to solve it.
But there are two equations and three unknowns
So im using divisibilty.
silent lantern
Uhh there are three equations
sacred junco
Not it that way man.
But i think i can solve this now
Big thanks for the help!
Now lets start 9th
silent lantern
No I mean u equate a and v, b and c and a and c
sacred junco
Thats what im doing.
sacred junco
Uh just contact a problem manager
Personal preference.=senpai or gomez or woog
So i found solutions @silent lantern
a=89
b=77
c=9
No sorry thats wrong.
silent lantern
It’s literally plug n chug
sacred junco
Holy hell what am i doing.
silent lantern
8 is too small
sacred junco
I had the habit too lol
sacred junco
nameception?
silent lantern
@sacred junco Entire write up. 0 errors for real this time
silent lantern
the 1 u last gave me
sacred junco
Kk
ornate crater
Need help with Groups/Rings/Fields. If anyone can help me please DM me 😦
swift shard
@ornate crater
You'll get a much better response on the server, a lot of people know abs alg here.
nocturne laurel
pliant cobalt
upbeat wren
Hmm okay. I'd show that a solution exists and then prove it is unique. 😃
upbeat wren
A start ...
Lemma: Let B be the set {x: x = sk + r} where s, r, and k are integers and s > 0 and 0 <= r < s. WTS B includes Every integer.
Does B include 0? Yes k = 0, t = 0 works.
Assuming B includes j, does b include j + 1?
For j we have a particular s, k, r. If r + 1 < s, then (s, k, r+1) works. If not r + 1 < s, r+1 Must equal s (since r and r+1 are consecutive integers there is no way s is less than one and greater than the other ... there is no way there is an integer s between consecutive integers r and r + 1), and s, k+1, 0 works.
So if it works for j, it works for j+1.
Similarly if it works for j, it works for j-1.
And we have a "two-tailed" induction so ALL integers > 0 and ALL integers < 0 and all integers = 0 work!
So all integers are in set B.
Uniqueness: suppose by way of contradiction s2, k2, r2 and s2, k2, r2 both work.
That is there exists c such that c = s2k2 + r2 and c = s1k1 + r1 ... with all the other restrictions.
shell ore
hello
can anyone help me with a few excersises
I've kinda forgot about them and they're due in 1 hour..
is the following true with all sets A, B and C:
$$A \cup C = B \cup C \Rightarrow A = B$$
stoic basinBOT
sturdy dirge
False.
shell ore
well, how do I write the proof for it?
swift shard
Imagine A and B are both subsets of C. Then A ∪ C = B ∪ C, but it's not necessary that A = B
dry acorn
whats i+1?
swift shard
It's 1 + i
dry acorn
whats that equal to?
upbeat wren
(error sorry!)
topaz idol
Ye I would say false
Draw a venn diagram and look at it
Would prove false by illustration
sacred junco
Can anyone help me out understanding vector spaces?
Like if something is in R^3 does that mean its not a vector space if one of the elements is not in R^3? Like in R^2?
If say the condition said V was a vector space in R^3?
silent lantern
@sacred junco a good idea to realise is that R^2 would be our normal cartesian coordinate system
and that its a vector space
with this visualisation, although it is hard to picture R^n, but i hope it eliminates on some of the definitions of x^i in R^n
sacred junco
i realize i worded my question a little strangely; I think I meant P^3 and not R^3
but thank you @silent lantern
silent lantern
ok
sacred junco
hi guys, is it ok to ask a question here? math related of course
sharp gulch
Fråga bara
sacred junco
quick question: if W is a subspace of R^2, it must contain the vector (0, 0)
im assuming no but i remembered a vector space must contain 2 proper subspaces which are the zero subspace and itself
sullen rune
yea needs the 0 vector
jovial spindle
for R^2 the zero vector would be (0,0) and if it's a subspace you have the zero vector is in W
sacred junco
what if the subspace is defined as (x, x) where x is greater than 0?
would it still have (0, 0)?
or if they do not specify can we just assume (0, 0) is there
jovial spindle
the 0 vector would always be present yeah
wait
no
it wouldn't be a subspace then
sacred junco
ooh
Vector space is a group
that makes things much more solid for me
thank you @jovial spindle
do you mind if i ask ye another question within the same veil?
jovial spindle
shoot it
sacred junco
You can use t!rep @Ping to give a person rep
t!rep @jovial spindle for being so helpful
unique vaultBOT
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sacred junco
my next question is about a vector space in the P_n group
if P_k is a subspace of P_n, is P_j a a subspace of P_k?
i thought not because P_k could be (1, 1 ,1 ,1) and p_j being (1, 1, 1)
i think i should note P_k is said to be the set of all polynomials of degree less than or equal to k with standard operations
im assuming based on how my text words the problem it actually is but i have no idea how to prove it
sacred junco
yeah
jovial spindle
sorry I don't necessarily know the notation but I know you're talking about the set of polynomials
sacred junco
i could take a picture for you to read the problem; i might of butchered my shortening of it
jovial spindle
well I'd think of it this way, you could have any ai set to 0 for any x^i where i>j
if that makes sense
sacred junco
ai?
jovial spindle
in abstract algebra we'd always just run down the list of things for a definition if we're proving a problem like this
ai as in, p_n is of the form a_nx^n + a_(n-1)x^(n-1) ... right
oh god that got butchered
sacred junco
im assuming you meant 'i' but if I set x^i where i is greater than j and i is 0...
