wdym?
#discrete-math
oblique dragon
So finally you get 3 * 10C2 right
oblique dragon
We do
oblique dragon
idts
granite forge
hm
oblique dragon
this formula doesn't work otherwise
granite forge
2, 3, 3?
oblique dragon
Nvm
I mean it counts, say 4, 3, 1 and 4, 1, 3 as different solutions
to x1 + x2 + x3 = 8
oblique dragon
(8 + 3 - 1) C (3 - 1)
it does
Think of the grade scenario
if you have A + B + C = 8
granite forge
yeah?
oblique dragon
3 people getting B and 1 getting C is different from 1 getting B and 3 getting C
granite forge
oh
true
fuck so now im confused
this is combination with repetition right?
cause 3+3+2 and order doesn't matter.
oblique dragon
(n + k - 1)C(k - 1) is the answer to in how many ways you can distribute k identical apples among n different people
granite forge
right
oblique dragon
Do you know how we derive the formula? Using balls and sticks and stuff
granite forge
yes I do
oblique dragon
right
So here x1, x2, x3 are different people
so x1 having 10 is not the same as x2 having 10
granite forge
but we want it to be the same
right?
cause we don't care about order
whereas in the grade problem
we did care
cause A is a grade itself, which is diff to B, and C, so order matters there
oblique dragon
well if we do care about the order this becomes kinda difficult to solve lol
granite forge
we don't care for our X1 X2 X3
oblique dragon
In my experience with combinatorics such questions do care about order
granite forge
so you are saying
oblique dragon
Since x1, x2 and x3 are 3 different variables
granite forge
10 + 4 + 4
is different to 4 + 10 + 10 ?
Oh
yeah we do care about order nvm
so what's wrong then lol?
OH
wait
im confused again
oblique dragon
You do
granite forge
that's what's confusing me
oblique dragon
Because you did this only for x1
oblique dragon
right
granite forge
that's what I was thinking first too
Oh
yeah
now that makes sense
when u explained the variables thing
right yeah
okay awesome
so
total would be
18 + 3 - 1 C 3 - 1
8 + 3 - 1 C 3 - 1
all combinations
wait
oblique dragon
and subtract from 190
granite forge
18 + 3 - 1 C 3 - 1
3 (8 + 3 - 1 C 3 - 1)
190 - 3*55 = 25
oblique dragon
190 - 3 * 55
granite forge
gotchu
oblique dragon
but this doesn't seem to be right hmm
granite forge
why is that
oblique dragon
Umm
granite forge
then it should be 190 ?
oblique dragon
so anyway I had counted 45 earlier
by taking x1 = 0 and so on
You get 1 + 2 + 3 + ... + 9
granite forge
im confused lol
oblique dragon
but yeah I suppose this was homework for you to learn the complement method
granite forge
did we do something wrong
oblique dragon
nah it isn't
granite forge
wait so u did let x1 = 0
oblique dragon
when x1 = 0 the other 2 can only be 9 and 9
granite forge
yeah
oblique dragon
and x1 = 1 the others 2 are only (8,9) or (9,8)
oblique dragon
ok no wait a sec
oblique dragon
yes
oblique dragon
wait a sec
granite forge
don't we need some odd num
granite forge
alr
oblique dragon
how do we get 45
when x1 = 0 you get 1, when x1 = 1 you get 2 solutions, when x1 = 2 you get 3 solutions, and so on
granite forge
im not too concerned about the numbers lol
granite forge
wait how
oblique dragon
and that's 55
oblique dragon
Yes
granite forge
@oblique dragon Do you think you could help with this as well please?
Here's what I've done:
I'm not sure if there is any flaws.
I think it's right tho.
If you could just point me at what's wrong I can try figure it out.
oblique dragon
I think it's right, but you also need to multiply by 5!
once you've chosen where you want your vowels, you also need to arrange them
Ah nvm, it's 22P5 - it's absolutely right then
sweet granite
Is there a way to write the problems out? Like 3^(2) for 3square?
Or is that not possible with the that symbol?
weary tiger
Is there a way to write the problems out? Like 3^(2) for 3square?
which one
granite forge
Ah nvm, it's 22P5 - it's absolutely right then
oh sweet as then. tyvm!!
vale blaze
hi, can anyone give me tips on how to study for discrete math? i’m having trouble even understanding the basics and it’s different from calculus and other maths im used to
vernal vigil
hi, can anyone give me tips on how to study for discrete math? i’m having troubl...
Ive found that reading and getting a basic understanding, and immediately practising works pretty well
vale blaze
Ive found that reading and getting a basic understanding, and immediately practi...
yeah i find that helps too but i always seem to forget information the next day. it’s harder for me to build a foundation of knowledge in this class to confidently move onto the next chapter
vernal vigil
yeah i find that helps too but i always seem to forget information the next day....
Then review after you sleep
vale blaze
Then review after you sleep
lol i’ll try that
sweet granite
What can be the bijection between the set of positive integers and negative integers?
brave rock
Write them down and think about how to send a positive integer to a negative one
Write (a few of) them out side-by-side if it helps
sweet granite
Write them down and think about how to send a positive integer to a negative one
So like 1=-1, 2=-2 etc?
brave rock
I'm really not interested in just giving you the answer. You should try determining it yourself.
sweet granite
I'm really not interested in just giving you the answer. You should try determin...
I understand
foggy pollen
So i need some help with drawing M as a graph. This is what I have so far, but im not sure if it's correct. Any help is greatly appreciated. Thank you.
slate dome
plz help
patent gull
So i need some help with drawing M as a graph. This is what I have so far, but i...
for a DFA you need a transition on every part of the alphabet so each of your states needs a transition for an A and B
warm quest
plz help
Dawg if you want help you're going to have to ask where you're actually confused.
grim karma
Hey guys, what do resources do y'all recommend for discreet math?
weary tiger
I drew out some base cases, and can't see any connction
winter prawn
can someone tell me why an order-homomorphism does not imply a meet- (or join-) homomorphism?
An explanation would be preferable, but counter-example will also do.
(Just began learning about lattice homomorphisms)
I have tried but am unable to figure out why?
(Also please tag me in the answer)
granite forge
Is anyone familiar with the Pigeon Hole Principle?
I am not sure how to approach this question
weary tiger
Hello, I cannot solve part c of this question for n >= 4
weary tiger
The game is to make each term of p be an odd number, since if only one of the terms is even p is even. so a1,a3,a5,...,an, must be even numbers. However, there are n-1/2 + 1 = (n+1)/2 odd numbers and only n-1/2 even numbers.
blazing rose
Im kinda stuck, just got into predicate logic
Image
How would I solve ii) ?
All Ive got until now would be to "for all" quantifiers and then i say ∀x ∀y (¬(x > y)
but how do I continue, since I have no clue how to express x should be real and y should be rational with the given predicates
actually y < x since im only allowed to use less(x,y)
cerulean radish
Doesn't it in the parenthesis say that you can use x < y though?
cerulean radish
Im kinda stuck, just got into predicate logic
Image
How would I solve ii) ?
All ...
