T(1)
#discrete-math
mystic moss
uh i think you can just have T(0)
shut fjord
this would represent no grapes
mystic moss
yeah, so 0 ways
mystic moss
oh wait you're right LOL
shut fjord
empty set rule I think
shut fjord
how would I be able to grab all the distinct cases?
mystic moss
i mean say i have k grapes
shut fjord
alright
mystic moss
how many can i give to quackers rn
shut fjord
like in one instance?
mystic moss
yeah in this bite
shut fjord
1, 2, or 3
mystic moss
yeah
and then for each case, how many possible ways are there to finish feeding the k-1, 2, or 3 grapes
mystic moss
wdym "for"
shut fjord
if I have k grapes he can only have 1, 2, or 3
mystic moss
oh you mean for the bite that is immediately next
i mean in total. we're finding subproblems here
mystic moss
so we want how many ways to finish feeding the k-1, k-2, or k-3 grapes
shut fjord
that means we can have k-1, k-2, and k-3
mystic moss
mhm
shut fjord
does this involve a summation?
mystic moss
no, now you recurse
shut fjord
I'm thinking..
if its OR
we can do addition
so like
T(k-1) + T(k-2) +T(k-3) ways?
mystic moss
mhm!
shut fjord
I sort of get it from a set stand point because we have OR
because duck can eat it in 3 different ways
1, 2, or 3
does T(k) calculate the ways for me?
They only wanted the recursive algorithm for this problem so maybe I'm over thinking it?
mystic moss
yeah you just need to find T(k) in terms of T(smaller than k)
shut fjord
oh wow
it does work
I just counted it from my n = 4 example
n = 4 --> T(4-1) + T(4-2) + T(n-3)
and I can literally count on my screen the # of ways from n = 1 to n = 3 and it does give me the ways of n = 4
thanks
bronze rain
This looks correct but you need to show it for all n.
noble wasp
how would i do that
bronze rain
Induction. You have the idea, just need to write it down now.
loud matrix
Could somebody help me out with a question about combinations
Reply as I prefer DM's instead
dusk stirrup
Not looking for the answer here, but is there an easier way to find the chromatic number of k11,15 without having to draw the graph out.
faint narwhal
yes
loud matrix
^ lol
dusk stirrup
What about it? I'm not looking for the chromatic number of k11,15.. I'm looking to find out if there is a solution that doesn't require drawing it out. One is the answer, one is a solution...
faint narwhal
Sorry, I was unsure if you wanted more information
So I left it at the bare minimum and then you could ask if you wanted more information
dusk stirrup
No worries.
faint narwhal
My hint is to maybe think about K_{m,n} in general. The 11 and 15 don't matter at all here
Is there an easy coloring you can give to K_{m,n}?
loud matrix
Anyone available?
<@&286206848099549185>
Any helpers can help me with this?
Just reply if you're available
granite trail
you ask to ask, post in the wrong channel, dont indicate that you've tried anything, and need an answer quick
Galaxy brain
weary tiger
So far using demorgans and seem to be getting no where any help with this one
Some of my work A' U A' ∩ C' U B'
A' ∩ (C ∩ B)
`means not
or compliment
They are not given :((
The universal set
What if we assume they are not empty? I am tempted to say false but I feel like there should be some kind of answer
This is all that is given
granite trail
Can it be false if he's being asked to prove the equivalence ?
weary tiger
I am almost 100% sure it is not false
granite trail
Well if you have (A u (C n B))' then that is the same as (A' ^ (B n C)') right. ?
weary tiger
yes
granite trail
Excuse my notation
weary tiger
yes I understand
For the right can I say
((A U B) upside down u (A U C))`
therefore A
A`
<@&286206848099549185>
hardy quartz
hello
i have this error when running my program in octave
this is my program
this is the error
this is my function
lofty crater
Am I allowed to cross post my question from #help-9? I would like to know if a series has an asymptote or upper bound, but I fear I lack the knowledge to determine it myself, so far.
lofty crater
Well, I got past the first hurdle. Now I must have messed up a number 2 where I should have had an $h$. The "Adjustment" in the following graph only works as long as $h=2$ (a half life of two units). The target end population $E$ works, as I can set it to any percentage of the initial population $N_0$ and the exponential decay curve fits. The somewhat poorly labeled $f(x)$ is the recursive function over which 25 iterations are made.
vital dewBOT
Roenbaeck
lofty crater
The graph I am fiddling with is here: https://www.geogebra.org/graphing/vnj3qsmm
The formula for "Adjustment" is obviously wrong, even if it works when h is 2.
lofty crater
Never mind, I found where I went astray. The adjustment should have been $(2^{\frac{1}{h}} - 1)E$.
vital dewBOT
Roenbaeck
lofty crater
The series $$N_n = \sum_{i=0}^{n} (N_{n-i} + C)(1-p)$$ looks like it has a linear asymptote, when $0 \le C < N_0$ and $0 < p < 1$. What kind of math would I need to verify this and possibly find the equation of the asymptote?
vital dewBOT
Roenbaeck
lofty crater
This is summing a series of exponential decays, but essentially the same.
It may be a coincidence and there is no asymptote, but if not, then I don't know how to prove that either.
lofty crater
The series $$N_n = \sum_{i=0}^{n} (N_{n-i} + C)(1-p)$$ looks like it has a linea...
For future reference, the asymptote exists and it is $$A(x) = C(x + 1) + \frac{N_0-C}{p}$$.
vital dewBOT
Roenbaeck
dire sphinx
what's the cartesian product of two sets of sets?
{{}, {0}} x {{}, {1}} = {<{}, {}>, <{}, {1}>, <{0}, {}>, <{0}, {1}>}
is this correct?
lofty crater
what's the cartesian product of two sets of sets?
That's correct
dire sphinx
nice, thank you 🙂
saw an example where they proceeded to take the cartesian products of the inner sets as well, i guess that's wrong then
naive gulch
I don't understand the definition of graphs, what would the third entry be ?
if f(x) was x ^ 2, then the first ordered pair is (x, x^2)
so does x^2 then become the x for the next ordered pair?
naive gulch
oh so like (1, 1^2) (2, 2^2)?
solid garden
could someone explain this? i understand the first part becaue k = 0 works with anything, but i dotn understand the second part
faint narwhal
what second part are you talking about
valid wave
Hi im confused with combinations
if im looking at possible name initial combinations
it would be 26^2
I get that because its 26 for first initial and 26 for the second
so its choices^number of "times"
?
valid wave
so im confused with the idea though
so its options^"holes"
ig?
how should i think about it?