jovial spindle
well then it's the 0 vector
0 vector on its own is trivially the subspace
of each space
you'll have additive identity, closure
oh woops
I mean to say if you consider the polynomials with degree higher than j
and make their coefficient 0
you see that polynomials of degree j are contained in those spaces
sacred junco
so k (0, 0, 0, 0) contains j (0, 0 , 0) so we can say j is contained in k?
er is a subspace rather*
jovial spindle
I'd believe so, I'm not used to the particular notation because I'm generalizing from abstract algebra
but I'd imagine it's isomorphic to the same groups
so yes
if someone else could chime in that'd be more helpful probably to double check
sacred junco
how might you define isomorphic in this scenario?
jovial spindle
I'd map the vector spaces in my head to how'd you represent P_n and see a 1-1 correspondence pretty quickly, so considering (a_2 x^2 , x_1 x^1 , a_0) mapping to p_2 = a_2 x^2 + a_1 x^1 + a_0
there'd be some more you'd determine specifically to see it was isomorphic and I might be missing something in looking at it as a vector space
sacred junco
sorry to interrogate you so; the text is dreadfully plain in explaining anything and i feel like taking everyone's bit of understanding will help me understand vector spaces
i am albeit shaky with the concept of vector spaces as i cant completely prove how a space is a vector space to begin with
such as something being in P_3 and lesser being a vector space but P_3 on its own not being a vector space
sacred junco
i meant P_2
sacred junco
P_2 is said to fail and the example my text gives is that if p(x) = x^2 and q(x) = 1 + x - x^2 you get p(x) + q(x) = 1 + x
so my vague understanding from there was that because the result is not in P_2 it is not a subspace
jovial spindle
hmm I'm just imagining in my head you could set the coefficient for the x^2 component to 0 so I don't see why that's the case for me either, hopefully someone who can see it swoops in
since by that same logic you could claim that there's no closure for any space of k
sacred junco
<@&286206848099549185> could one of you explain to me how the set of all P_2 is not a vector space but the set of all P_4 and less is?
silver solar
It all has to do with the fact that the coefficient of the highest order term needs to be non-zero in the first case, with the restriction lacking in the second case, making it a vector space
ornate crater
Groups-Rings-Fields Topic: So I understand that for a set to be a GROUP, it must follow certain properties(closure, association etc.). But what I don't get is how do I know during validation of these properties if i must use addition or multiplication for example (a+b=x) or (a.b=x) to prove these properties
runic arch
A group has only one operation, so it can either be addition or multiplication
The context should make it obvious
static sparrow
it can be something else too
runic arch
yes totally
sacred junco
t!rep @silver solar
unique vaultBOT
🆙 | Charlatan, you can award more reputation in 0 hours, 24 minutes and 45 seconds.
silver solar
Why thank you good sir
vast vessel
yw
sacred junco
So a question
Let's say I have some arithmetic progressions
px+m
Like
5x+3
11x+2
1/p natural numbers will be members of a set created by the progression of p
If we were to count the number of naturals in the set less than n we could approximate this by n/p
I know that at most this will be off by 1
But for two or more progressions
Where the sets are joined
silent lantern
\frac{N-m}{p}
That
But for the intersection of two progressions, find a formula that finds their intersection
sacred junco
I just want to know the maximum error
silent lantern
Cough Chinese remainder theorem
sacred junco
Yeah I know
silent lantern
It’s still floor or ceiling functions
silent lantern
*floor
sacred junco
A set of progressions
sacred junco
What would be the maximum error I could expect from the frequency of occurrence in the total distribution
shell ore
can anyone help me understand borel sets and open sets
or "simple" borel sets and "simple open sets"
I think this is the definition of "simple open sets" but I'm not 100% sure
$$\bigcup_{i=1}^{n}B_i$$
stoic basinBOT
warped jay
im having trouble understanding the twin prime conjecture
does anyone have a good understanding of it
vast vessel
what about it? @warped jay
warped jay
i think i understand it, correct me if i'm wrong
What the 70 million means is that, if you have two consecutive prime numbers of which the gap between these two prime numbers is greater than 70 million (say the gap is x), there is a finite amount of times that this can occur, by this I mean, there are only n more times that pairs of consecutive prime numbers with this gap x can occur.
Say two consecutive prime numbers have a gap of 100 million between them, the amount of occurrences of these pairs is finite so it might occur only 10 more times (to infinity).
torpid peak
@vast vessel
vast vessel
what
@torpid peak what does 70 million and 100 million have to do with anything?
The twin prime conjecture is merely the statement that there exists infinitely many twin primes.
silver solar
You're wrong @torpid peak
Zhang's 70 million prime gap theorem just states that there are infinitely many pairs of primes that have a gap of 70 million between them (or less)
It doesn't make assertions on bigger prime gaps
warped jay
so does that prove the twin theory conjecture?
exotic pumice
The current record is 246
Which means there are infinitely many pairs of primes at most 246 apart
This does not prove the twin prime conjecture, for which 246 would have to be replaced by 2
warped jay
they came down to 6
silver solar
Assuming a certain conjecture tho
warped jay
yea
not assuming any conjectures its come down to 256 or something
sorry this was already mentioned lol xd
leaden peak
Is there a term for reducing a number to the smallest member of its equivalence class mod n? For example reducing 26 mod 3 to 2? I keep wanting to say simplify, or reduce to simplest form, but that can't be the right way to describe it.
tame comet
Finding the unique modulus
Is a better way
Think of it similar to saying something like
Cos(90) = cos(450)
@leaden peak
leaden peak
Hmm, that works as a term, at least better than simplest form. Come to think of it, would you also use unique as the term for this situation in trigonometry?