You need to negate before declaring y
So for all x in R, not (for all y in Q, y < x)
blazing rose
So for all x in R, not (for all y in Q, y < x)
yea but I cant just say for all x in a specific set of numbers
I have to define in some other way that x is real and y is rational
cerulean radish
Pretty weird
Well
In general instead of "for all x in A, P(x)" you can say "for all x, x in A implies P(x)"
blazing rose
they only gave us the predicate integer(x) for this task
cerulean radish
Ah, the universal set is R, so you just have to define what it means for y to be in Q
blazing rose
exactly
cerulean radish
For all x, not (for all y, if there exists a, b such that integer(a), integer(b), less(0, b) and by = a, then y < x)
blazing rose
if there exists a, b such that integer(a), integer(b), less(0, b) and by = a
this is the definition of rationals then?
like the other way around
cerulean radish
Right, that's equivalent to saying y is rational
fringe turret
The concept of tautology and fallacy includes "regardless of the value of its variables" in their definition. Does it means functions represent functionality as there names?
def tautology(x):
return True
def fallacy(x):
return False
Logically speaking these make sense, but in the example the instructor is using $P + \bar{P}$ and $P \cdot \bar{P}$.
vital dewBOT
tbhaxor
fringe turret
So I need to know is my example also valid or not?
fallow dune
What issue are you having? It's pretty straightforward.
damn, j cuz something straightforward to u doesnt mean it is to me :/
lean rover
Let Φ be the formula ∀x∀y(x>1∧n=x×y→∃z(x=2×z)). What is the formula obtained by substituting Φ[x + z/n]? Only make the variable renamings that are absolutely necessary. I don't really undestand what I am being asked. Would the answer look something like: ∀w ∀y (w>1 ∧ x+z = w×y -> ∃w(w = 2×a)?
fickle orchid
Im new to taking discrete math and i was instructed to find the general solution for this recurrence relation, i guess my problem is with finding the characteristic relation for this reccurence relation..if someone one could please help with the method of approach.
lethal fog
$\mathbb{F}_{\text{un}}$
vital dewBOT
sakka ismail
torpid mango
im supposed to use that if n is a positive integer i can use this to find the answer in mod 19
not sure how :/
fierce granite
Quick question, is 9mod3 = 0
pale epoch
not sure how :/
look at the theorems you covered in class, there is one due to Fermat that might be helpful
pale epoch
Quick question, is 9mod3 = 0
you should be able to check this yourself
torpid mango
but how do i use the fact that 35 to the power of 35 is 18n -1 mod 18
when i have to decide the remainder in mod 19
pale epoch
what does fermats little theorem say?
pale epoch
yes
torpid mango
and p does not devide a
pale epoch
so you see there is a p-1 (and not a p) in the exponent
so just write that down for the question you are asked and see what it tells you there
torpid mango
but what confuses me is that we are swapping from mod 18 to mod 19
why is it true for both mod 18 and mod 19
i still have to solve for mod 19
so im not sure how counting in mod 18 to find out that 35^35 = -1
helps me in mod 19
fierce granite
you should be able to check this yourself
IDK number theory, I am a physicist who took group theory and I have met this and am lost :(((((
pale epoch
what is your definition of mod then
fierce granite
n|(a-b)
pale epoch
so does 3 divide 9?
pale epoch
why is it true for both mod 18 and mod 19
why would it be?
just look at the theorem again, what is a here? and more importantly what is p?
what does it then say
pale epoch
yes but isn't it 0 - 9 /3
why are you dividing here
the relation is (should be) symmetric
so yes, the alternative question is "is -9 divisibile by 3"
which is true iff 9 is divisibile by 3
fierce granite
9mod(3) as far as I know is saying if we divide 9 by 3 we get a remainder of 0
but I have only just realised the difference between congruence and = for this type of equation...
I think that is where my confusion was
pale epoch
9mod(3) as far as I know is saying if we divide 9 by 3 we get a remainder of 0
yeah kinda
but the actual math formulation is as you said it
you check if a number is divisible by another number
which doesnt actually require you to perform division
pale epoch
its the standard formulation
fierce granite
mmm
pale epoch
the actual actual math formulation is via group/ring theory
fierce granite
so what is the difference exactly they between $ a \equiv bmod(n)$. and $a = bmod(n)$
vital dewBOT
INEEDANAME
pale epoch
the first one is what most people would write
the mod defines an equivalence relation, so you use \equiv instead of =
the second one means the same thing but i dont think you would usually see this
its a bit weird to write things like 9 = 0, you know
fierce granite
ye lol
So I see at the start of this groups book : $3 \equiv 24mod(7)$ because $ 3 - 24 = (-3) \cross 7$
vital dewBOT
INEEDANAME
fierce granite
I don't understand why they have brought -3 into it
pale epoch
because 3-24 is negative?
fierce granite
so a is the remainder of b/n, but in the book it says n|(a-b). This is confusing me, the fact that there is multiple ways of interpreting it
like 24/7 = 3R3. What is the significance of the -3
weary tiger
I don't understand any of this
pale epoch
well, the former way only works for the smallest representative
you also have $24 \equiv 31 \pmod{7}$
vital dewBOT
Lochverstärker
fierce granite
But 31/7 is 4R3
How have I managed to turn one of the simplest things into this 😅
Ill go away and think about it
thanks for the help 🙂
torpid mango
can i reduce exponents in modulo ?
like is -1 as an exponent for mod 19 equivalent to 18 as an exponent in mod 19 :S?
pale epoch
no
weary tiger
yes
pale epoch
its not that simple, try some examples
try some small examples
maybe say mod 5
and try i dunno, 3^10 mod 5
fierce granite
**Lochverstärker**
Ye so congruence is different... I was thinking of 31 mod(7) = 3
pale epoch
then compute 3^9, 3^8, ... mod 5
and see when you get 3^10 again
then try to relate this to fermats little theorem
(look at exponent laws)
if you mix in negative exponents it gets a bit weirder too in some cases
weary tiger
pale epoch
Ye so congruence is different... I was thinking of 31 mod(7) = 3
usually you dont use mod like that, at least in mathematics
but yes, you could check if both numbers have the same remainder when dividing with remainder
the issue here is you have to fix an algorithm to perform division with remainder
the other definition is independent of that
fierce granite
I have come across somet called the (Zn,+) group that is my motivation for understanding the definition of = and congruence.
I think the = version is used here
weary tiger
= 4
coral parcel
(Zn,+) is just another notation for the thing we're talking about here.
fierce granite
why is there a small n on the second line before y
coral parcel
Whether one writes use \equiv or = is a bit context and audience dependent. People tend to be scrupulous about writing \equiv in introductory works, to be sure the reader distinguishes it from "these are literally the same number". However, when one has some confidence that the reader knows what going on, people often feel free to just write = and consider the "(mod n)" implicit, with the view that the numbers actually refer to group elements and 9 and 0 do both point to the same group element.
coral parcel
If you know some group theory from another source, you may already know the group as the cyclic group with n elements.
torpid mango
so if i use fermats little
and try to use it in mod 19
i get the 35 to the power of -1 which i dont want
coral parcel
so (Zn,+) is apparently defined as x+y = (x+y) modn
That's one way to realize it, yes. Mathematicians tend to favor a different definition that gives an isomorphic group in a more involved way, but generalizes more easily to other similar situations in abstract algebra.
torpid mango
i am really at a loss on how to use mod 18 to help me solve the remainder in mod 19
fierce granite
That's one way to realize it, yes. Mathematicians tend to favor a different defi...
