Can you see whether it's onto and why?
#discrete-math
drifting sail
wintry schooner
oh okay but how would I actually show my working out
to explain that it is not 1-1
drifting sail
The definition of "1-1" is that if f(x)=f(y) for some x,y in the domain, then x=y.
wintry schooner
hmmm so what about the x^2
does that not affect whether it is 1-1
f(x, y) = x^2 y
drifting sail
in some sense the "x^2" is exactly why the function is not 1-1.
if you had e.g. g(z)=z^2 defined on R, then that is not 1-1 since it's an even function, i.e. g(-z)=(-z)^2=z^2=g(z).
oh okay but how would I actually show my working out
as I said, it's enough to show that there are two preimages with the same image.
wintry schooner
oh alright I guess I understand it a bit now
and if my understanding is not wrong I think that this function is not onto because of the R^2 ?
weary tiger
@mystic moss hey I have a question. How come the angle between b1 and b2 is b1 - b2?
I never really understood that part
versed summit
pallid lintel
is the empty set the complement of a set
take natural numbers, then it is empty set + natural numbers
steady kernel
The Empty set is the complement of the universe
mystic moss
@weary tiger the angle between p_1 and p_2. (image credits go to the internet here) if RS was the x-axis, and P was p_1 and Q was p_2, we have that b=b_1-b_2.
weary tiger
@mystic moss like i feel like b = b1 - b2 would be b2
mystic moss
b_1 is <PSR, and b_2 is <QSR
weary tiger
so then b is PSQ?
mystic moss
yes, exactly
mystic moss
just an example to show that i was able to put the first 6 points in separate groups, but the seventh point had to room with someone else
weary tiger
ok god this question is so hard
mint stag
Is there any better way to solve this recurrence equation?
mystic moss
@weary tiger don't worry, just ponder about it for a while and you'll get it. i suggest reviewing trig on the coordinate plane, perhaps
weary tiger
yeah like i don't get what you mean by
and so it's basically saying that at least one angle b must be less than pi/6, or 30 degrees
@mystic moss did you mean 0<= b <= pi/6?
mystic moss
yes, exactly
weary tiger
in general does that mean that ai and aj are both in the same region? @mystic moss
mystic moss
@weary tiger rather, it means that if a_i and a_j are in the same region, then we know 0<=b<=pi/6
weary tiger
and how do we know if a_i and a_j are in the same region?
mystic moss
pigeonhole
weary tiger
right true
silver ravine
Hello?
I have a combinatorics and probability problem. The major part is combinatorics that's why asking here!
Question: Six people, including A,B, and C, form a queue in a random order (all 6! orderings are equiprobable). Consider the event "B is between A and C in the queue". What is its probability? (The order of A and C can be arbitrary, but B should be between them)
I am not able to understand how the counting is happening here! Here was one of the reply which I don't understand!
For A_B_C:
When A is in the first position, C has 4 possibilities.
p1= C at 6th position: B has 4 possibilities hence no. of outcomes=24
p2= C at 5th position: B has 3 possibilities hence no. of outcomes=3*3!=18
p3=C at 4th position: B has 2 possibilities hence no. of outcomes= 2*3!=12
p4=C at 3th position: B has 1 possibilities hence no. of outcomes= 1*3!=6
total =60
Similarly, computing values for other positions of A we get:
-
When A is in the second position, C has 3 possibilities. total=36
-
When A is in the third position, C has 2 possibilities. total=18
-
When A is in the fourth position, C has 1 possibilities. total=3! = 6
A total= 60+36+18+6=120
For C:
Similarly we can compute for the format C_B_A
C Total= 120
Final total =240/720
Can anyone explain!?
drifting sail
and if my understanding is not wrong I think that this function is not onto because of the R^2 ?
@wintry schooner nono, the domain has nothing to do with it. In fact the function is onto: for each z in R you can find (x,y) in R^2 such that f(x,y)=x^2y=z. For instance (x,y)=(1,z) works.
Is there any better way to solve this recurrence equation?
@mint stag what's shown there is the standard algorithm to solve these things, using the characteristic polynomial just like in second-order ODEs. But: you know how in ODEs you can convert second-order into a first-order system? Here you can do the same.
vital dewBOT
bastian.uwu:
Compile Error! Click the 
reaction for details. (You may edit your message)
drifting sail
for recurrence relations this algorithm is often far clearer than the one involving the characteristic polynomial (which is clearer in the case of ODEs) to me; perhaps it is for you as well.
of course they're equivalent and in fact you prove that the characteristic poly approach works by treating it as a linear system first, again just like in ODEs
pallid lintel
what programming language should i learn if i want to do something with polynomials and graphs, and make lots of numbers
mint stag
Oh, that's nice. Thanks! 😄
serene basalt
maybe sagemath with python
silver ravine
Can anyone help me with my problem above?
quaint star
@silver ravine Can I give a much simpler explanation?