@tame comet
tame comet
The point is that there are only a certain amount of unique terms for the moduli.
For numbers mod n, for instance
the main values of the remainder are {0, 1, 2, ... , n-1}
Modulus does allow freedom with it
i.e (8 is congruent to 5 mod 3 is equal to saying 8 is congruent to 2 mod 3 based on the definition)
It's not simplify, per se. It's more like principal modulus (the value within that remainder set)
leaden peak
I see
silent lantern
guys
how do we show that the last non zero digit of a factorial is not eventually periodic
silent lantern
<@&286206848099549185>
safe sage
"This can be deduced from the fact that the sequence can be obtained as a fixed point of a morphism."
source: https://oeis.org/A008904
not sure what that means but it may be helpful to you
sacred junco
How is everyone so smart
silent lantern
@safe sage it’s supposed to be elementary
Without any group theory cuz I don’t know them
lost hawk
Is pi × 10 to the Power of infinity rational or irrational number?
wide shuttle
erm... let's be safe and say irrational
nevermind that you cant take 10 to the power of infintiy i dont think
unless you apply the partial sum of a geometric series to the infinity sum of a geometric series
1+x+xx+xxx+xxxx.... + x^n = $$ \frac{(x^(n+1)-1}{x-1} $$
stoic basinBOT
Rendering failed. Check your code. You can edit your existing message if needed.
wide shuttle
$$ \frac{x^n * x - 1}{x-1} $$
stoic basinBOT
stoic basinBOT
wide shuttle
which is the same as $$ \frac{-1}{x-1} $$
stoic basinBOT
wide shuttle
so if you say they are equal at infinity, then x^(n+1) = x^n * x = 0,
so, then it would be pi * 0... which would be rational, but i don't think that its genereally accepted to raise a number to infinity
wild zinc
the infinite sum doesn't converge
wide shuttle
of course for abs(x)>1, yeah
wild zinc
|10| > 1 :^)
wide shuttle
ik
its just that you can sum divergent series, and even by that method, it doesn't really make sense to take pi * 10^infinity and try to determine its rationality
full summit
can someone please help me with thi
swift shard
Basically, how many ways can you group the elements together such that a and c are in the same group, such that d and c are in the same group, but c and e are not in the same group?
Before anything else, it's fairly clear that a, c, d will always reside in the same group, and e will not reside in this group. Thus, the places for b are all we can choose.
b can go with a, c, d
b can go with e
b can go by itself.
Three different possible groupings, three different possible equivalence relations.
full summit
thank you for the reply but what exactly is a equivalence relation? from what i understand its this
i dont understand this^ relates to the question
swift shard
Equivalence relations, on the general set, is a grouping of the set such that every element appears exactly once in some set
You may have heard "reflexive, symmetric, transitive" before
full summit
yes
swift shard
So an equivalence relation on {a, b, c, d} may be {(a, c) (b) (d)}
In this, a and c are equivalent.
full summit
oh if it were a~b then it would be {(a,b)(c)(d)}?
swift shard
Ya ya. In the question, the only three possible equivalence relations are
{(a, b, c, d), (e)}
{(a, c, d), (b, e)}
{(a, c, d), (b), (e)}
full summit
oh okay thank you
if b wasnt there then it would just be one equivalence relation right?
mint sorrel
i need help please, i cant understand this
the first one is countable because there is an injection from {0,1} --> N
is this right?
sacred junco
WHat is 1+1
swift shard
@mint sorrel
The number of FUNCTIONS from {0, 1} to N. That is, how many ways can you map 0 and 1 to the naturals?
This is countable for a similar reason the rationals are countable
sudden bridge
I might be misunderstanding something, but does “mapping 0,1 to N” mean you need to be able to uniquely create all natural numbers with some input 0,1?
That’s straight up impossible isn’t it?
mint sorrel
i dont understand this very well, but i am assuming there are only 2 ways map {0,1} to N
sudden bridge
But you haven’t accounted for 2
swift shard
Another may be
f(0) = 18869
f(1) = 813952980428
mint sorrel
so there are infinte ways to map them to naturals?
sudden bridge
Oh so you just have to be able to create some natural integer, you don’t have to be able to answer f(0;1?) = 168?
mint sorrel
and for part b it is not a valid function since all domain elements arent mapped to codomain?
sudden bridge
It would be finite if you assume natural numbers are finite
swift shard
Wat
But they're not
mint sorrel
and for (b) it is not a valid function since all domain elements arent mapped to codomain?
swift shard
There are infinite ways, in fact there are countably many ways to do the mapping in a)
For b) consider the mapping
f(1) = 1
f(2) = 1
f(3) = 1
f(4) = 1
...
Such a mapping might be
0100111100110000...