Ok, thanks for the help. I am sure as I progress it will all become clear...
torpid mango
i can definetly figure out what 35 tto the power of 35 is in mod 18
i just cannot take the step after that, transfering what 35^35 is in mod 18 to help me in mod 19
so any help would be appretiated
im supposed to get the remainder 6
for my problem
coral parcel
To use Fermat's little theorem here, you need first to evaluate 35^35, that is, the exponent in your power tower -- modulo 18, not modulo 19.
So there you're looking at (18+18-1)^35 modulo 18.
But that's the same as (-1)^35 modulo 18, which is easy.
torpid mango
but so far when ive been using fermats little theorem
let me give another example
2 sec
coral parcel
i just cannot take the step after that, transfering what 35^35 is in mod 18 to h...
Oh, I see you already got what I wrote.
torpid mango
this is a question i could actually solve
on the same topic
i feel like the other one is harder
coral parcel
i get the 35 to the power of -1 which i dont want
This is exactly what you should get.
Your next step should then be rewrite 35^(-1) to (35 mod 19)^(-1),.
And from there all you need to do is test all of the numbers from 0 to 19 to find the one that is the inverse of 16 (mod 19).
(Or use the extended euclidean algorithm, if you're feeling really systematic).
coral parcel
Yet another way to compute 16^(-1) mod 19 would be to rewrite it to 16^17 mod 19, again using Fermat's little theorem, and then you can use exponentiation by squaring to get it in a handful of multiplications.
Or: 16^-1 == (2^4)^-1 == 2^(-4) == 2^14 (mod 19). And then you can again use exponentiation by squaring, if you don't remember that 2^14 = 16384.
torpid mango
so after some food break
i now have the inverse of 16^-1 is equal to 6
so i now have 35^6 right ?
coral parcel
What, no.
torpid mango
.<
coral parcel
The inverse was modulo 19, so that's the bottom exponentiation.
I assumed you had already seen that 35^(35^35) == 35^-1 (mod 19) because 35^35 == -1 (mod 18).
coral parcel
No.
When you want to compute 35^blah modulo 19, and you're using Fermat's little theorem, want to find blah modulo 18.
35 is the same as 16 modulo 19, but it is not the same as 16 modulo 18.
coral parcel
That looks right at least at the short glance I took.
So your first step for 35^(35^35) is to compute 35^35 modulo 18. You can't use Fermat's little theorem directly there, because 18 is not prime. However, since 35 == -1 (mod 18), we have 35^35 == (-1)^35 (mod 18), and raising -1 to powers is easy.
torpid mango
but like in general, in all the other questions im just using the modulo given
so like if i am to decide the remainder when dividing by 19
i just use fermats little theorem where p = 19
but this one is different
because i get stuck with a negative exponent
im sorry, ive only been to uni a few weeks
coral parcel
What do you mean "get stuck"?
You just managed to compute 35^-1 modulo 19 just fine.
torpid mango
whats the difference between the 2 alternatives
cause in the one where i assume im correct
the 1222^2403
i use the little theorem
just with p=601
coral parcel
but like in general, in all the other questions im just using the modulo given
Did those other questions have a power in the exponent position?
torpid mango
no
they did not
OH
WAIT
maybe
do i understand now
OH
wait
SO ITS
little theorem
2 times ?
cause i have 2 exponents?
or am i stupid still
xD
coral parcel
More or less. Almost. Except that at the upper layer you're working with a different modulus, one that won't be prime, so Fermat's little theorem won't work there.
torpid mango
oh oh oh wait
what i have to do
a exists in modulo 19
the exponent
exists in modulo 18
coral parcel
Yes.
torpid mango
but modulo 18 is a part of modulo 19
in the bottom
so what i calculate
inside the exponent
is still part of mod 19 equation
?
coral parcel
I'm not exactly sure what you mean by "part of" there. It's a step on the path that will eventually lead you to a result of the final mod 19, but on that step you're actually looking for an intermediate result that's mod 18.
torpid mango
so p-1 in the formula
just means, calculate this in modulo p-1
not the actual like same number as p ?
thats whats confusing me
idk if im making any sense at all
i mean in this formula
p is the modulo
but p-1 is like, calculate the exponent in mod p-1 ?
coral parcel
I see what you're saying, but please give some time to respond!
fierce granite
🤣
coral parcel
Fermat's little theorem indeed says $a^{p-1} \equiv_p 1$. But that's not what you're directly \emph{doing} in these problems. Instead we're using this calculation
$$a^{b(p-1)+c} = a^{b(p-1)} a^c = \underbrace{a^{p-1}a^{p-1}\cdots a^{p-1}}_{b\text{ times}} a^c $$
and therefore as a consequence of the little theorem we have
$$a^{b(p-1)+c} \equiv_p 1\cdot 1\cdots 1\cdot a^c$$
So what we need to do when we see $a^{something}$ is \emph{first} to find a way to write the $something$ as a multiple of $p-1$ plus some $c$. That's why we're now looking for mod $p-1$, and it's only \emph{after} we've found that that Fermat's little theorem even applies.
vital dewBOT
Troposphere
coral parcel
^ @torpid mango
coral parcel
Ok sorry.
torpid mango
i think whats really hard for me to understand
is that in the same equation
there exists p19
and p18
coral parcel
What do you mean by p18?
coral parcel
I don't understand the notation "p18"
torpid mango
since we solve the exponent in p18 it is -1
what i mean with p=18
is that im calculating in modulo 18
and then in modulo 19
but the fact that i can use the calculation in mod 18 for mod 19 confuses me
coral parcel
Okay, all that looks correct. (I was confused by you setting p=18, because the letter p is generally used for primes, which 18 isn't, and in any case p already has a meaning here).
torpid mango
i just dont know why im calculating the exponent in modulo 18
well
i understand it if i have a simple equation
like the first one
this one is alot simpler for me to understand
coral parcel
but the fact that i can use the calculation in mod 18 for mod 19 confuses me
I think the underlying point is that we have rules like
$$ a\equiv_p b \text{ and } c\equiv_p d \implies a+c \equiv b+d $$
$$ a\equiv_p b \text{ and } c\equiv_p d \implies ac \equiv bd $$
but there IS NOT a rule that says
$$ a\equiv_p b \text{ and } c\equiv_p d \implies a^c \equiv b^d $$
vital dewBOT
Troposphere
coral parcel
Fermat's little theorem effectively says that \emph{if $p$ is prime} then we have
$$ a\equiv_p b \text{ and } c\equiv_{p-1} d \implies a^c \equiv_p b^d $$
vital dewBOT
Troposphere
torpid mango
i mean i think i know how to solve it now
with your explanations
i just hope i understand it xD
so if i only have one exponent
i can just apply the theorem directly
but if i have a power in the exponent
i need figure out what the expression in the exponent is in mod p-1
i think your final explanation works very well for me
this one
im playing around with it now
coral parcel
Note that when you said "2403 = 4·600 + 3" you were also "figuring out what the expression in the exponent (namely 2403) is modulo p-1".