Likw making 3 element lists with 5 numbers
5 options for numbers with 3 "holes"
so 5^3
3^5 seems like it would make more sense tho
idk
im confusing myself
thinking about it
errant bear
you have n many options for the first "hole", and if you have replacement, you still have n options for the second "hole", and same with the 3rd hole, so you have n * n * n options total
valid wave
ahhh gotcha
options^hole
for some reason I think thinking about pigeonhole theory when thinking about this
valid wave
hi if anyones up
when looking at an equation such as the second
whats the point of the k-1
so we dont go out of bounds?
im not quite following that part
im assuming its something with that?
could someone elaborate
unreal stump
You can't choose the option you already chose
So,for first you have n options
For second you have (n-1) options
For third you have (n-2) options and so on
You get a pattern that for kth,you have (n-k+1) options
valid wave
ah ok
so when considering that
crap
im just struggling to understand that n-k+1 part
so with a size of 3
and 3 options
we have (3 - 1 ) x (3 - 2) x (3 - (3 - 1))
unreal stump
You notice a pattern and say that's (n-k+1)
valid wave
but isnt it being "selected" again
unreal stump
we have (3 - 1 ) x (3 - 2) x (3 - (3 - 1))
It should be 3(3-1)(3-2)
valid wave
because the last one is still 1
unreal stump
First one you have 3 choices
valid wave
and second to last is 1 aswell
unreal stump
Second to last should be 2
valid wave
OHHH
crap
did it wrong
ahhhh
3 x (3 - 1 ) x (3 - (3-1))
gotcchaaaa
makes sense
last question if you have time
what about n < k ?
unreal stump
Not possible
i.e,You get 0
Because you can't choose a 3 tuple out of 2 things
If the elements have to be distinct
Because you have 2 options to choose first one
And 1 option to choose 2nd one
After that you have no options for 3rd
unreal stump
No,it will still give you 0
Because when you are going from positive to negative, you encounter a zero term somewhere
Like here,using the formula you get
2(2-1)(2-2)=0
opal cedar
I am dealing with sets of strings and are asked what the shortest string over alphabet {1} that is not in { e, 1, 1^2, 1^3, 1^5 }^3. How do I go about expanding a set with an exponent?
the exponents refer to repeated characters, so the set actually looks like { e, 1, 11, 111, 11111 }^3
weary tiger
@opal cedar ok so we are looking for the smallest n such that 1^n is not in that set
opal cedar
yes, my confusion comes from what do you do when the set is noted as { e, 1, 1^2, 1^3, 1^5 } ^3
weary tiger
so what that means is
opal cedar
the set itself being raised to the third power
weary tiger
you are getting a new set that is
all the elements such that it is the product of 3 elements in the set
opal cedar
so does this end up being if the set is A, A x A x A
weary tiger
yeh so in this case its just
all the strings 1^{i}
such that i is a sum of 3 of 0,1,2,3,5
including repeats
if that makes sense
opal cedar
sets can have repeats?
weary tiger
im explaining this rlly badly sorry for that
but imagine you have {e, 1, 1^2, 1^3}^2={e,1,1^2,1^3,1^4,1^5,1^6}
its just the new set made by picking two elements in the set and combing them
opal cedar
I end up with this huge set
weary tiger
basically you are doing AxAxA and then getting all the 3-tuples (a,b,c)
and converting them to abc
loads of repeats
you can do it quite simply by just looking for low n where 1^n isnt in the set
should get n=14
opal cedar
['', '1', '11', '111', '1111', '11111', '111111', '1111111', '11111111', '111111111', '1111111111', '11111111111', '111111111111', '1111111111111', '111111111111111']
is the set
with no repeats
weary tiger
yeh
opal cedar
I get n=15 with the empty string
weary tiger
(1^5)^3 = 1^15
so smallest element not in it cant be length 15
these are the lengths of the elements in it
shut fjord
Having trouble thinking of a recurrence relation for this problem
I believe it is a dynamic programming problem but how would I express minimizing all of the different operations we can do?
To summarize what the question is asking for, basically they want word A to be the same as word B using the 3 operations in bullet form, and they can have different lengths
weary tiger
Yes thank you I solved it with a derivitive of what you did A` is universal set - A and then the right side I used set difference
hidden totem
can anyone help with providing a proof for the following combi identity?
$\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$
vital dewBOT
CosmoVibe
hidden totem
i also accept answers here if anyone wants some rep: https://math.stackexchange.com/questions/4021206/proof-of-sum-k-0n-binom2kk-22n-2k-2n1-binom2nn-frac
mystic moss
To summarize what the question is asking for, basically they want word A to be t...
what subproblems have you thought of
shadow pond
Hello
viral patioBOT
valid wave
wait so
im doing this question
but under the explanation it says
that n! is
but I thought it was just up to the (n-k+1)
where did that (n-k) and (n-k-1)
and all that come from
hidden totem
@valid wave you're misreading that bit
it is giving you the definition of just regular n!, not the permutation formula in the first screenshot
they are writing out all of the factors from n-k down as well so people can see why it cancels out after dividing. it's not the most intuitive way of writing it admittedly, but that's just wikipedia writing style i guess
naive gulch
Can I use the logical equivalency laws for truth tables to prove unions for the distributive law?
or can I just prove that for the bottom question they're subsets of each other?
I'm not too sure how to go about it that way
minor dagger
just show they are subsets of each other
pick an arbitrary element of one and show it is in the other
and vice versa
naive gulch
pick an arbitrary element of one and show it is in the other
like an actual element like 5 ?
minor dagger
no
naive gulch
or a ∈ A
minor dagger
some arbitrary element like (x,y)
yeah
if it is arbitrary and you show that it holds then it can be said it will hold for all elements
naive gulch
So saying that a ∈ A for A x (B ∩ C) and also a ∈ A for (A x B) ∩ (A x C) would prove it?
if we have arbitrary element a in A that's in both
minor dagger
if you are taking an element of the Cartesian product it needs to be an ordered pair
naive gulch
Oh wait yeah
It can only be one element where the Cartesian product is like A^3 right
naive gulch
wait sidetracking but if set A = {a} and A x A x A
would we be using {empty set, a}
as our pair?
minor dagger
you would have (a,a,a) as your ordered pair
but we wouldnt be using a finite set
unless you were looking to show it on a finite set
naive gulch
ah okay thats what I put
minor dagger
use the set of real numbers
naive gulch
naive gulch
yeah sorry I totally sidetracked because I realize I might've done another problem wrong
So in the case of what I was saying before I can use (x, y) and I need to prove that (x, y) arbitrary elements ∈ A for both the cartesian product
minor dagger
for sets A,B,C?
minor dagger
you're gonna wanna assume that (x,y) is an element of one and show its in the other
and vice versa
naive gulch
by saying of one do you mean the entire left hand and right hand side or set A / B etc
minor dagger
$(x,y)\in A \times (B \cap C ) \Rightarrow (x,y)\in (A \times B) \cap (A \times C)$
vital dewBOT
Star_
naive gulch
Ah okay, why would it not be x, y, z because we have A B and C all in play?