I hope you can recognize that looks a lot like binary, and binary is countable
plain shore
Is infinite binary countable?
swift shard
Not in both directions, so infinite binary with a decimal place is not
But
10000... → 1
01000... → 2
11000... → 3
00100 → 4
Ect
mint sorrel
thank you kaynex life saver
plain shore
Ah yeah, I get yah. Thanks
mint sorrel
i have tried to do euclidian algorithm but cant seem to prove that they are same, for gcd(a,3a+b) always getting remainder
plain shore
The Euclidean algorithm hinges on the fact that gcd(a, b) = gcd (a, b mod a)
If you are allowed to use this result then gcd(a, 3a + b) can be manipulated to gcd(a, b) because 3a + b = b modulo a
mint sorrel
how does 3a+b = b mod a?
wide shuttle
a = nm, b =nj,
3a + b = 3nm + nj
the greater common divisor is n
not of rigor, but that's a basic version
wide shuttle
ye, but the gcd of 21 and 5 are still 1
swift shard
mod "splits" over addition and multiplication. So
na + b (mod a) = b
plain shore
x+y mod n = x mod n + y mod n. This is a property of mod
So 3a + b mod a = 3a mod a + b mod a
Mod is the remainder of a division: a mod b is the remainder when a is divided by b. So 3a mod a must be zero because the remainder when you divide 3a by a is zero because it goes in perfectly.
So 3a + b mod a = 0 + b mod a = b mod a
If that makes any sense?
lusty siren
Could someone help me out with this problem
Is equivalent relations of a set the same as partitions of a set
sullen rune
nah it's a set of pairs of elements
where the pair (a,b) means a=b, and this set of pairs has to follow the 3 requirements of an equivalence relation
lusty siren
Could you elaborate a bit more because I’m confused on this part
I get each pair has to be reflexive, symmetric and transitive but how do I know that the pair meets those requirements
Like could you explain it given a set of {1,2,3}
Or {x,y,z}
swift shard
@lusty siren
Yes, it's the partitions of a set
lusty siren
Im so confused how can the partitions of a set be equivalent relations
swift shard
{(1),(2),(3)}
{(1,2), (3)}
{(1), (2,3)}
{(1,3), (2)}
{(1, 2, 3)}
I believe are all of them
Anyway, something like {(1),(2,3)} means 2 is equivalent to 3, but 1 isn't equivalent to either
Equivalence relations will group the elements up into sets of "equivalent elements".
lusty siren
So {(1),(2),(3)} means that none of the elements are equivalent
I think I get it now though
thank you for saving me from a headache
swift shard
Np. Feel free to ask if you have anything else!
lusty siren
Isnt {(1),(2),(3)} the same as {(1,2,3)}?
swift shard
No, because {(1,2,3)} means all elements are equivalent
lusty siren
Oh ok thanks mate
silk pilot
I stumbled upon transfinite numbers and apparently Natural numbers and Rational number have the same cardinality as Real numbers. What about an ordered pair like (R,R)? Would that have the same cardinality as R?
I have no knowledge of infinite set theory aside from wikipedia stuff
wild zinc
|R^n| = |R|
sturdy dirge
|R^N| > |R| or |R^R| > |R| ?
wild zinc
|R^N| = R, |R^R| > |R|
silk pilot
Does |N^N| > |N| work?
wild zinc
ya
silk pilot
so then is |R|=|N^N|?
wait, if there are infinite primes then N can be partitioned into subsets of primes which can each be partitioned by prime indexes infinitely. If N has the lowest possible cardinality, then each infinite partition would = |N|. Wouldnt this imply |N^N| = |N|?
vast vessel
What does "N^N" mean?
vast vessel
What does that mean?
N^k for natural k is fairly well defined
But what do you mean for N^|N|?
silk pilot
Coordinates on a |N| dimensional hyperplane I guess
I always understood R^k that way
Idk I probably shouldn't even attempt this sort of math
Considering I managed to fail algebra
wild zinc
N^N is sequences of natural numbers
using decimal expansions it's pretty easy to construct a surjective map from N^N onto [0,1]
which is enough to show |N^N| > N
silk pilot
But cant Q be surjectively mapped to [0,1]?
pulsar orchid
Hey i am sorry if im asking in the wrong channel but id like to know how to approach my situation
i am supposed to check the following statement for reflexivity, symmetrie and transivity
im not entirely sure if thats correctly said
R:= 'not0' on IN x IN
sturdy dirge
Why |R^N| = |R| ?
pulsar orchid
that gets me even more confused
the statement i have to proof on is written as i typed it in here
pulsar orchid
i think i asked the question in the wrong channel sorry :D
stoic basinBOT
Rendering failed. Check your code. You can edit your existing message if needed.
haughty falcon
so, fiddling with the magic square of squares problem I noticed that the following equation has pairwise solutions that grow exponentially with a factor of roughly 6. and I have no idea why
$$a^2+7^2=2b^2\newline
17^2+7^2=213^2\newline
23^2+7^2=217^2\newline
103^2+7^2=273^2\newline
137^2+7^2=297^2\newline
601^2+7^2=2425^2\newline
799^2+7^2=2565^2\newline
3503^2+7^2=22477^2\newline
4657^2+7^2=23293^2\newline
20417^2+7^2=214437^2\newline
27143^2+7^2=219193^2$$
stoic basinBOT
haughty falcon
I'm only considering coprime solutions in this, and am skipping the 1^2+7^2 and 7^2+7^2 solutions (the latter of which is trivial and not coprime)
upbeat wren
So here's a quick theorem that might help. The center of every 3x3 magic square is the average of both sets of opposite corners (and consequently both the non-center squares in the middle row and middle column). That is to say if a magic square of squares exists for a 3x3, then there are 4 triples (a, b, c) where a^2 + b^2 = 2c^2 where c is a constant for all four triples.