Either A is between B and C, or B is between A and C, or C is between A and B. Since each of these is equally likely, the probability that B is between A and C is 1/3
silver ravine
Hey the explanation is what I thought of! But I want to understand the combinatorial approach! Actually my combinatorics a bit weak so wanna practice on it! Btw thanks a hell lot for your explaination!
quaint star
Np, care to say exactly which combination you are having a hard time understanding? Because there's a lot of cases. Say the first place where you are lost.
silver ravine
So in p1 I think its 4! So 24 other than I don't understand why we are having all them being multiplied by a number and 3!
mint bane
h e l p
this what i have so far
ive considered trying strong induction but idk what i'd gain from it
vital dewBOT
mint bane
my b haha
unreal stump
Assume Gn is of that form for all n<=k
And show G(k+1) is also of that form
The G(n-1) term will simplify and you get your result
mint bane
so try strong induction...? or am i interpreting that wrong
unreal stump
Yes
mint bane
aight will do thanks for the tip
rotund tartan
can somone teach me ratios in dm
mint bane
like... fractions?
errant bear
no, r a t i o s
mint bane
also drake (or anyone) idk what im doing wrong but going with strong induction just takes me down kind of the same road idk
this is what i set up but it’s pretty much almost the same as the last thing - trying to expand the G(k-1) will put a G(k-3) so im obv not gonna do that
mint bane
my last approach was to try and make the 3^k+1 - 2^k+1 into 3^k -2^k and find equivalence that way - with strong induction i have more flexibility on the other side of the equation but idk how i'd use it i guess
unreal stump
Say till k , $G_k=3^k-2^k$
Now consider $G_{k+1}$
$G_{k+1}=5G_k-6G_{k-1} \implies
G_{k+1}=5(3^k-2^k)-6(3^{k-1}-2^{k-1})
\implies
G_{k+1}=(5-2)3^k-(5-3)2^k
\implies
G_{k+1}=3^{k+1}-2^{k+1}$
valid wave
😦
vital dewBOT
DrunkenDrake:
valid wave
anyone have an idea?
unreal stump
The equivalence class is set of all lists with 3 elements which have 1 even number
mint bane
thank you sm @unreal stump i really need to learn to play around with these things more
valid wave
wdym @unreal stump
quaint star
Are you leaving out divisions here? Please fix it @near root
near root
I already figured out why it wasn't true due to the fact that I was ignoring lots of scenarios. Thanks anyway
heady sedge
Hey I'm having trouble converting english statements to premises so that I can put them in a truth table
It is often said that some rights are universal. A little reflection, however, shows that this cannot be the case. For rights always result from an agreement amongst people. But agreements by their very nature are contractual, and so rights must then also be contractual. What is contractual, however, cannot be universal, since contracts vary from one society to another and since they are also liable to change even within a society.
mystic moss
@heady sedge where exactly are you having trouble?
heady sedge
I cant seem to figure out the conclusion
mystic moss
the conclusion is that no rights are universal
it's trippy because the wording in the beginning is a bit convoluted, but do you see how?
heady sedge
yeah I can see that now, I'm guessing that this is invalid since If that is the conclusion, there is a premise that says What is contractual, however, cannot be universal
wait I got confused again nevermind 😅
chrome gull
"For rights always result from an agreement amongst people." One may disagree with this premise, but the rest of the argument seems to flow logically from it, at least to me. The reason why it flows because universal rights definitionally are not contractual rights, so if you take as a premise that the only rights are contractual then there are of course no universal rights. At least, this is how I'm reading it. Curious to hear other takes.
(Yeah I'm also puzzled why this is here. I wanted to make some comment about math initially and almost got sucked in culture war 😦 ; this logic question is very loaded, yikes. Seems like as good an opportunity as any to learn to think more carefully though.)
mystic moss
perhaps it would be more #proofs-and-logic
if we're going to consider this as philosophy for a second--what do rights mean if they do not result from an agreement between people?
chrome gull
Yeah, that's where I think it becomes very tricky indeed, especially if you rule out theology -- where do universal rights come from? How do we know what the universal rights are? And so on -- hard questions with very real political implications. This is why its very culture war adjacent to me, and am trying to carefully avoiding making statements near that.
So it just seems like a minefield of a logic exercise to throw out to students, lol. Either a very devious or a very oblivious prof.
Even discussing it on this level feels like a dirty thing to do in this server 😦 -- there are better spaces for it. (Like bringing something unclean into the temple of math.)
Let's talk about recurrence relations! I wanted to thank @drifting sail for that nice insight -- usually my approach to solving recurrences is to tear out my hear keeping track of the indexing (if Wolfram does not spit out the answer, lol), so I will try the suggested approach next time.