minor dagger
no
naive gulch
Sorry for the confusion I'm definitely feeling a little slow in the brain right now but thanks for the help
minor dagger
no
naive gulch
okay yeah then I get why we aren't using 3 that makes more sense
minor dagger
If you want to show it you could break it down a bit
try using set builder notation
$\left { (x,y)\in \mathbb{R}|x\in A \wedge y\in(B\cap C) \right }$
vital dewBOT
Star_
minor dagger
like this
minor dagger
if you do this a few steps further you will get to the result you want
short steeple
i'm setting up an inductive proof of this problem (my question is if induction is the right way to go about it):
Let $n$ be an arbitrary positive integer. What
is the least positive integer $m$ for which $K_n$ is a minor of
$K_{m,n}$?
vital dewBOT
Snodlop
short steeple
actually.... i think i'm finding a formulaic method for this
i'll probably be back here with a more formative question soon
ok i have a question about this: a part of the problem that i didn't include (because it's sort of implied) was that m must be some function of n. however, can't m just be 2? if Kn has to be a minor of Km,n, can't i just reduce it down to two vertices and their edge?
to that end, why not 1?
oh wait
kn has to be the minor
oh i'm a dummy lol
hang on
yeah lol
it would have to be n-1 then, right? the only real way to turn the bipartite graph into a complete graph would be to contract edges, so you contract n-1 edges to get the first n-1 vertices of Kn
weary tiger
Can someone help me prove or disprove this proposition? I’m stuck after this step
weary tiger
@haughty garden isn't it ¬q V ¬q?
haughty garden
Oop yeah
shadow pond
hey guys
i got a question about setting up my proposition
i am having trouble converting the english into a statement
weary tiger
Yeah I did that to give me T v (¬q ^ ¬q) which then gave me T v ¬q
but them im stuck after that
haughty garden
The result follows
vital dewBOT
opengangs
weary tiger
ohh okay, thank you so much
rough zenith
Can someone tell me what I did wrong here?
obtuse lance
@rough zenith don't cross post your questions
@weary tiger why did you delete your question
haughty garden
@rough zenith
Hint: your index starts at n = 1
shadow pond
how can i condense this using laws of equivalence
first thing i did was apply the negation
but then i'm stuck
i have (p ^ ~q) ^ (p || q)
haughty garden
Distribution law looks like the next logical step
naive gulch
How should I approach this problem?
I think I can prove the ^-1 part but I don't know how to explain why they switch places
g and f
obtuse lance
compose them all together
and see why
think of it this way, you put on your socks, then you put on your shoes, what do you take off to undo this, do you remove your socks or shoes first?
naive gulch
Ooh that makes sense I understand the concept I think I just need to work on formally explaining why that works
obtuse lance
like I said, write it out
put all the f,g,f^-1 and g^-1 written and make them all compose to make the identity function
naive gulch
the identity function being g^-1 o f^-1?
vital dewBOT
Merosity
obtuse lance
you could write this as $(f^{-1} \circ f)(x) = id(x)$ if you wanted
naive gulch
$(f^{-1} \circ f)(x) = \id(x)$
vital dewBOT
Stanford
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
Merosity
obtuse lance
it doesn't like a space at the end before the $ for some reason
naive gulch
Ah yeah I've seen that I think in my textbook
So using the two functions and its two invertibles we can get to the identity function, but I'm a little confused on why we need to do that?
obtuse lance
you write it that way so you can drop the x and just write $f^{-1} \circ f = id$
vital dewBOT
Merosity
obtuse lance
confused in what way
specifically $(f \circ g)^{-1}$ is the inverse of $f \circ g$ but you're wanting to know what it is in terms of $f^{-1}$ and $g^{-1}$
vital dewBOT
Merosity
naive gulch
so because we want to put it in terms of f inverse and g inverse we want to drop the x via what you sent above?
obtuse lance
no
we can do it with x
maybe that's easier
they already told you the inverse is to flip them right
$g^{-1}(f^{-1}(f(g(x)))) = x$
vital dewBOT
Merosity
obtuse lance
this is what you want to show
or alternatively $g^{-1} \circ f^{-1} \circ f \circ g = id$
vital dewBOT
Merosity
naive gulch
Ah okay thats what you meant by compiling them together and getting the identity, I wrote that down but wasn't sure how it connected sorry for the confusion 😓
obtuse lance
it's ok
so start from here $g^{-1}(f^{-1}(f(g(x))))$
can you simplify this at all?
vital dewBOT
Merosity
naive gulch
yeah f inverse (f (x)) is just x so it would just be g inverse g(x) which simplifies down to x right?
I sorta know how to use texit but it'd take me ages so I'll write informally for now
naive gulch
yeah if it was f inverse(g inverse(f(g(x)) it wouldn't work
obtuse lance
cool
humble bridge
How do I solve this with induction? I've dealt with recursive functions like T(n) = 2T(n/2) etc, but in this problem particularly, I have two recursive terms on the right hand side. What am I supposed to do differently?
like, what should I assume T(n/2) or T(n/4) as?
weary tiger
<@456226577798135808> why did you delete your question
cuz no one was answering
honest forge
I am confused by what they mean for 'so no negation is to the left'
honest forge
So I'd start by doing something like P(x)="x can learn new tricks" then negation would be Ǝx~P(x)
naive gulch
Like instead of leaving it as ¬(x < 3) it would be (x greater than or equal to 3)
I think they just don't want you to say "Some old dogs cannot learn new tricks", instead of using not they want you to phrase the sentence where its a negation without the not
honest forge
express the statement using quantifiers means make it: P(x)= x can learn new tricks, and ƎxP(x)
The negation of the sentence would be, some dogs cannot learn new tricks
or do i say x cannot learn new tricks
weary tiger
There are 6 couples: 6 wives and 6 husbands. They want to sit around a round table
How many ways are there to sit them so that every man sits opposite his wife?
can someone pllllllz answer this
pale epoch
what have you tried
weary tiger
@pale epoch #help-7｜zen1thxyz
fluid remnant
If there are two undirected graphs that have the same number of vertices and edges, and the degree of every vertex is the same, is that sufficient to prove that the graphs are isomorphic?