Want To Prove: Let a 3x3 square be given with magic sum M and entries: x1, x2, x3, y1, y2, y3, z1, z2, z3. The center number is the average of the opposite corners.
Note that x1 + y1 + z1 = M. x3 + y3 + z3 = M, y1 + y2 + y3 = M, x1 + y2 + z3 = M, and x3 + y2 + z1 = M.
Add the last three equations and subtract the first two:
y2 + y2 + y2 = M.
x1 + y2 + z3 = y2 + y2 + y2
x1 + z3 = 2y2 or (x1 + z3)/2 = y2 😃
haughty falcon
that is what I did to even consider the equation above 😉
the interesting question was: "why do the solutions for a^2+7^2=2b^2 appear pairwise and is there a formula to generate them?"
wild zinc
my guess is there's some sort of way of generating one solution from another, and it just so happens that there are two smallish solutions that aren't from generating eachother
notably if (a,b) is a solution, then (3a+4b,2a+3b) is a solution
(3a+4b)^2 + 7^2 = 9a^2 + 24ab + 16b^2 + 7^2
2(2a+3b)^2 = 2(4a^2+12ab+9b^2) = 8a^2 + 24ab + 18b^2 = (3a+4b)^2+7^2 - (a^2-2b^2+7^2)
however since (a,b) is a solution, a^2-2b^2+7^2 = 0
and hence (3a+4b)^2+7^2 = 2(2a+3b)^2
so (3a+4b,2a+3b) is a solution iff (a,b) is a solution
(to get iff, note that the equations showed that (3a+4b,2a+3b) was a solution iff a^2-2b^2+7^2 = 0, which is equivalent to (a,b) is a solution)
using euclid's algorithm it's not too hard to show that gcd(3a+4b,2a+3b) = gcd(a,b)
as for showing these are the only ones
there may be some way involving inverting (a,b) -> (3a+4b,2a+3b) but I'm not too sure
but certainly this gives a way of characterising the solutions you've found into two families: (a,b) = (17,13), (103,73), (601,425), (3503,2477) etc. and (a,b) = (1,5), (23,17), (137,97), (799, 565) etc.
@haughty falcon
haughty falcon
❤ @wild zinc
wild zinc
for more stuff about this and similar equations look up pell equations
haughty falcon
gonna do that
blissful venture
Looks like pigeonhole principle
wide shuttle
Is that a pigeon?
plain matrix
Can I eat it?
sacred junco
$$ 666^{666} = 1 \text{} mod{13} $$
stoic basinBOT
sacred junco
=pup calc {\sqrt{Subfactorial{Hyperfactorial(0!+0!)}}}!+0!
sacred junco
;-;
vast vessel
?
sacred junco
crude thicket
Hey guys. I need to find the negation for a) x>0 and b) (x<0) or (x>0)
do I need to change x>0 to x<0? or how do I negate something like this
wild zinc
x <= 0
sterile shadow
If I take a number to the power of 3 and its irrational, does it mean it was irrational before the multiplication? If so, how do I prove it?
stuck lintel
I mean, assume it's not irrational and that it can be written in the form p/q where p and q are integers. Then it means (p^3)/(q^3) is irrational which is clearly not true.
i legit just wanna learn how to solve the general case tho :/
sharp flame
Can someone help? "If k|a and k|b, then, again arguing directly from the definition you gave in (i), show that lcm(a, b) = k * lcm(a/k, b/k)."
Definition of lcm: "For a, b in Z, both nonzero, the least common multiple of a and b is a positive integer m = lcm(a, b) such that a|m, b|m, and if n is any integer such that a|n and b|n, then m|n."
i've already shown that a| (k * lcm(a/k, b/k) and that b | (k * lcm(a/k, b/k))
So it's "if n is any integer such that a|n and b|n, then (k * lcm(a/k, b/k)) | n" that i'm stuck on
stuck lintel
Let $$a=ka_1$$ and $$b=kb_1$$. Lcm$$(ka_1, kb_1) =$$ $$k$$Lcm$$(a_1, b_1) = k$$Lcm$$(\frac{a}{k}, \frac{b}{k})$$
stoic basinBOT
sharp flame
Thanks--will think/go back to working on that
Well.. that approach is so much better than mine
I forgot you could factor out common factors
stuck lintel
Okay i think i figured it out mudkip
sharp flame
(Thanks!)
stuck lintel
np
full summit
what is remainder of 2^123 +e^121 when divided by 11??
sterile shadow
1 * 2^(n-1) +3 * 2^(n-2) + 5 * 2^(n-3)+ .... = 2^n+1 + 2^n but how do I prove it? I tried to type, for example, 5 as (2+2+1) and multiply one by one but I just can't find why.
stuck lintel
When you have a combination of an arithmetic and geometric sequence, multiply by the common ratio and cancel it out with the original
sterile shadow
Not sure if I get it, what do you mean by common ratio?