I'd prefer to say it this way: In order to use Fermat's little theorem to answer
a) what is the remainder of a^(whatever) modulo p?
you always switch to solving a sub-task:
b) what is the remainder of (whatever) modulo p-1?
The difference betwen the two exercises is that if there's only one exponent, answering (b) is a quick matter of arithmetic, whereas if (whatever) is itself an exponentiation, then answering (b) begins to take more thought and tricks of its own.
torpid mango
ah
i think i understand now
now im getting the correct thing aswell
heureka
i love you friend
uni math is hard >.<
i hope im cut out for computer science
i think i finally get it
torpid mango
Note that when you said "2403 = 4·600 + 3" you were _also_ "figuring out what th...
this explanation works very well for me
anyway
im really greatful for the help
helps me alot
torpid mango
Note that when you said "2403 = 4·600 + 3" you were _also_ "figuring out what th...
are you a teacher in some courses 😄 ?
coral parcel
No.
torpid mango
i have reached this stage now
so now i just figure out a number that multiplies with -3 from 0-18 so that i get 1 ?
and that number is the inverse and therefore the same ? in mod 19
coral parcel
You've already done that once: remember that -3 == 16 (mod 19).
coral parcel
i now have the inverse of 16^-1 is equal to 6
Here.
coral parcel
Yes.
torpid mango
yay!
the number that i want to find the inverse for in mod x
has to be relatively prime with x right ?
so in my case gcd(19,16) is 1
and therefore 16 has an inverse in mod 19
right ?
coral parcel
Yes.
However, since 19 is prime, the only numbers that don't have inverses are multiples of 19.
torpid mango
oh right !
so no need to be scared of inverses as long as i keep track of what modulo im in if i have nestled exponents
cause like you said, when im working with the p-1 expression
i was not working in mod 19 anymore
fierce granite
@coral parcel
coral parcel
What seems to be the problem?
fierce granite
Can you please assign me perma studying...
I keep wasting time in the disc channels 🤣
coral parcel
How about the self-assignable ordinary studying role?
fierce granite
Eh where, hold on
coral parcel
"Channels & roles" -> scroll down
fierce granite
👍
lone linden
"Channels & roles" -> scroll down
Light mode 
vital dewBOT
SyntaXErroR
cerulean radish
Why is {∅} not a subset of {{∅}}
Because ∅ is not an element in the latter
cerulean radish
or why is the power set of {{∅}} only {∅, {{∅}}}
The power set of {a} is always {∅, {a}}, so, naturally, the power set of {{∅}} will be {∅, {{∅}}}
blazing rose
Because ∅ is not an element in the latter
but isnt the empty set an element of every set?
faint sphinx
The empty set is a subset of every set, but not necessarily an element of every set.
coral parcel
{Ø} is not a subset of {{Ø}} because there is something that is an element of {Ø}, but is not an element of {{Ø}} -- namely Ø.
blazing rose
The empty set is a *subset* of every set, but not necessarily an *element* of ev...
ahh right my bad
blazing rose
The power set of {a} is always {∅, {a}}, so, naturally, the power set of {{∅}} w...
why is {{∅}} not a subset of {∅, {{∅}}}
cerulean radish
Because {∅} is not an element in the latter
blazing rose
ohh
its so confusing
ok so whenever I talk about subset of I basically only look at the elements inbetween the set of the LHS right?
cerulean radish
Right, and check if they are present as elements of the other set
worn wolf
its so confusing
What are the elements of S = {u,v,w}? What are the subsets of S? What is 2^S?
blazing rose
What are the elements of S = {u,v,w}? What are the subsets of S? What is 2^S?
elements are u,v,w
subsets {u}, {v}, {w}, {u,v}, {v,w}, {u,w}, {u,v,w}, {empty set}
worn wolf
no, not {\emptyset}
cerulean radish
Don't put the empty set inside {}
blazing rose
P(S) = {{u}, {v}, {w}, {u,v}, {v,w}, {u,w}, {u,v,w}, empty set}
oh yea
only empty set without the brackets
like this?
worn wolf
Ok, then you know how things work
cerulean radish
Yes
blazing rose
tysm guys
worn wolf
why is {{∅}} not a subset of {∅, {{∅}}}
Now just replace u=\emptyset, and v={{\emptyset}} here
and it will become clear
worn wolf
why is {{∅}} not a subset of {∅, {{∅}}}
how can you make {{\emptyset}} a subset though?
What do you need to add to S ={\emptyset, {{\emptyset}}}
blazing rose
if in the second set there would be {∅} as an element?
worn wolf
Yes
blazing rose
tyyy
worn wolf
It can be both a subset and an element, e.g. S = {{\emptyset}, {{\emptyset}}}
It's an element because of the second element, and a subzet because of the first here
blazing rose
It's an element because of the second element, and a subzet because of the first...
yeee
lyric tundra
I'm srsly confused here, wouldn't there be no partial order because it doesn't satisfy the condition for antisymmetry?
brave rock
I'm srsly confused here, wouldn't there be no partial order because it doesn't s...
Are you sure it doesn't satisfy the antisymmetry condition? To disprove antisymmetry, you should try to produce a counterexample.
lyric tundra
(0,2)R(1,5) but not (1,5)R(0,2)? Based on the condition
coral parcel
That's not a counterexample to antisymmetry, though.
lyric tundra
Huhm
coral parcel
It would be a counterexample to symmetry, but nobody says the relation is symmetric anyway.
lyric tundra
But they aren't equal either no?
brave rock
Write down the definition of antisymmetry.
Once you've done that, write down its negation. This will tell you what you need to find in order to prove it is false.
lyric tundra
If aRb and bRa then a=b from what I recall
brave rock
You neglected the quantifiers, but this is fine.
lyric tundra
💀im confuzzled rn
brave rock
Now what is the negation
brave rock
This is not correct.
Because you ignored the quantifiers, you have forgotten about them, and what's more you have not negated the implication correctly.
I'm going to just correct you here, but you clearly need to review some logic here.
The negation of antisymmetry is "there exist a and b such that a R b and b R a and a =/= b"
lyric tundra
I see
brave rock
So if you claim that the relation is not antisymmetric, you must prove this.
You will now note that your supposed counterexample is not correct.
lyric tundra
But still, I fail to see how it could be anti symmetric on the given elements of A and the condition
brave rock
Well you can simply check every pair.
As it happens it's not hard to see if you notice that a+b uniquely determines (a,b) in this specific set.
Then it's an equivalence relation since divisibility is an equivalence relation on the naturals.
lyric tundra
(0,2)R(1,5) but not (1,5)R(0,2)? Based on the condition
I mean, from this example I proposed it isn't symmetric nor it could be antisymmetric either wouldn't it?
brave rock
No
I don't know what your confusion here is exactly. You have not provided a counterexample to antisymmetry.
brave rock
The negation of antisymmetry is "there exist a and b such that a R b and b R a a...
This is what you would need to prove in order to disprove antisymmetry.
lyric tundra
Im rackin my brain rn but I got nothing. How is it antisymmetry exactly?
The elements in the set plus the condition makes me think it isnt tho
coral parcel
If aRb and bRa then a=b from what I recall
You correctly (except for forgetting the quantifiers) reproduced the definition of antisymmetry here.
lyric tundra
You correctly (except for forgetting the quantifiers) reproduced the definition ...