(I didn't lift that from a homework question--the question is just "are these graphs isomorphic?" and I noticed that they have those properties in common.)
fluid remnant
"degree of every vertex"?
did I use the wrong term? Every vertex has three edges connected to it, in both graphs.
fluid remnant
Yes
pale epoch
the answer is not
fluid remnant
Ah, thank you
clear sierra
How did they get from here to here?
clear sierra
<@&286206848099549185> pls
past iris
x_1x_2x_3 can be considered as numbers
E.g for the first minterm which 2
The binary equivalent is 010
So first digit x_1 should be zero, second digit x_2 should 1 and third digit x_3 should be zero
(x_1)'(x_2)(x_3)'=010=2
clear sierra
Omg thank you so much!!
uncut matrix
just wanted to link this video as it is by far the best explanation to the well ordering principle I've ever come across and something I've been struggling with for a few days - thanks again for the help you guys always provide me btw! (maybe I'm just dumb lmao)
https://www.youtube.com/watch?v=DeMEQXoB7T4
vagrant elm
Hello! I have a question about question 1. here:
I am not sure how to compute the intersection between the two automata's languages
I am only given this diagram so I assume I first have to figure out what L(M1) and L(M2) is, which I did here:
But I'm not sure what to do now
fallow pawn
how do i find the probability of getting a hand of poker card that contains no card that can be divided by 3 and cant have the same suit? the rank and suit can appear twice
pale epoch
Hello! I have a question about question 1. here:
technically you don't need to figure out the accepted languages
given two FSA you can construct one that accepts their intersection
in this case the automata are more or less identical
they only differ by accepting states
the accepting states are disjoint however (you can also see this in the languages you wrote down)
vagrant elm
Yeah, so how would the diagram look? I'm very new to this and we have only been shown how to construct the union
How would I construct the diagram?
pale epoch
its essentially the same way
except for the fact that you have to change the accepting states
vagrant elm
Can you point me to a guide or something?
pale epoch
do you have notes on the union?
vagrant elm
Not really, prof just went over it really quickly
All I have is this
From a slide
(the M that's cut off says M2)
pale epoch
yes, thats the general way
you can do the same for intersection
except
wait what is that letter
anyways, so what this construction does essentially is simulate both automata at the same time
you have to do the same thing for the intersection
vagrant elm
that's a q
pale epoch
but in the union case a state is accepting if either the first state is accepting in the first automata
or the second state is accepting in the second automata
so (q_i, r_j) is accepting if either q_i is accepting in M_1 or r_j is accepting in M_2
in the case of the intersection, you have to change it such that a state (q_i, r_j) is accepting if and only if both q_i is accepting in M_1 and q_j is accepting in M_2
[also a side note you forgot to mark the starting state in your picture]
vagrant elm
i'm lost
pale epoch
then go through the union case again until you understand it first
there are probably youtube videos about this
vagrant elm
I actually looked for some but I couldn't find a good one. I'd love a recommendation though, or perhaps a link to a website describing it
pale epoch
Sipser's Introduction to the theory of computation gives this proof idea
and this proof
vagrant elm
Thanks, I'll take a look tomorrow since it's late here. Appreciate your help!
balmy hornet
I wanted to clarify something on my end. So if two sets T1={1,2,3} and T2={3,4,5} and we are trying to find the |T1 U T2|. It would just be {1,2,3,4,5}? Since there are no negative then I do not need to worry about changing any numbers from neg to pos. This would also apply to symbols Δ ,∩, and minus symbol?
errant bear
I wanted to clarify something on my end. So if two sets T1={1,2,3} and T2={3,4,5...
your answer is correct if you dont mean the cardinality of T1 U T2 but wdym about changing from neg to pos
balmy hornet
your answer is correct if you dont mean the cardinality of T1 U T2 but wdym abou...
Since | T1 U T2 | would make all the values inside be positive since it is the absolute value of it. So the answer would be | {1,2,3,4,5} | which is then just {1,2,3,4,5}
errant bear
how do you define cardinality 
balmy hornet
how do you define cardinality <:hmmm:773801269532819466>
its like a of a set is the number of elements it contain
errant bear
notation wise
balmy hornet
notation wise
just |S|?
errant bear
so do you see the confusion
balmy hornet
so do you see the confusion
so then |{1,2,3,4,5}| (sorry for the delayed message)
errant bear
is {1,2,3,4,5} a set? (yes)
so letting S={1,2,3,4,5} makes it clear that it isnt clear what you mean
@balmy hornet
usually, if you want to take the abs val of elements of a set, youd just define a new set
so like, let S be a set, then define T = {|x| : x in S}
faint narwhal
Wow. That's super clear. Thanks for the explanation!
balmy hornet
<@269605444996038658>
Sorry ok, let ,me refer back to the labeling from how i asked the problem. I do no know why i decided to put in S so now it is confusing me. my mistake. If T1={1,2,3} and T2={3,4,5} were unioned then T1 U T2 = {1,2,3,4,5}. |T1 U T2| = {1,2,3,4,5} ( T = {|T1| and |T2|} )
balmy hornet
Wow. That's super clear. Thanks for the explanation!
making me nervous lmao
errant bear
this still doesnt rlly make sense, |T1 U T2| means the cardinality of the set T1 U T2, unless you specify otherwise. anybody who sees that without you speciffying that |S| := the set of the absolute value of elements of S.
balmy hornet
this still doesnt rlly make sense, |T1 U T2| means the cardinality of the set T1...
I can show you the problem? I made that problem with different numbers so i can figure out the other one for myself. Maybe im missing something
balmy hornet
Im going to post a part of it so its not bombardment
Consider two sets S1 = {1, 2, 3, 4, 5} and S2 = {1, 3, 5, 7, 9}. Compute the following: |S1 ∪ S2| =
errant bear
so why are you taking the abs value
did they defin |S| to be the set of the absolute value of the elements of S?
balmy hornet
did they defin |S| to be the set of the absolute value of the elements of S?
no? like they do not directly say that but also to denote its cardinality by writing |S|
im sorry im alot more confused than i should be
errant bear
if they denote cardinality as |S| for a set S
then its asking for the cardinality of the set S1 U S2
balmy hornet
i dont get the point of this problem
like
problem 1 : S1 ∪ S2
then Problem 2 : |S1 ∪ S2|
@errant bear
faint narwhal
|S1 U S2| is the cardinality of the set
limpid root
my choice is the first answer, but im not 100%. can someone prove me right or wrong?
cobalt latch
can a euler circuit have bridges?
can a euler path have bridges?
winged sun
Hello everyone!
Quick, probably very simple question:
When you want to check combinations out of a set, where some elements are identical, how does that work?
For example set has 6 elements, with 2 out of them identical. And you want to know how many subsets there are. Is it just 5 C 3?
Thinking about it I would think it's perhaps 6 C 3 divided by something
stray reef
it's not a set if it has duplicates
winged sun
Right!
How would you phrase it?
Say there are 6 balls two out of which are identical, and we want to see how many different ways we can choose 3 out of them
stray reef
i'd split into cases based on how many of the identical balls we pick
winged sun
And then divide the cases by the permutations of those, then sum them up?
stray reef
idk what division you're talking about
say we have 6 balls which are all differently colored except 2 of them are red
we make three cases: no red balls, 1 red ball, both red balls
winged sun
And how do you count the combinations of each case?
stray reef
you need to choose 1, 2 or 3 balls respectively from the 4 pairwise distinct balls that remain
winged sun
I need to calculate how many possibilites there are of throwing 3 six sided die in a way that the results are sorted upwards (so like 1, 3, 6) for example
Any Tips?
errant bear
do you mean, like the first die has to be lower than the second, and the second lower than the third
digital wadi
Φ intersection {Φ} = null/void right?
stray reef
"null/void"?
do you mean the empty set
if so, then yes this is true for any set Phi
digital wadi
yes. that's what I meant :)
BUT, due to me being greek, I mistook that symbol of empty set (crossed zero) with a φ...