Like 2 in this example I guess?
stuck lintel
Yeah (or 1/2 works too)
Here I’ll show you the process in a bit, I’m eating lunch right now
sterile shadow
enjoy your meal
stuck lintel
okay im back @sterile shadow
so basically, let the value of the sequence you're trying to find be 2^n * x. Then $$x = \sum_{k=0}^{\infty} \frac{2k+1}{2^{k+1}} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \dots$$ right?
stoic basinBOT
Rendering failed. Check your code. You can edit your existing message if needed.
stuck lintel
oof
sterile shadow
big oof
stuck lintel
anyways, when you factor ot the 2^n you get 1/2 + 3/4 + 5/8 + ...
so let that be x
and so 2x = 1 + 3/2 + 5/4 + 7/8 + ...
so if you subtract 2x - x what do you get?
sterile shadow
I get x
stuck lintel
._.
sterile shadow
let me work with taht for a while
stuck lintel
em
okay so look at it this way:
2x = 1 + 3/2 + 5/4 + 7/8 + 9/16 + etc.
x = 1/2 + 3/4 + 5/8 + 7/16 + ...
sterile shadow
ok I get series
stuck lintel
so if you cancel out numbers with the same denominator you get....
stuck lintel
...............
sterile shadow
im not looking
stuck lintel
bruh
sterile shadow
waht did you eat?
stuck lintel
a tuna sandwich at the school cafeteria
sterile shadow
nice
stuck lintel
😄
haughty falcon
@wild zinc I finally did the calculations myself and indeed (3a+4b,2a+3b) is a new solution (the inverse is (3a-4b,-2a+3b), if I made no mistake). You might have noticed that this doesn't actually require the 7 to begin with, so we can actually use it for any triplet (a,b,c) (with a and b interchangable). this means we can use the following rules to generate as many solutions as we like, to try and use in our magic square of squares:
(a,b,c) is a solution to a²+b²=2c², then also valid solutions are:
(b, a, c)
(a,b,-c)
(-a,b,c)
(an, bn, cn)
(3a+4c,b,2a+3c)
(3a-4c,b,-2a+3c)
and the trivial solution we can start with is (1,1,1) and we can build up our solutions from it as one huge tree
wild zinc
nice
stuck lintel
:P
sterile shadow
@stuck lintel and why do we need to make the series 2x -x, but it doesn't work with just x?
or there is a way but it's more complicated
?
wild zinc
you can find the sum from k = 0 to infinity of k x^(k-1) by doing
d/dx (x^k) = kx^(k-1)
so d/dx (sum of x^k) = sum of kx^(k-1) as sum of x^k converges uniformly on (-1,1)
so d/dx (1/1-x) = sum of kx^(k-1)
so sum of kx^(k-1) = 1/(1-x)^2
then since x is independent of the sum (ie we're summing using the variable k)
sum of kx^k = x/(1-x)^2
stuck lintel
aka it’s more complicated
wild zinc
ya I didn't get around to saying that bit woops :^)
proven topaz
@stuck lintel op
stuck lintel
wat
haughty falcon
@wild zinc in case you're interested: I managed to boil it all down to a few lines of code that generate lots (mayb all?) coprime pairs:
(I basically got rid of c and instead considered the pairs (a,b) themselves, as they are what is used to calculate c in the first place. I then have a progression with the matrix you found, another one with the same matrix, but a and b are swapped, and finally one where a and b are swapped and b is flipped to negative. the latter will create the two strands we previously observed, however if we don't swap before doing it, the values will collapse back to small values after doing it twice)
void f(long long a, long long b, int count){
printf("%I64d %I64d\n",a,b);
if(count<maxcount){
long long s= 4*sqrt((a*a+b*b)/2);
f(3*a+s,b,count+1);
f(3*b+s,a,count+1);
f(-3*b+s,a,count+1);
}
}
int main(int argc, char* argv[]){
f(7,1,0);
return 0;
}```is the order of operations the same for both?
past summit
dusk wadi
How do 13
I thought looking at n^7 - 77 and the Fibonacci numbers mod 7 might work but any other places to start?
sacred junco
hey i have a proof that says that there are more integers between 0-infinity than there are real number between 0-1
anyone want to take a look at it? it's rather short
stuck lintel
they're both the same size infinities
sacred junco
this video says there are more between 0-1 than 0-infinity
and I've a proof of the opposite being true, using the same method of counting
i've never heard that they're the same size. In what sense is that the case?
stuck lintel
they're both countable infinities so they're the same size
sacred junco
how? i've learned that reals (0-1) is uncountable infinity
wild zinc
there are more reals between 0 and 1 than integers
shrewd garden
go for it
sacred junco
if u put a 1 to 1 relation between the reals from 0-1 and from 0-infinity
you can do it like this:
0,1 = 1
0,2 = 2
0,3 = 3
....
0,11 = 11
0,12 = 12
....