Tbf, we weren't exactly told what about the quantifiers when we were given the definition on the relation
coral parcel
A different but equivalent formulation would be:
We can a relation antisymmetric if there doesn't exist two different a and b in the domain such that aRb and bRa.
Therefore, in order to claim that the relation is not antisymmetric, you need to know that there are two such a and b in the domain.
brave rock
Im rackin my brain rn but I got nothing. How is it antisymmetry exactly?
I made a slight typo in the last message. To be crystal clear: you would have to prove this following in order to disprove antisymmetry:
there exist a and b such that a R b and b R a and a =/= b
brave rock
Tbf, we weren't exactly told what about the quantifiers when we were given the d...
Forall.
It's saying "for all a and b, ... etc"
coral parcel
It's possible that our friend with the untypeable name hasn't even learned what quantifiers are.
lyric tundra
Sorry, havent renamed my nickname, I've gone past quantifiers so I do actually have an idea on quantifiers
scenic current
can any1 help me with my porblem/
brave rock
steady geyser
Can't find it in my textbook, what does the notation A^C mean?
next sail
Suppose that P(x) and Q(x) are predicates, and that D is a set that is defined in the image above.
I'm trying to prove that (∀x∈D, Q(x))→(∀x∈U,P(x)→Q(x)) but I'm kinda stuck at just being able to show that ∀x∈D,P(x)→Q(x). How do I continue this proof?
I read from a book that a statement of the form "∀x∈U, P(x)→Q(x)" can always be rewritten in the form ∀x∈D, Q(x), where D={x∈U | P(x)} (basically D is the truth set of P(x)). Conversely, a statement in the form of ∀x∈D,Q(x) can be rewritten as ∀x,x∈D→Q(x). So that's what I'm trying to prove
weary tiger
Determine the number of binary strings of length 16 in which the pattern 100 occurs exactly
four times.
weary tiger
Is the answer (8C4)x2^4 - 60?
oblique dragon
Why 8C4?
and if you're excluding the ones that have five 100s, you'd be subtracting 6x2 = 12 right?
rigid oriole
Is the answer (8C4)x2^4 - 60?
Its not obvious to me at a glance what this is supposed to be.
There's also a double counting trap here which idk if ur accounting for
weary tiger
I got the answer its ok
past tartan
yo guy
i have an ucoming discrete mathematic mid term test
and my lecturer allow me to bring in a cheat sheet ( A4 paper size) - could u all tell me what should i put in there from your understanding of the course
brave rock
We have not taken your course
Protip: there is probably a list of intended learning outcomes somewhere for your course. Put those on your sheet.
safe steppe
Hi guys, can someone please help me with the following problem, this is what I have got so far but I've been stuck for some time and dont know what to do next?
blissful shell
you could make x^(k+1) into x^k +(x-1)(x^k)
safe steppe
How would I show that thats divisible by x-1
Because the plus/minus outside of the (x-1) drives me crazy, I keep on trying to get the whole thing multiplied by (x-1) so I can prove its divisibility
blissful shell
(x-1)x^k is definetly divisible by x-1
what's left is just x^k - 1
which is also divisible by x-1 from the induction hypothesis
rigid oriole
you could make x^(k+1) into x^k +(x-1)(x^k)
no idea what you're writing tbh
oh holy fuck no wonder i was so confused
x^{k+1}
rigid oriole
x^(k+1) - 1 = (x^k) - 1 + (x-1)(x^k)
I feel like u meant to write this
which would make much more sense
worthy siren
it has been proven that if a | b and a | c then a | mb+nc for all integers m and n.
setting a = x-1, b = x^k - 1, and c = x^(k+1) - 1.
it is assumed that a | b.
so by showing that a | mb + nc for some integers m and n. and also a does not necessarily divide n.
you can conclude that a | c. which is your preposition for Induction step.
Answer: setting m=-1 and n=1. as we know x-1 does not necessarily divide 1. then
-> mb + nc = -( x^k - 1 ) + x^(k+1) - 1 = x^(k+1) - x^k = x^k ( x-1 )
and it is obvious that x-1 | x^k (x-1)
so we have proven that x-1 | x^(k+1) -1.
safe steppe
*Answer*: setting m=-1 and n=1. as we know x-1 does not necessarily divide 1. th...
Ohh, thanks alot, this helped me alot!
worthy siren
Ohh, thanks alot, this helped me alot!
You're welcome.
untold flicker
Hello!
I have an assignment that asked me to find the general solution to two congruence equations of the form ax congruent b (mod 37). I solved for the general solution, but the second part of the question is asking me to identify the relationships between the two classes of solutions.
I am nto quite sure what the question is asking, can anyone point me in the right direction?
<@&286206848099549185>
weary tiger
Can someone link me resouces to understanding what these counting techniques should be applied?
$$\text{1: } n^k$$
$$\text{2: } \frac{n!}{(n-k)!}$$
$$\text{3: } \binom{n}{k}$$
$$\text{4: } \binom{n+k-1}{n-1}$$
$$\text{5: } \binom{n+k-1}{k}$$
Before I thought that 1) should be applied when youa are counting permutations where order doesn't matter
The 2) is for when you are counting permutations where order does matter
3) is for combinations where order does matter
4) is for combinations whereorder doesn't matter
I have no idea how 5) should be applied
Apparently 5 is the number of k-element multisets that can be made from a n-element set.
vital dewBOT
<:F_button:1095679234497843251>
haughty garden
- is used to count when you don’t care about duplicates; each slot has n choices and there are k slots to fill. Each slot is distinguishable and each object is distinguishable.
- you don’t want duplicates but you care about the order.
- you don’t want duplicates but you don’t care about the order.
- and 5) are exactly the same. You want to determine the number of ways to place n indistinguishable objects into k distinguishable slots
To show that 4 and 5 are the same, you can use the binomial symmetry property
weary tiger
Thank you
high scroll
can anyone help me with this question?
I've tried expanding it manually to see how I can turn the summation on the LHS to the summation on the RHS, but I've run into some problems
everything makes sense, until I find out what the 2 remaining terms are
I found them to be -a(0)b(0) and a(n)b(n-1), but the RHS in the question has a(n)b(n) instead of a(n)b(n-1)
and I double checked multiple times but don't think I did anything wrong
meager prawn
Your sum on the last line, as written, should not go up to n-1, but n-2
Because it currently includes an (bn - bn-1), terms that shouldn't exist because an bn was never taken from anywhere
meager prawn
I found them to be -a(0)b(0) and a(n)b(n-1), but the RHS in the question has a(n...
Then in adding and subtracting what it takes to make the sum go up to n-1, you should arrive at the desired answer
Exercise:
Prove Abel's transform using actual sigma notation and not handwavy ... computations, that led to you making this mistake
high scroll
Then in adding and subtracting what it takes to make the sum go up to n-1, you s...
dude yeah ur right
damn ok yeah I got to the answer
can believe I didn't see that
thanks man
meager prawn
can anyone help me with this question?