But I presume the same with what I though Phi applies to empty set just as well.
(Set are amazing, but sometimes you can feel you're going insane with set within a set within a set etc)
stray reef
$A \cap {A} = \varnothing$ is true even for nonempty $A$ because no set contains itself as an element
vital dewBOT
Ann
stray reef
$\varnothing \cap A = \varnothing$ is also true for any set $A$
vital dewBOT
Ann
hardy quartz
so i have a function that looks like this rn+1 = rn/2
how would i change that to rn = whatever
am i not remembering some highschool maths things ?
vast narwhal
multiply both sides by 2 (to clear out denominators, can make problem easier to read and compute)
fallow pawn
can someone explain this? i dont quite understand
winged sun
Hello, I'm quite challenged by an exam question:
\newline
The alphabet A consists of 16 letters, including x. \newline
a) How many words of length k are there, that include x exactly twice?
\newline
I came up with $15^{k-2} * (k-1)^2$.
The reasoning behind this is to first calculate all the words of length k-2 without any x's, then multiply by all the positions an x could take, so k-1, and multiply by the positions the second x could take, another k-1. This solution makes sense to me.
\newline
My friend has a different solution ${k \choose 2} * 15^{k-2}$. His reasoning is to first figure out all the possible positions for the x's, and then multiply by all the possibilities of the remaining spots. Also kinda makes sense.
\newline
Could someone explain who is correct and why?
naive saffron
Your friend is correct here
vital dewBOT
Wey
naive saffron
In your reasoning, you said the number places to put the second x's is actually k since you now have k-1 objects and k gaps to place the second object
further more, you over counted
winged sun
But wouldn't it also be k-1 because if it's to the left or to the right of the first x, it would be the same position
I'd be very curious to hear about the overcounting!
naive saffron
ok let me give an example
say a word is aaxaaaaxaa (im too lazy to write 16 letters but the point is the same)
you can obtain this by putting in the first x then the second x, so you get aaaaaaaa -> aaxaaaaaa -> aaxaaaaxaa, or aaaaaaaa -> aaaaaaxaa -> aaxaaaaxaa
so the same word aaxaaaaxaa is counted twice
winged sun
ahhhhh
naive saffron
now to account for this, you can just divide by 2
winged sun
that's awesome
naive saffron
but also, this division by 2 will also fix what you mentioned about two x's being adjacent
naive saffron
lmao it's more just experience
anyways
so what you should've gotten was $\frac12(k-1)k\cdot15^{k-2}$
winged sun
Could you explain how it would fix the adjacency problem?
vital dewBOT
Whoever
naive saffron
oh because we counted a word like "axxa" exactly twice like I said before, but we divided by 2 so we only counted it once
winged sun
we're counting it more than twice though right?
Aren't we counting it 4 times?
ax1x2a ax2x1a, and then the same thing by placing the other one first?
naive saffron
well
no
if you want to get axxa, the first x can only be placed at 1 spot
but the second x has two positions to be in
namely to the left or right of the first x
ok so
try writing out all the 4 ways you claimed there are
and you'll realize that some of them are the same
winged sun
Really confusing
winged sun
i think i get what you mean
naive saffron
ok
winged sun
Crazy though to me
That you just see it immediately
Could you help me with the rest of the questions?
so b and c
b) How many words of length k are there, that have at least one x?
My solution is
$16^k-15^k$
vital dewBOT
Wey
winged sun
And I thought it would be pretty obvious, but my friend has a different one again
He started again with the positioning of the x
naive saffron
yes
winged sun
And has
naive saffron
there are 16 different places the x can be
winged sun
$16^k-k$
vital dewBOT
Wey
naive saffron
winged sun
Is mine correct?
naive saffron
both seem wrong
winged sun
Here's my logic
16^k is all the words of length k
15^k all the words without an x
so i just subtract that
winged sun
Ok, I wasn'T really getting his explanation anway
But good, because this makes sense to me
and c, which I also believe I have right is:
How many words of length k <= 16 are there, whcih don't contain a single letter twice
$sum_{k=1}^{16} {16 \choose k}$
vital dewBOT
Wey
naive saffron
well you also need to account for the different ways to rearrange the letters
naive saffron
yeah
winged sun
Ok nice
Awesome
So my job is to think abit about the double counting thing you mentioned earlier
I have a really tricky one
That apparently only gave 5/60 points
The following describes a new game of chance.
There are 7 red, 5 blue, 3 purple balls. The moderator now throws a 4-sided dice for each red ball, an 8-sided dice for each blue ball, and a 12-sided dice for each purple one. He then writes the respective results on the balls.
Next, all 15 balls are put in a box, mixed, and pulled out one by one. How many possible arrangements of balls can there be?
I started by figuring out the possible positions of balls
So I multiplied the red and blue positions ${15 \choose 7} \cdot {8 \choose 5}$, the purple ones take the remaining ones.
vital dewBOT
Wey
winged sun
And then I tried to continue but got dumbfounded by the added complexity of the balls possibly having the same numbers
naive saffron
ok look at the red balls in a line
then you get a list of numbers all between 1 and 4 right?
you can think of this as a number written in base 4
so it's precisely how many 7 digit numbers there are in base 4
naive saffron
there's probably another way to do it, but this is the first that came into my mind
winged sun
Now we figure out the psoisibilites for the blue and purple ones
and then we multiply the above result by those three numbers right
winged sun
so it's precisely how many 7 digit numbers there are in base 4
wait isn't this just 4^7?
naive saffron
no
oh wait yeah
it's probably just 4^7
numbers having the same number doesn't matter
winged sun
But we would be double counting right
naive saffron
i don't think so
ok this is proof that i don't have enough sleep so im going to leave here, i have already made 2 mistakes
i think this is correct
winged sun
Haha okay have a good sleep, thanks alot anyway
I'll go take a break and continue tomorrow as well
fallow pawn
can someone explain the why behind this for finding # of terms
quaint star
See how the signs alternate between + and -
For even numbers, you want a - and for odd numbers you want a +
Now you don't know whether n (the number of the last line) is even or odd
But if n is even, then (-1)^(n+1) will be -1, so you get a -
And if n is odd, then (-1)^(n+1) will be +1 so you get a +
@fallow pawn
So it's just continuing the pattern, but you don't know whether the last line should be + or -
When you actually do it, you don't need to remember that. Just alternate the signs.
fallow pawn
@quaint star is there a way to say how many total terms there are when it is n set?