The analogy between treating DE and recurrences is very nice. I know that there are all sorts of analogies, but haven't made them so explicit in my mind. I guess this is the kind of thing I would learn if I worked through generatingfunctionology?
unique herald
Can someone explain this
how does 2). even make sense
I don't understand the proof
steady kernel
If a,b > 0, then, their product ab > 0. Use a = b
forest zenith
yeah they wrote that in a really dumb way. they could have used a better example for contrapositive
@unique herald do you understand proof by contrapositive in general? if so then i think you can ignore this example since it's just written weirdly. do you get the example used here? https://en.wikipedia.org/wiki/Proof_by_contrapositive
they should have written a line like what enigsis put
formal tulip
Q={(q,X)|q∈Q1,X⊆Q2,X^q =X}
I'm looking at concatenation with finite state machines, would someone be able to help me translate this?
shut fjord
Answer is Θ(n^(3/2) * log n)
but I am only getting n^2 * log(n) * sqrt(n)
the outer loop is Θ(n), the 1st inner loop is Θ(logn) the 2nd inner loop is probably wrong, I wrote Θ(n*sqrt(n))
can someone explain what the 2nd inner loop is? its based off m which is floor of sqrt(n)
solemn dome
my professor is finding the full CNF for this, but I dont understand when to use false or true to solve it
in the previous example he used V true, but this time he used V False
how do i know when to use false vs true to convert to a full CNF
quaint star
Show the previous example? Since you are using OR here, you must use false. If you used true, it would just always be true. Whereas, if you used AND, you would have to use true.
solemn dome
ok i have another question then because i dont really understand what you're saying
how do i know when to use or
if im just given the intial "not A and B"
how do i know what to do with it
quaint star
I don't know exactly when to use and/or. I'm just explaining why when you add an extra AND, you must use a true with it. And why if you add an extra OR, you must use a false with it.
Can you clarify what you didn't understand?
solemn dome
i guess i just dont understand how to start solving the problem
because i dont see how he went from the given problem to the next line
quaint star
But do you understand why or is accompanied by false?
solemn dome
no i dont
quaint star
A is not equivalent to A or True. Because A or True is always true.
True is always true, so A or True is always true
But A is equivalent to A or False. False is always false, so A or False is true precisely when A is true and it's false when A is false
solemn dome
ill be honest im trying so hard to understand the wording
but this is so complicated to me
ok i get it now
i see what you're trying to say
but how does that help me know if to use false or true for solving the problem
quaint star
You didn't show the entire solution, so I don't know. I've never worked with CNFs. I just wanted to explain that part.
elfin egret
Hello everyone, I'm stuck at a counting problem
Can anyone provide a hint or sketch for what method I should use to prove this?
drifting sail
The analogy between treating DE and recurrences is very nice. I know that there are all sorts of analogies, but haven't made them so explicit in my mind.
@chrome gull A good book that works through this analogy is Saber Elaydi's "Introduction to Difference Equations". The first 4 or so chapters read very much like your typical ODE course
pallid belfry
Does this look correct?
errant bear
you can stop at the second to last line, and say that the RHS is greater than 2^(n+1) + 1, which is what you want
also, this claim is false as n = 0 fails
frank zinc
wait that prompt
does anyone think that partitions between all possible intervals within some width could be a key to solving this problem
pallid belfry
@errant bear Gotcha, would you mind looking over one other proof of mine?
errant bear
errant bear
just send it weeb
errant bear
y do you have the last 3 lines, is that to prove your factoring was correct?
vital dewBOT
pallid belfry
y do you have the last 3 lines, is that to prove your factoring was correct?
@errant bear Yes
errant bear
hm, well its a tiny bit confusing since it just looks like you just went in a circle, i mean its fine just write like "we know this because: " after the 4th to last line or smth
sick swallow
there is a simpler way to prove this
pallid belfry
What is that?
sick swallow
u can just do 8(sum of i, from 1 to n) -5n =4(n+1)n-5n=RHS
pallid belfry
Oh
interesting
Is the way incorrect in anyway or is that just a simpler way to prove it
pallid belfry
oh so it would be my induction hypothesis? @sick swallow
sick swallow
yea oops
chrome gull
@chrome gull A good book that works through this analogy is Saber Elaydi's "Introduction to Difference Equations". The first 4 or so chapters read very much like your typical ODE course
@drifting sail Thanks!
zealous wedge
How do I go about this problem? I don't really know where to start tbh
somber trout
zealous wedge
no I haven't, thats the part I don't know how to solve.