Cultural note:
Intuitively, it's the equivalent of a discrete IBP
high scroll
what's an IBP?
meager prawn
Integration by parts
high scroll
oh yeah
meager prawn
Helps to remember it
plain pulsar
q(0) = 0
q(1)= 1^2
q(2)= 1^2+3^2
q(3)= 1^2 + 3^2 + 5^2
q(4)= 1^2 + 3^2 + 5^2 + 7^2
Find a closed-form expression for q(n), where n is any integer greater than or equal to 0.```no idea how to find a closed form expression
can someone help me
haughty garden
You want to compute the sum of (2k - 1)^2 from k = 1 to n. Expand to find the sum of 4k^2 - 4k + 1; each sum has a well known closed form
plain pulsar
oh i see
sharp wave
I'm trying to prove that this statement is valid
I understand the logic behind this, but I'm not sure how to properly prove it line-by-line
I know all the rules of inference and law of propositional logic as well
Right now I know that not q is supposed to make the conclusion of the first line false
but i dont know how to convey this properly strictly using the rules of inference and laws of propositional logic
keen cave
guys I don't understand discreete math at all can you please help me....
I am confused why is it saying p only if q when it is q only if p
because the result is q
and p is the predecessor
<@&286206848099549185>
also what's the difference between p -> q and q -> p they are just flipped but I know that when P is F then it is always true in the first statement (p - > q)
livid marsh
Anyone who knows annuity
tight vessel
placid carbon
also what's the difference between p -> q and q -> p they are just flipped bu...
It’s just a weird quirk because sometimes english is sometimes hard to translate into formal logic. https://www.khanacademy.org/test-prep/lsat/lsat-lessons/logic-toolbox-new/a/logic-toolbox--if-and-only-if
vast basin
@kindred nymph i am once again requesting your assistance:
i have some work here for a problem and im not sure why its wrong but it clearly is... my thought process was essentially stars & bars the spots into the numbers and find the total choices for each number and then the possibilities for the operator.... clearly im wrong tho
kindred nymph
Stars and bars'ing the operators is definitely the first thing I would try. Let me check your work. @vast basin
So the leading digit cannot be 0 for any number, not just the first
Your construction would allow "40 + 04 = 44" for instance
You have to split this up into 2 possibilities, three 2-digit numbers, or some arrangement of 3,2,1-digit numbers, or 1,1,4.
For the 2-digit numbers, each is going to be 9*10 possibilities, for the numbers. For the 3,2,1 case the 3 is 9*10^2, and the 1 is 9 possibilities (so overall we get the same 9^3*10^3 behavior, similarly for the 1,1,4 case.
That's the only error I spotted, and you made it in both main cases, (2 operators vs 1 operator), but I stopped looking after I found it.
vast basin
Your construction would allow "40 + 04 = 44" for instance
oh shit u right thanks
let me adjust and resend
hmm so
this still isnt right
vast basin
@kindred nymph i accounted for the fact that in both cases the first digit of any number cant be a 0, so thats 3x digits main case 1 and 4x digits in main case 2? the rest are allowed to be however we want tho i think? but i might be overcounting the ways to count the ways to put the digits? because the 9-possible digits have to go first??
kindred nymph
The particular arrangement of the digits into numbers is dictated entirely by the stars and bars.
You should have 4^2 for the number of operators in the second case @vast basin
vast basin
You should have 4^2 for the number of operators in the second case <@38070869742...
oh yes
also: do i add the two main cases??
or multiply???
because with adding idt its right :(
sharp wave

I already solved it but thanks
However I’ve never seen proof by contradiction used to validate an argument before
I thought that was just for theorems
Not sure how that works
kindred nymph
because with adding idt its right \:(
It should be added
meager prawn
However I’ve never seen proof by contradiction used to validate an argument befo...
What's the difference between the proof of a theorem and an argument ?
kindred nymph
@vast basin I think your stars and bars calculation is incorrect
For the 3 operator case there are 5 digits so 4 places an operator can go, and so there are 4 choose 3 arrangements
For the 2 operator case there are 6 digits so 5 places, 5 choose 2.
sharp wave
What's the difference between the proof of a theorem and an argument ?
I'm probably wrong, but thereoms were statements like "the sum of an even and an odd number is odd" while arguments those logical expressions
at least thats what i understand
ive never used proof by contradiction in logical expressions before
vast basin
<@380708697422299146> I think your stars and bars calculation is incorrect
yeah so it is, someone else has the answer (which concurs with yours):
10*(9^3*10^3)*4
+
4*(9^4*10)*4^2
but im not 100% on why precisely that is the answer, something about the 9-possible digits being placed already and you can only place the 10-possible digits with S&B?
oh
i was doing the snb wrong
you S&B the OPS AND THE DIIGITS?????
WHAAAAT
kindred nymph
What were you attempting to SnB?
fringe turret
Confused between logic, reason and logical reasoning for proving arguments of a proof statement.
Since logic and reasoning are intertwined, I am not sure which is exactly what. And if you are given premise and conclusion in the proof. Which is used in the intermediate arguments to reach the conclusion from premise.
I would rather ask you search "being logical means" and "being reasoning means". Both of them contain "reason" and "logic" words in them.
If you are proving something, are you providing reasons or logics or your claims?
I am stuck in this loop since two days 🤦♂️
It started with line "Logic of rules specify the meaning of mathematical statement"
vast basin
@kindred nymph so with this.... i dont know how to go past the 2nd one? because with 2 yellows, how do i count???
do i count 3!*2! ???
am i even coutning 3! on the first one correctly???
primal bronze
I'm trying to solve 9^100 mod 21. I tried using Euler's formula but that wont work since 9 and 21 are non-coprime. I googled and saw that Chinese Remainder Theorem could be used, but we don't use that here and they don't even teach it. How would I solve that?
brave rock
I'm trying to solve 9^100 mod 21. I tried using Euler's formula but that wont wo...
Hint: try calculating a few powers by hand. E.g. what is 9^2, 9^3, 9^4 mod 21? Try just these and see what you find.
vast basin
@kindred nymph i tried this... it no worky :(
primal bronze
Hint: try calculating a few powers by hand. E.g. what is 9^2, 9^3, 9^4 mod 21? T...
Thank you, solved. I found out that by climbing up the powers they are 9^2≡18, 9^3≡15, 9^4≡9, 9^5≡18, and the cycle continues until you reach 9^100≡9. I only need to write the answer more formally. I know there are congruence groups (not sure in english). Maybe I could use those.
I also have to find out why a cycle like that even exists with the powers.
kindred nymph
@vast basin this is a little bit more complicated to count. There are three places A can go, and four that D can go. One of the places D can go A cannot go, and one of the places A can go D cannot. So you have to account for all 3 possibilities, that A is in the special spot for D, vice versa, and both.
But also you have to take into account that A and D could appear as a repeat as well.
We know at least one A and at least one D, but we haven't yet ruled out two or more.
Hope this helps
willow halo
what are the rules for combining same qualifiers? for example, can I simplify (∀xP(x)) ∧ (∀xQ(x)) to
∀x(P(x) ∧ Q(x)), given that ∀xP(x) and ∀xQ(x) are true?
vast basin
@kindred nymph by any chance, does this apply somehow to what you were saying?