like 3 sets, there are 7 terms, and 15 terms for 4 sets
im having trouble identifying a pattern to calculate how many terms exist in n set
quaint star
It's how many ways can you choose one set from the n sets + how many ways can you choose two sets from the n sets + ... + how many ways can you choose n sets from the n sets
Which is actually equal to 2^n - 1
Another way to think of it is, there are n sets, and for each set you have two options (include it or don't include it), so that gives 2^n possiblilities. The -1 is because we don't consider the possibility where we exclude all the sets.
fallow pawn
so giving n sets. can i answer this like there will be (2^n)-1 total terms because there are two options when choosing everyset, thus 2^n. Then 1 is being subtracted to take out the overcount where all sets are counted more than once
@quaint star
quaint star
There is no overcounting in the sense that you are counting the same thing more than once
2^n includes the option where your choose not to include every set in your intersection, we don't want to include that, so we subtract 1
fallow pawn
gotcha, and about what you said earlier regarding + -
given the same logic and m has to be even. i do not know the signs to both of these conditions right?
since for the 1st one, 2 set results in -
and the second condition, an even number can be halved into two odd numbers and also two even numbers
@quaint star
quaint star
So, do you know the sign if you don't know whether it is even or odd?
quaint star
I don't understand what you mean about the special case
But, that's it, you don't know because m/2 can be even or odd
fallow pawn
I also don’t know for the condition right?
Wait no I do
I got it mixed up lol
Thx a lot for the clarification
slow jewel
I hate discrete maths
round geyser
is discrete morse theory a hot topic of research
umbral aspen
It depends on what you mean by "hot topic" @round geyser
umbral aspen
I would say judging by basic searches of peer reviewed articles, not comparatively to other branches of mathematics. Probably doesn't have billions of dollars thrown at it.
cobalt latch
can a euler circuit have bridges?
can a euler path have bridges?
fallow pawn
how can i determine if this is an overcount, undercount, or accurate
faint narwhal
do you know what the inclusion exclusion principle says
fallow pawn
do you know what the inclusion exclusion principle says
yea
faint narwhal
Then just compare it to what is written
fallow pawn
Then just compare it to what is written
is there a more sufficient way to determine if it is over under or accurate count without comparing the sum of each set intersection
faint narwhal
What
You just compare it to what inclusion-exclusion tells you
what they have written down there is missing a set of terms
fallow pawn
wait it would just be an under count right since missing set-3 terms
am i just overthinking it
mystic moss
can a euler circuit have bridges?
can a euler path have bridges?
What have you tried
weary tiger
Can someone help me with proving this using logic laws, I keep getting to T and T but that isn’t a logic law
cobalt latch
What have you tried
idk it just ask if they can have bridges or not in general and draw an example to support it
winged sun
Would appreciate if someone could have a look at my solution to the following exam question:
The following describes a new game of chance.
There are 7 red, 5 blue, 3 purple balls. The moderator now throws a 4-sided dice for each red ball, an 8-sided dice for each blue ball, and a 12-sided dice for each purple one. He then writes the respective results on the balls.
Next, all 15 balls are put in a box, mixed, and pulled out one by one. How many possible arrangements of balls can there be?
So I multiplied the red and blue positions ${15 \choose 7} \cdot {8 \choose 5}$, the purple ones take the remaining ones.
vital dewBOT
Wey
winged sun
Then I multiply that by $4^7, 8^5, 12^3$.
vital dewBOT
Wey
lost pivot
What’s the difference between a partially ordered relation, a < B and B not < a? Aren’t both the same?
winged sun
I think in this case it's just because it should be B not <= a, no?
lost pivot
a partial order doesnt require every pair of elements to be comparable
So does, B not < a, imply these elements are not comparable?
lost pivot
I think in this case it's just because it should be B not <= a, no?
@winged sun how does the equality make any difference here?
stray reef
you could still have a < b
lost pivot
you could still have a < b
Right, now I get it. So b not < a is true for the A’s such that a <b and the A’s that are not comparable to b, right? Edit : not comparable to b
hardy yoke
∃a∃b(P(a) ^ P(b)) can a=b here?
If yes then how can I make it so that ∃a doesn't include b from ∃b.
glass condor
why not $\exists a (\exists b \neq a (P(a) \xor P(b)))$ ?
vital dewBOT
ConfusedReptile
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
hardy yoke
beautiful, thanks
tame glen
Can anyone help with this question
Prove... DeMorgan's theorem
( A ∆ B )¹ = A ∆ B¹ = A¹ ∆ B
naive gulch
A little confused on this, I said a) is countably infinite since for every positive integer there is there'll be an integer that can't be divided by 3
winged sun
I would also think that a is countably infinite 😮
naive gulch
:o I'm just not sure what else I need to prove to say its definitely countably infinite
naive gulch
Ah, bijection is one to one and onto?
naive gulch
Ah thanks that makes sense!
Yeah I knew it was a subset just wasn't sure if I could just directly say that :o
ty
slow jewel
Help
weary tiger
Can I get help on this?
livid drum
@weary tiger I find it strange that step 2 has an extra close parentheses.
Anyways, why do you believe it’s between 3-4?
weary tiger
Do you know if the answer is b for this problem @livid drum
livid drum
@weary tiger Yes, besides the parentheses issues, there are no mistakes.
uncut matrix
hey everyone, I'm doing some quantifer questions and I ran into some problems:
for x ranges over all living things,
R(x): x comes from the 23rd century, S(x): x likes to eat pizza
they wanted me to write the statement: nobody from the 23rd century dislikes eating pizza
my answer was ¬∃x(R(x)->¬S(x)) but the solution given was ∀x(R(x)->S(x))
I don't understand the solution, because while my solution can be simplified to ∀x(R(x) and S(x)) (for all x that come from 23rd century, everybody likes eating pizza), their solution simplifies to ∀x(R(x) and NOT S(x)) which implies for all x that come from the 23rd century, none of them likes eating pizza, unless im computing this wrong
livid drum
@uncut matrix you are computing this wrong. Your original answer is incorrect.
¬∃x. R(x)⇒¬S(x) means "their does not exist a (living thing) x such that if R(x) then S(x)".
What you instead mean to communicate is ¬∃x. R(x)∧¬S(x). This means "there does not exist a x such that R(x) and S(x)".
Carefully examine the differences between these two.
Carefully examine ∀x. R(x)⇒S(x).
And also ∀x. R(x)∧¬S(x).
They are not equivalent.
@uncut matrix
On why you might be making these mistakes?
I want you to consider the following sentences:
-
"Their exists an x which satisfies P(x), such that Q(x)".
Symbolically, this is ∃x. P(x)∧Q(x); this is NOT ∃x. P(x)⇒Q(x). -
"For all x which satisfy P(x), Q(x)."