0,201 = 201
etc
this means that all numbers from 0-1 has a corresponding integer number
but I didn't use all integers
as an example 10,20,30 etc isnt used
vast vessel
t!yt how to count past infinity
unique vaultBOT
shrewd garden
what about 0.01
shrewd garden
the reals are uncountable, it's been proven 😉
vast vessel
btw this goes in more like #proofs-and-logic
sacred junco
ok ty
sacred junco
@shrewd garden what if you counted like a mirror:
0,1 = 1
0,01 = 10
0,2 = 2
0,201 = 102
wouldn't every number be represented then?
shrewd garden
the problem with that is that it counts all rational numbers between 0 and 1
and that is countable
what integer would represent 0.π, where it's 0 followed by the digits of π? Clearly there's no end to π so how do we find a closed form solution of what number it corresponds to?
shrewd garden
well it's 0.314159..., and by your method it'd be some number that's ...951413
shrewd garden
exactly. so it's uncountable
sacred junco
then both are uncountable
shrewd garden
0.π is a real number, it exists. But for it to be countable under your method there is a natural number that corresponds to this one, but no such number can exist
sacred junco
oh, my point was that both integers and reals are uncountable then
shrewd garden
I think you're really misinterpreting what countable means
sacred junco
i might be
shrewd garden
:P
sacred junco
😄
said in another way: there is a way to find a 1-to-1 relation between integers and reals in 0-1
they're equally "huge"
0.333333... becomes ...33333333
etc
...33333 = 1 / 0,...3...3...3...3...3..
sterile shadow
How do I solve this? I guess it's doable by induction but I don't know how 
static sparrow
You could ln() both sides, and since ln is increasing over R+, you would havfd to compare (n+1)ln(n+1) and nln(n+2) instead
This can be done by studying a function which to t>=0 associates (t+1)ln(t+1)-tln(t+2) and by doing the usual analysis stuff to show that it only outputs positive values
@sterile shadow
sterile shadow
I haven't got enough tools introduced yet in order to do this (even logarythms haven't been formally implemented) so there must be other way 😦
sacred junco
Can [a + a ...] } a times = a mod a
jade temple
Yes. a mod a = 0 mod a. And every term in that sum is divisible by a.
odd nexus
@sterile shadow you can use bernoulli
shrewd garden
that looks right, yeah
warped jay
alright.. still confused cuz i think im allowed to move the brackets
which would change the distributive law
pale condor
silver solar
Most generally speaking, if you have two functions, say f and g, that are proportional to each other, then f(x)/g(x) should be (approximately) constant, and that constant is called the constant of proportionality
sacred junco
is a popular problem
sacred junco
all primes bigger than 3 are one away from a multiple of 6 so that's pretty constraining
once you get up large enough the differences will have more larger gaps not just consecutive ones separated by 2, 4, and 6, it's inevitable I guess I'm just being a wet blanket
sacred junco
Have you heard of Dirichlet's theorem?
dense turtle
yes
shell ore
can anyone help me with a problem
in number theory you can divide a number up to a^2 +b^2 + c^2 + d^2
is there some efficient way to compute a,b,c,d?
half fable
what
shell ore
@half fable you can write any number as a^2 +b^2 + c^2 + d^2, for example 17 you can write as 0^2+2^2+2^2+3^2 = 17
half fable
you can?
shell ore
ah thanks, I couldn't find the name for it
k nice I still have 2 hours to write an algo for this
indigo valley
(x,y) ∈ (some number, some number)^2
can anyone explain what this means?
i understand ∈ means the element "is in set"
but not sure why the question has a "^2" there
sacred junco
Oh, that's the Cartesian product. (2,3), for example is a member of R x R or R^2
indigo valley
hmm
i dont really know number theory so im having understanding that
ill just plug in arbitray values like the ones in the question: (x,y) ∈ (0, .5)^2
so (.1,.1) is an element of the set because both x and y values are between 0 and .5?
sacred junco
no that's not how it works
I think the problem is (a,b) is not an interval like [a,b] or [a,b) that kind of thing
(2,3) represents a point
so (3,-2, 1.5) ∈ R^3
so each entry is a real number, and there are 3 of them
the set theory aspect of it is that R is the set of real numbers
hopefully that clears it up a bit?
indigo valley
i see what you mean
its wierd in the context of this question though
its one of those no outside help ones so im wary of screencapping it
sacred junco
sounds like you don't understand the notation being used to even ask the question in the first place
so that's strange
indigo valley
yea im skipping this one lol
wild zinc
I think you are right
but it's bad notation
(0, 0.5)^2 in this context should only really be referring to the open interval (0,0.5)
cartesian product of that with itself
(0,0.5) being an ordered pair makes (x,y) in (0,0.5)^2 be very simple
and also wouldn't make sense with the example (0.1,0.1) in (0,0.5)^2
it seems like (x,y) refers to an ordered pair and (0,0.5)^2 refers to cartesian product of an open interval
sacred junco
Find all points of horizontal or vertical tangency
r=4sin(t)
0 to pi
@vast vessel
vast vessel
pings
sacred junco
No idea
vast vessel
Just plug the equation for r into the equations for x and y
And take derivative with respect to theta
vast vessel
=tex y=r\sin(\theta)=4\sin^2(\theta)
stoic basinBOT
vast vessel
=tex \frac{dy}{d\theta}=~?
stoic basinBOT
sacred junco
8sincos
vast vessel
So those are points that possibly have horizontal tangents
Since the dy = 0
Now do same for x
To find vertical tangents
sacred junco
I'm going to see prof idk anything going on here
vast vessel
Okay then
So how do you usually do these problems?