(im ngl i barely comprehended what you said)
(but this seems like it might be answer to the problem)
(but idk im very confuzzed)
kindred nymph
You probably do have to use inclusion exclusion
plucky wedge
what are the rules for combining same qualifiers? for example, can I simplify (∀...
just takes a quick search
fair sundial
To get A+ in this class, it is necessary for you to get Final grade on the midterm exam.
is the one above different from this:
To get A+ in this class is necessary for you to get Final grade on the midterm exam?
im still new to discrete math...
coral parcel
That's really more a pitfall of the English language than of mathematics. The first sentence is a slightly stilted way of saying
Getting Final grade on the midterm exams is a thing that is necessary for getting A+ in this class.
whereas the other is a (just as stilted) way of saying
Getting A+ in this class is a thing that is necessary for getting Final grade on the midterm exams.
So the two sentences are making different claims about what is necessary for what.
vast basin
given the prompt and this problem, how do i count? help! (maybe @kindred nymph, if you're around)
problem (as text):
If there are n=8 Zorches in group Z, how many different reflexive and symmetric relations from Z to Z are possible?
kindred nymph
@vast basin this is equivalent to asking for the number of set partitions of a 8 member set.
vast basin
what's a set partition? (i may have missed this in lec, but i dont remember learning it)
kindred nymph
A set partition is where you take a set and split it into several smaller disjoint sets.
So {a, b, c} can be partitioned into
{{a}, {b}, {c}}
{{a, b}, {c}}
{{a, c}, {b}}
{{a}, {b, c}}
{{a, b, c}}
vast basin
ah okay
kindred nymph
vast basin
so, one of the other problems asks this:
Two Zorches either freeb together {{1,2}} or don't {{1}, {2}}. The two possible freebs relations are {(1,1), (1,2), (2,1), (2,2)} or {(1,1), (2,2)}, but it is simpler to just write the sets.
To add a 3rd Zorch, we can add it inside an existing set, or add it in as a set itself: {{1,2,3}}, {{1,2}, {3}}, {{1,3}, {2}}, {{1}, {2,3}}, or {{1}, {2}. {3}} are the five possible freeb relations
How many different freeb relations are possible for 4 Zorches?```i thought this is asking about counting the disjoint sets?
because i made this spreadsheet to count:
kindred nymph
Exactly.
Great job, you managed to derive the bell number recurrence on your own 😄
vast basin
okay
so im wondering, how does this apply to the original problem
asking about the # of reflexive & symmetric relations?
because the number given is 268,435,456
which is $2^{(\frac{n*(n-1)}{2})}$
vital dewBOT
aidanlok
vast basin
and im not gonna lie, what the hell
the "2^" part i kinda get
its choosing whether the "pair" from the cartesian product S x S is in the set
but where the hell does (n*(n-1))/2 come from
kindred nymph
So.
If you have mFn for any m and n
And the relationship is symmetric, then you have nFm
So we only count the ones that are different
That's why you have n(n-1)/2
That's the nth triangle number
vast basin
So we only count the ones that are different
the ones that are different in what sense?
kindred nymph
If aFb then bFa, so if we choose aFb as true, then bFa must also be true, we are not free to choose.
vast basin
the pairs from S x S must come as a pair
okay
so why does that also include the reflexive?
kindred nymph
aFa doesn't have a symmetry to exclude
Maybe it will help if you draw a grid, label the rows a-d, columns a-d and for each grid space write rFc "row F column", so the top left is aFa, the one below it is aFb and so on
Then draw a line down the diagonal
The squares below the diagonal, we are free to choose, this is n(n-1)/2
The ones on the diagonal are set because they are reflexive
The ones above are set by the ones below
vast basin
okay
so say we simplify down to a 2x2
{a,a} and {b,b} are reflexive
if we choose {a,b}, we are required by symmetry to pick {b,a}
this i understand
quick side question: where does n(n-1)/2 come from? the area of a triangle is 1/2bh but both the base and the height are n i thought???
kindred nymph
This isn't a proper triangle though.
The diagonal is jagged
It's missing a little
kindred nymph
It's missing n/2 total squares
Half of each square on the diagonal
1/2 n^2 - 1/2 n = n(n-1)/2
vast basin
ahhhh i see
okay
so if im understanding correctly, the orange colored are the reflexive, and the black colored are the ones we need to choose for symmetric, right?
kindred nymph
Exactly
vast basin
okay
kindred nymph
And the white ones are determined by the black choices
vast basin
yes
okay
so dumb question: why is the count for the reflexive ones 2^(n*(n-1)) according to the answers?
i wouldve thought its 2^n
considering we have n reflexive pairs
vast basin
im not sure i follow
kindred nymph
Reflexive only means that the diagonal is true
But now both black and white squares can be freely chosen
kindred nymph
So if we have absolutely no requirements, then we have 2^n^2 choices
vast basin
yes
kindred nymph
We lock the diagonal we have 2^(n^2 - n) choices
Because there are n diagonal entries
kindred nymph
Reflexive means aFa is always true
vast basin
the problem says:
If there are n=4 Zorches in group Z, how many different reflexive relations from Z to Z are possible?
kindred nymph
We can't choose it to be true or false
kindred nymph
So a relation that is reflexive, we have to make sure aFa, bFb, etc are true, but the rest are allowed to be whatever, aFb can be either true or false
kindred nymph
👍 glad to hear
vast basin
okay okay
kindred nymph
vast basin
this all makes sense now
thanks so much for all the help sir
i very much appreciate it
if you got like a kofi or somethin ill buy you coffee for all the help you give me lmfao
ur singlehandedly savin my ass rn in this class
fleet slate
can someone bully me
kindred nymph
Your pfp is not protected by copyright, and you're bad at choosing the correct channel. 
vast basin
weary tiger
can someone bully me
stfu
fallow dune
yesterday my professor put 30 questions on the test
while the class period is 1 hour 15 mins
30 proofs btw
faint sphinx
fallow dune
basically what im saying is
no1 finished
💀
man istg these teachers setting us for failure
weary tiger
man istg these teachers setting us for failure
they don't care. they're just setting up their bank accounts for the next pay check
weary tiger
i wish i got into a better college
look for good professors. It's not necessarily the college
fallow dune
that is true
blissful shell
uh is this mean A must be an empty set?
obtuse lance
not necessarily
weary tiger
uh is this mean A must be an empty set?
Suppose that $A = {1,2}$ and $B = {1,2,{1,2}}$
vital dewBOT
<:F_button:1095679234497843251>
blissful shell
oh yea thats true, how did i miss that, thx anyway
blazing rose
A={1} is not a subset of its power set right?
blazing rose
Yea
brave rock
Yes indeed, you are right
blazing rose
I’m confused because we had a statement in our last lecture which we had to prove or disprove, it was A is subset of its power set for any A
And our prof used the counterexample A = {{empty set}}
Isn’t that a bit too complicated to prove that?
I mean sure it’s working but {1} would be enough I guess?
brave rock
Not really. In fact under the usual definition of 1, your examples are equal.
In some ways your prof produced the simplest possible example without referring to something that isn't defined.
blazing rose
In some ways your prof produced the simplest possible example without referring ...
Right I see
Empty set is the simplest to use for lots of things I suppose?
In set theory
brave rock
Sure
blazing rose
Okay thanks a lot
vernal vigil
For Transformations of Polygons using matrices, can anyone explain to me how to analyse the before and after and determine what the transformation was?