Symbolically, this is ∀x. P(x)⇒Q(x); this is NOT ∀x. P(x)∧Q(x).
uncut matrix
thanks for taking your time to carefully construct an explanation for me
I'm going to look into it
tender hearth
weary tiger
Can someone help me with proving this propositional, I have most of the steps but I’m just stuck on the last one that says t and t
uncut matrix
@weary tiger not 100% sure but
p and (¬p or q) = (p and ¬p) or (p and q)
which is F or (p and q)
simple nova
Can anyone help me with this one
Can't use DeMorgans or anything
Has to be rule of infernce
weary tiger
@uncut matrix I'm a little confused what you mean by that?
What step are you referring to?
uncut matrix
during implication (3) you're trying to apply NOT on 3 terms but you haven't properly simplified it
there's some property which is like a and (b or c) = (a and b) or (a and c)
so before replacing -> on step 3 you should simplify it properly first idk
lemme give it a try
ok so lemme send a photo
This is what I got but idk if its 100% correct
simple nova
After can you assist me Beyond?
uncut matrix
sorry I haven't gotten to inference yet buddy
I'm still learning quantifiers myself rn
moving onto sets and induction after
not really sure how I'd do it without using demorgan's
weary tiger
@uncut matrix thank you, I realized that I forgot to use the distribution law
uncut matrix
np
weary tiger
@uncut matrix I have one more question, on the very last step, would -P v T be the domination law?
Heres what the domination law is
livid drum
@simple nova
I will show you a proof heavily translated from my proof system of choice.
You will then have to convert it to yours.
simple nova
@livid drum wow bro, thanks I'll look at it now
and ask questions if they happen
What does DEM LEM and all that mean
livid drum
No problem but i think its a bad answer. Because to know how to convert it to yours, is to know the answer to begin with.
DEM is short in my system for demonstration.
It opens a a new branch in the proof environment.
LEM is short for lemma. It corresponds to something which is to be demonstrated.
livid drum
yes. To demonstrate the Negation of X, you may assume X and continue.
in this case, to demonstrate the negation of the negation of p, we may assume the negation of p, and continue.
(This system is equivalent to a natural deductions envirnoment)
livid drum
Right.
So in order to prove neg neg p, we assume neg p.
Then we claim to be able to prove p imp neg q.
For from the assumption of p, you can combine it with the neg p to arrive at a contradiction. You can now conclude whatever you wish.
So we conclude neg q. This is the neg eliimination step.
Now that we've proven p imp neg q,
this itself contradicts our assumption that neg (p imp neg q).
And again we can conclude anything.
So this is another neg elimination step.
This allows us to prove neg neg p.
So the whole point of that was to then be able to conclude p.
After all our goal is to get p and q.
So we;ll need p today and q tommorow.
neg neg p implies p is an axiom of classical logic.
and we can modus ponens (ie, imp elim rule) to get p.
We repeat this procedure to acquire q, and thus we can get p and q (using the and intro rule).
This shows one direction of the equivalence.
weary tiger
@livid drum do you know if the last step is a domination law? #discrete-math message
The -pvt step
livid drum
Sure, the -pvt to t is a domination law, sure.
weary tiger
So then is the negation just ignored? Because the domination law is pvt=t
livid drum
its not just for one paritcular proposition,
it's impoartantly for any proposition.
Maybe use captial letters to indicate any propisitonal formula.
Then it's A v t = t
neg p is one particular prop that A can be.
weary tiger
Oh okay I understand that now, thank you
simple nova
@livid drumafter the first assumption, then you get not not p
is that just called double negation?
Little confused about 1.4 and 1.5
livid drum
i only got not not p by virtue of steps 1.5 and 1.2.
One says apple, the other says negation apple.
So i can conclude anything i want from them.
So i conclude not not p.
Okay so let's do those in english starting from 1.3.
1.3 says "I claim i have a proof of not not p."
1.4 says "we are allowed to assume not p."
1.5 says "I claim to have a proof of p imp not q"
1.6 says "we are allowed to assume p"
1.7 says " from 1.6 and 1.4, we have arrived at a contradiction. therefore we may conclude whatever we wish. In particular we conclude, not q"
(so now we have finished the proof of p imp not q)
1.8 says "from 1.5 and 1.2, we have arrived at a contradiction. therefore we may conclude whatever we wish. In particular we conclude, not not p"
(so now we have finished the proof of not not p)
simple nova
so not q is the reaction to absurdity right?
i think that is what they call it
1.7 is a reaction to absurdity
right?
livid drum
I guess sure. yea it makes sense to call it a reaction to absurditiy.
although i havent heard that before. Sounds cool though
simple nova
@livid drumConfused how you got .22 from .20 and .21
What rule can you do to get to that jump
weary tiger
Hello everyone, I hope you all are well. I have a brief question with the topic of rules of inference. When you make assumptions, is it allowed for one to make the assumption for False to be true? like [FALSE]
echo bolt
Whether it's "allowed" or not depends on the system that you're working in, but I don't think I've ever seen it forbidden and I've written formal proofs in a couple of different systems (though I'm not an expert by any means). The important things to note if you do that are 1) you can prove anything if you assume false [in "normal" logic, at least] and 2) this will just get you theorems of the form "FALSE => something", which are probably not that useful, since you'll (hopefully) never be able to prove false!
See https://en.wikipedia.org/wiki/Ex_falso_quodlibet and related pages for further reading.
slow jewel
How would you do this question?
stray reef
inclusion-exclusion, probably.
livid drum
@simple nova Maybe it isn't fair to say that 1.22 arises because of 1.20 and 1.21--it doesn't.
1.22 is the assertion that we can prove the claim "neg (p imp neg q)". We may happen to use previous steps inside that proof, but that's yet to be seen.
As we read the proof one line at a time, we have no reason to doubt 1.22.
It is yet to be demonstrated.
If a stranger comes to you and says: "I have proof of god from PA", you should then hear him out.
You might have problems with particular steps in his proof, but you cannot object to his initial claim, at that time, without even hearing him out.
1.23-1.25 are the steps of this proof.
1.23 says "we may assume p imp neg q"
1.24 says "because we have p, and p imp neg q, we may conclude neg q"
1.25 says "because we have q and neg q, we may conclude anything-- in particular we may conclude what we want to show: neg (p imp neg q)".
And this completes that lemma.
surreal pawn
Is there a way to solve this problem without having to write out all the sample space?
weary tiger
why is this not considered reflexive but considered symmetric and antisymmetric?
coral raven
2*2 =/= 1
weary tiger
doesnt R={(1,1)} only tho?
weary tiger
it says a set of positive integers in the question
i dont think the R refers to real numbers
rigid oriole
$2\cdot2\neq1\implies(2, 2)\not\in R$ is a counterexample
vital dewBOT
Shuri2060
rigid oriole
oh nvm shouldve read up fully
A relation is reflexive iff $(x, x) \in R$ for all $x$ in the set it's defined on.
weary tiger
im confused of what R would consists of because im seeing contradictory answers online
weary tiger
the answer key says it is not reflexive tho
rigid oriole
Of course it isn't.