Also why're we in #elementary-number-theory
vast vessel
🔨 😡 
sacred junco
U
vast vessel
sacred junco
take the derivative of a triangular number with respect to the number of prime factors
vast vessel
noble jay
if a and b are relatively prime, with a= (product of primes )^x
and b = (product of primes )^y
why does xy=0
sturdy dirge
More context, it seems absurd.
spare thistle
@noble jay a=z^x , b=z^y (where z>=2). Relatively prime means gcd(a,b)=1. In order for gcd(z^x, z^y)=1 to be true, either x=0 or y=0
@hardy flare AOPS intro to number theory or for something for people with more experience reading textbooks, An Introduction to the Theory of Numbers by Hardy
hardy flare
Ok
spare thistle
I'm probably missing something simple here.. but how do I solve for a (x,y,z) pair that satisfies a*x-b*y-c*z=1, given a,b,c? a,b,c,x,y,z are all integers
I know how to do it for a 2-variable diophantine equation using Euclid's gcd algorithm , but how can I do it for 3?
fading magnet
tex \sum_{d \leqslant n} \frac1d M(\frac{n}d)$
How to calculate this?
M - Mertens function
spare thistle
Is there a bot that displays latex?
fading magnet
violet light
stoic basinBOT
fading magnet
neon comet
I need help with the commas in set theory or whatever its called
p has no condition here ? correct?
only q has the condition Z and cant equal to 0?
so does this mean that p could be literally anything?
<@&286206848099549185>
wheat swallow
neon comet
what?
vast vessel
lmfao
hardy iron
p and q both have the condition Z
vast vessel
#❓how-to-get-help, rule #4: If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once.
Please read the rules: #❓how-to-get-help!
hexed atlas
$p$ ranges over all the integers, and $q$ ranges over all the integers satisfying $q\neq 0$
neon comet
lets say i wanted p to be allowed to be anything
hexed atlas
Define "anything"
vast vessel
double dollars @hexed atlas
neon comet
lets say C and R
hexed atlas
@vast vessel Hard to break habits :P
vast vessel
:P
neon comet
How would such syntax add into that
hexed atlas
Any complex number?
hexed atlas
$$X = \Bigg{ \frac{p}{q} \mid p \in \mathbb{C}, q\in \mathbb{Z}, q \neq 0 \Bigg}$$
neon comet
basically my question can get boiled down to what ends the comma
stoic basinBOT
hexed atlas
Yeah
neon comet
ok
hexed atlas
They both mean "such that"
neon comet
so you have comma separated p and q here with their C and Z
hexed atlas
, here just means and
hexed atlas
All $\frac{p}{q}$ such that $p$ is in the complex numbers and $q$ is in the integers and $q$ is not equal to $0$
Heck
All $$\frac{p}{q}$$ such that $$p$$ is in the complex numbers and $$q$$ is in the integers and $$q$$ is not equal to $$0$$
stoic basinBOT
neon comet
if the p is alone between | and qEZ how come it must follow Z?
hexed atlas
Oh, you removed $\in \bC$ but not the $p$
Well, that doesn't really mean anything
$p$ has to live somewhere
neon comet
yeah
hexed atlas
If you don't specify where it lives, it might live in the set of all chairs for all we know
neon comet
So basically if the p would be alove in between the commas it doesnt have any own condition and it has to follow the next comming rule?
alone*
stoic basinBOT
Rendering failed. Check your code. You can edit your existing message if needed.
hexed atlas
You get what I mean
neon comet
sorry no im new to this
hexed atlas
You can read it as "p and q in complex numbers"
hardy iron
$$p,q \in \b{C}$$ means that $$p \in \b{C}$$ and $$q \in \b{C}$$
stoic basinBOT
neon comet
ok
hexed atlas
Thanks Hatena
hexed atlas
In what sense?
hexed atlas
Just unnecessary, still mathematically correct
hexed atlas
No worries :)
swift shard
Not sure about the hint, maybe it will come up! Let's follow the method of induction for now
First, gotta prove the base case
hexed atlas
Hmm, unless I'm mistaken it can be done directly
Using modular arithmetic
Express the power of four as a power of five perhaps, to use the hint
Noting that the hint essentially says that $$\amod{5^2}{4}{21}$$
stoic basinBOT
Rendering failed. Check your code. You can edit your existing message if needed.
fossil cape
hmm
Stupid phone
Anyway
I have this but I don't know how I'd relate this back to the base case
actually
i think i might just forget how exponent laws work
hexed atlas
@fossil cape Did you try the direct approach I mentioned above?
hexed atlas
fossil cape
oh god
hexed atlas
❓
hexed atlas
Basically, if things are relatively prime, you can just substitute
fossil cape
are 25 and 4 relatively prime?
hexed atlas
The thing you are substituting needs to be relatively prime with the modulo
fossil cape
oh
hexed atlas
Here, 5 and 21 share no factors, and are hence relatively prime
fossil cape
aya
hexed atlas
Also 4 and 21
fossil cape
this is hell
hexed atlas
Dw, you can do it!
In the world of modulo 21, 4 is exactly the same number as 25
So wherever 4 appears, you can replace it with 25
fossil cape
Is this right
hexed atlas
Yep that looks great!
fossil cape