An example for reference
kindred nymph
@vernal vigil you can select any two pairs of points on the triangle, and set up a system of equations to solve. Let the initial points be denoted with subscript i and final with subscript f. $p_{n,i} = (x_{n,i}, y_{n,i})$
Our transformation matrix $A$ acts on the initial points $P_i$ to produce our final points $P_f$.
\begin{align*}
P_i &= \begin{pmatrix}
x_{1,i} & x_{2,i} \
y_{1,i} & y_{2,i}
\end{pmatrix} \
AP_i &= P_f \
A &= P_f P_i^{-1}
\end{align*}
vital dewBOT
OmnipotentEntity
kindred nymph
This is probably better in #linear-algebra though
mighty flame
Notation question but, if I have an interval [a,b] can I just say let S be a set and S = [a,b]. And would this mean that saying x in S is equivilant to saying x in [a,b]
faint sphinx
Just say "Let S = [a, b]". Or just write out [a, b]
worldly pine
Hi is there a discrete math tutor
spare shuttle
Hello, I am learning about "semantic consequence" in propositional logic and I noticed that its definition seems to make use of a quantifier, with the domain of quantification being all models. Are quantifiers introduced as part of the metalanguage, similar to how metavariables are? Thanks.
https://en.wikipedia.org/wiki/Logical_consequence#Semantic_consequence
coral parcel
The metalanguage for talking about propositional logic is (unless one explicitly specifies something else) the same informal or semi-formal mathematical language we use to talk about other mathematical subjects, such as algebra or real analysis. That definitely lets us quantify over the objects our theory is about.
spare shuttle
The metalanguage for talking about propositional logic is (unless one explicitly...
Ok, that makes sense. I do find it interesting that, for propositional logic, the model-theoretic account of logical consequence requires quantifiers. By comparison the proof-theoretic account uses syntactic consequence that just requires a decision procedure (no quantifiers).
coral parcel
Model theory generally puts higher demands on the metatheory than proof theory does.
This is mostly by design: the purpose of proof theory is to define valid arguments with as few technical prerequisites as we can get away with.
In contrast, model theory is more intuitively meaningful, at the cost of needing more machinery.
We can do proof theory in a relative vacuum, but the most convincing reason to care about it I know is that it happens to be sound with respect to the model-theoretic semantics.
frigid lion
this is the answer key for one of my assignments. my question is about part B. I know that if the number starts with a non zero digit then it's a negative number and zero is postive. I don't understand how they got that the number is positive because it's lower than (444444)9.
grand barn
I just need confirmation that I am answering what they are asking for. I'm practicing for my midsemester. I'm doubting my understanding of what the question is asking me to do
coral parcel
The question only seems answerable if we assume that orange and yellow are equally likely, and green and blue are equally likely, and indigo and violet are equally likely.
But then it's just a matter of writing all the probabilities as multiples of, say, P(violet), and then use the fact they must sum to 1 to discover what P(violet) was in the first place.
grand barn
The question only seems answerable if we assume that orange and yellow are equal...
thanks man.
delicate skiff
Hello, I wonder if there’s a very simple and specific way or certain methodology to comprehend truth tables, I’m suffering with this
vestal bronze
they are tables, i don't think you don't understand them, it must be something else
you mean like making a table in the first place maybe?
delicate skiff
you mean like making a table in the first place maybe?
No like, what defines P in a table for example, how do I know if it’s T or F in the first place, and in the rows it’s in, like rows themselves are what’s complicated I can’t tell how P and Q functions in a two row table, or three row table
p, q, r
Sry, that’s the three row I meant
delicate skiff
How?
vestal bronze
you have some sort of statement, like p → (r & ~q) for example
the table has the same information as that entire statement
you don't know if q is F or T
it's just a super large unpacked equivalent of a short statement
not statement, but like formula idk what's the right word
delicate skiff
When my professor put an example of truth tables she define them in only T and F examples in truth tables only, that’s why I can’t tell
delicate skiff
not statement, but like formula idk what's the right word
I think I get the concept
Conceptually it’s pretty comprehensible but sets themselves are undefinable to me yet
vestal bronze
the whole table just says P→Q
it's rules for what you get from different inputs
like P→Q is code for "something with truth table of TTT, TFF, FTT, FFT"
delicate skiff
If p is true then q is also true if p is true then q is false if p is false then q is true if p is false then q is false
So it’s
Consecutive, successive
vestal bronze
that's not what that table says
delicate skiff
Not specifically the table
vestal bronze
If p is true then q is also true if p is true then q is false ...
it's a contradiction
delicate skiff
I’m reminiscing some ideas my prof aforementioned before because they’re the only ones I understood I need to revisit the courses
delicate skiff
If p is true then q is also true if p is true then q is false ...
it's a contrad...
Yes
vestal bronze
basically there are 16 truth tables, if you have 2 variables
the last column is different, and could be anything
it would be a truth table
and it would correspond to some formula
if the formula is ~P, the table would say FFTT in the last column
if the formula is Q&P it would say TFFF
that's literally all it is
FFFF could be like (P & ~P)
or anything, there's no obvious simplest formula
delicate skiff
Appreciated.
vestal bronze
instead of P → Q you could say P ≤ Q, like if F is 0 and T is 1
that's never done, but that's the meaning
vestal bronze
Appreciated.
i think that one thing kinda explains most of it
"in logic we use → instead of ≤, because it's not supposed to be numbers"
plucky wedge
Hello, I wonder if there’s a very simple and specific way or certain methodology...
For a truth table, it is useful for figuring out the truth value of a compound proposition i.e (p V q) given the truth value of it's propositional variables(p, q), it also useful to see if two compound propositions are logically equivalent, meaning when given the same propositional variables with the same truth value they produce the exact same truth table, or the exact same truth value in all cases. So if we want to prove if two compound propositions are logically equivalent then we start out by listing all of the variables in a table
p q
Now lets say we want to prove the commutative law that says p V q is logically equivalent to q V p.
First we need to check for every possible case that p and q could be. The number of possible non repeating cases we can have is 4. We can have (p = T, q = T), or (p = T, q = F), or (p = F, q = T), or lastly (p = F, q = F). If we try to make more outcomes we will repeat ourselves so these are the only four cases.
p q
T F
T T
F T
F F
Now given each possible case we can add our compound propositions to the truth table and find the output of each case.
p q p V q q V p
T F T T
T T T T
F T T T
F F F F
Note: each row is it's own case and so we are using that current variable truth state for a specific row for the compound's proposition truth value in that specific row, for example we'll use the truth states of p, q in the first row and input it into the compound proposition p V q which would result in the truth value T and that would go in row 1 since that is the case we used(I think that's what you were asking?). Now you can see that p V q and q V p has the exact same truth table or column making them logically equivalent. You will get the same output when using the same inputs for those two compound propositions. A way to know the number of non-repeating cases you'll have is by taking the possible states of each variable which for us is 2 , each variable can only be T or F, lets call this number of possibilities x. Then we take this to the power of the amount of variables we have which is 2, lets call this number n. From this we can see that number of possible cases is x^n(2^2)(possible states of variables^number of variables). If we have the variables p, q, r our possible cases would be 2^3, which is 8.
Hopefully I explained that good, was a good way for me to recall some things I learned also
vestal bronze
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keen cave
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