Because (2, 2) isn't in R
for it to be reflexive, (1, 1), (2, 2), (3, 3), .............. all have to be in R
vital dewBOT
Shuri2060
rigid oriole
^
weary tiger
oh so for reflexive we check in the domain of all possible pairs and compare it to R
vital dewBOT
Shuri2060
rigid oriole
What you do for checking for reflexivity is different from the other 2 in a sense if that's what you mean
weary tiger
yea so for symmetric and transitive do we only consider the stuff in R
rigid oriole
well that's only because you are proving those implications
wispy matrix
im about to begin Mathematical Proofs: A Transition to Advanced Mathematics (4th Edition) (Chartrand, Polimeni, Zhang)
anyone have any advice to suggestions or want to join
weary tiger
I've been stuck on this for far too long, can anyone push me in the right direction. I'm now sure how I can use the definition of proper subset here.
weary tiger
Any help would be appreciated
mystic moss
im about to begin Mathematical Proofs: A Transition to Advanced Mathematics (4th...
join how?
vital dewBOT
slimvesus
terse ginkgo
how do u multiply combanations
naive gulch
How would I go about solving iii)?
Would it relate to hypothetical syllogism?
Is there a specific law I can use? Since I know b is a multiple of a and c is a muliple of b, it makes sense since b | c and a | b then c is a multiple of a
haughty garden
Start with the definition of divides
flat inlet
How would I go about solving iii)?
if a|b and b|c, then a = nb, b = mc. Then a = nmc and so a|c
don't overthink these problems, just write out definitions and poke them until what you want pops out
spark forge
A college has a rectangular center of grass called “the quad” that is 100 yards by 400 yards. They want to put in a sidewalk across the diagonal that will costs $1,000 per yard. How much will the sidewalk cost to the nearest dollar?
robust mango
@spark forge draw a rectangle, use the pythagorean theorem to find the length of the diagonal
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hearty crane
no hang on, he might have a point
young patio
robust mango
For part (d), aren't R and T also not antisymmetric?
Another question: Let R be a relation on the set A( the set A consists of at least 3 elements). If R is symmetric, then R∩R^(-1) is not empty. ( True or false ? ) If we take the empty set which is a subset of A, we know it is symmetric. And hence R^(-1) is also symmetric and empty, and R∩R^(-1) is empty.
The book says that the statement "If R is symmetric, then R∩R^(-1) is not empty" is true, but isn't my counter example correct? and that would mean the statement is false
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spark forge
Find the radius of a cylindrical fire hose that holds a volume of 314 ft³ and has a height of 400 ft.
20.2 ft
0.5 ft
324.65 ft
24.17 ft
ivory falcon
Can anyone help me simplify a boolean expression?
naive gulch
How does cubing work when mod is included
if a = 3 (mod 19)
and we have a ^3
can we do 3 ^3 mod 19?
coral raven
yes
weary tiger
how does finding reflexivity, symmetry, and transitivity change when the relation is on a set X x X as opposed to just X?
coral raven
they're exactly the same, they just apply to the elements of X x X instead of the elements of X
solid garden
can someone explain how to show that
R = {(x, y) : 5|(x - y) ⊆ Z × Z
is an equivalence relation
minor lake
nvm im got it holy shit im damb
tranquil jungle
For B, I know that you can't use a specific example because it would be loss of generality, but what about A and C? I'm not sure and I think A might be using the element and subset wrong and C the last statement seems odd
This is the theorem and proof
tranquil jungle
mhm
faint narwhal
you should think about A more and figure out exactly how they're using element and subset wrong though
Does C actually prove the statement that needs to be proved?
tranquil jungle
I have no idea
for A, it's the first statement, x belonging to the power set doesn't mean that it has to be in the normal set
the statement after that reverses the proof for x belong in B, which I think should be the correct one
fresh rapids
Could someone give me some hints on how i would even start the proofs for these?
This one I dont even know where to start.
This one just seems so obvious that i dont know how you would prove it "properly."
[2, 3] ⊆ (1, 4)
Do you know what ⊆ means?
tranquil jungle
one element, so that would mean P(B) has only 1?
fresh rapids
Subset, yeah?
faint narwhal
one element, so that would mean P(B) has only 1?
no
faint narwhal
Subset, yeah?
What does subset mean?
faint narwhal
So you don't really know what subset means
tranquil jungle
How could 3 not be in P(A) if it was an element in A?
fresh rapids
I guess it would be if every element in the set is in the other
faint narwhal
How could 3 not be in P(A) if it was an element in A?
If A is the set {1,2,3}, what is P(A)?
faint narwhal
I guess it would be if every element in the set is in the other
be more rigorous. What does it mean for A⊆ B?
faint narwhal
So it looks like { {}, 1, 2, 3, {1,2}, {1,3}, {2,3}, {1,2,3}} ?
tranquil jungle
yes
naive gulch
Did I mis interpret this question? Its asking for c in 0 to 18 closed but I didn't really get something relevant to that?
faint narwhal
@tranquil jungle Isn't that a bit weird? Why is everything like {1,2} a set except for 1,2,3?
More specifically, is it true that 1 ⊆ {1,2,3}?
tranquil jungle
Oh, 1 and {1} aren't the same
faint narwhal
Did I mis interpret this question? Its asking for c in 0 to 18 closed but I didn...
You can reduce mod 19. For example if your answer was 33, you can say that 33 = 14 (mod 19)
tranquil jungle
Oh, A is not using correct notations
I can see that in C
it should be {x} belonging in P(A)
not x belonging in P(A)
faint narwhal
Right
tranquil jungle
C looks like it has everything correct, but it doesnt show that that P(A) is a subset of P(B)
faint narwhal
C looks like it has everything correct, but it doesnt show that that P(A) is a s...
exactly
tranquil jungle
ty Zoph
naive gulch
You can reduce mod 19. For example if your answer was 33, you can say that 33 = ...
Didn't I already reduce mod 19 when I cubed? Little confused by what you mean
faint narwhal
hm
how should I put this
for example, if you're trying to solve part b), you get that c = 24 (mod 19) right
so c = 24 is a valid solution
since 24 = 24 (mod 19)
However, c = 5 is also a valid solution. Since 5 = 24 (mod 19)
@naive gulch
also nice pfp
naive gulch
ah I think I get it?
And thanks :) SBY lovers
So I got 1439 (mod 19)
would c = 2 then?
faint narwhal
Uh check your calculations, that's not quite right
naive gulch
1439 or the c part?
faint narwhal
the c part
naive gulch
I thought it was 1+4+3+9 = 17 and the difference is 2
faint narwhal
Uh
I'm not sure why you're adding the digits together? Maybe you're thinking of the rule for divisibility by 9, but that doesn't work here
naive gulch
oh wait lmao yeah I flipped it up thinking about the divisibility by 9 oops
wouldn't c be 33 then
which is over 18 or did I do my math wrong again
naive gulch
ah yeah we can reduce more