@somber trout sorry for late response
vital dewBOT
Orange905:
vital dewBOT
Orange905:
zealous wedge
oh I see, since they give you a_1 as 1/2, you plug that into the equation to solve for a_2
and so on
so a_2 would be 3(1/2) + 1 which is 5/2
analog vault
Can someone help me figure out how to prove if these are true or false?
delicate ridge
as with all beginner's formal logic, translate the symbols into english
if A(x) means person x owns a car
and B(x) means person x owns a bike
then the first one says "if everyone owns a car, then there exists a person such that they either own a car or a bike"
and the second one says "if there exists a person that either owns a bike or a car, then there exists a person that owns a car"
steady kernel
Ask your question
vital dewBOT
TedNowKaczynski:
pale epoch
🤔
gritty crescent
I was trying to prove that the axiom of regularity, together with the singleton set axiom rules out the possibility of set containing themselves.
I don't understand the proof I saw online correctly.
unreal stump
Can you share that proof?
unreal stump
Did you understand why {a} and a are disjoint?(If that is confusing,take {a,b} and a)
gritty crescent
Well, {a} and a are supposed to be disjoint by the axiom of regularity.
unreal stump
Now,with axiom of pairing,you make a new set B,such that it's only element is A
So,You end up with a singleton set
unreal stump
Let's say you have a number 2
Now,make a set A such that its elements are 2 and 3(basically {2,3})
2 is in A
Any doubts?
(Now replace 2 with your set)
a will be the only element in {a}
gritty crescent
Yes, so far so good.
unreal stump
So,is {a} a singleton?
gritty crescent
Yes
gritty crescent
Yes, correct.
I understand up till here.
How does the "definition of disjoint sets" lead us to conclude that a is not in a?
unreal stump
So a and {a} have no common elements
Which means a (from {a})cannot be an element of a
gritty crescent
Hmmm, I see.
unreal stump
Empty set axiom: empty set exists
Axiom of pairing: given two sets x,y a set containing x and y exists
Axiom of Union: given a set x, the set which is the union of all elements in x exists
Axiom (schema I think) of something: given a first order sentence you can make subsets of a given set
Axiom of regularity I think: it says like any set contains an element disjoint from it I believe. Basically exists so you can’t have a set in itself.
Axiom of Extensionality: two sets are equal iff they contain the same elements.
Axiom of infinity: an infinite set exists (you make N from this)
Axiom of powerset: the powerset of any set exists.
Axiom (schema I think) of replacement: basically says images of functions are sets.
Axiom of choice: the axiom of choice lol.
This might be useful
karmic nymph
to my knowledge, there is only a bijection if a function is one to one and onto
onto meaning that no elements in the codomain are "left out"
here is says that there is always a bijection if finite sets have the same number of elements
but
what if a is mapped to 1 and b is also mapped to 2
stray reef
what if a is mapped to 1 and b is also mapped to 2
?
karmic nymph
what did u not understand
glass condor
there is only a bijection if a function is one to one and onto
that's the definition of a bijection, yes.
karmic nymph
yes
glass condor
what if a is mapped to 1 and b is also mapped to 2
...you mean, exactly like it is on the picture? 🤔
stray reef
a bijection can map a to 1 and b to 2 no problem
stray reef
that's exactly what i pointed out
karmic nymph
If you look at the example above
Its one to one, and onto. But what if a is mapped to 1 and b is also mapped to 1
stray reef
just because a bijection exists doesn't mean every function between your sets will be one.
it is very easy to create a non-bijection even between two sets of the same size.
glass condor
Then it's not a bijection, because it's not one-to-one. Not every function between two sets of the same size is a bijection, naturally 🤔
karmic nymph
so the slide is incorrect?
stray reef
the slide is correct
glass condor
...why do you think so?
stray reef
again
just because there is a bijection between your sets
doesn't mean every function between your sets will be bijective
karmic nymph
the slide says
If finite sets have the same number of elements, then there is always a bijection between
them
stray reef
yes
karmic nymph
but i can easily see a function that there wont be a bijection?
stray reef
you're looking at a non-bijection and asking "why doesn't a bijection exist?"
sick swallow
That is not what it’s saying
stray reef
this is like looking at a map that shows you several routes from A to B, going off into a dead end, and asking "why can't i get to B????"
sick swallow
There exists a function is not equivalent to for all functions
karmic nymph
does the slide mean, if finite sets have the same number of elements, then a function is possible where it would be a bijection?
stray reef
wording
karmic nymph
ok
glass condor
a bijection exists $\iff$ the exists a function between the sets such that this function is a bijection.
vital dewBOT
ConfusedReptile:
glass condor
it does not mean "every function between the sets is a bijection".
karmic nymph
right i understand
ty
waiitttt
so an N number of infinite sets have the same size if there is bijection between them???
stray reef
??
karmic nymph
if i have 2 infinite sets
and there is bijection between them
they are of the same size?
the two sets
stray reef
yes by defn
for infinite sets bijections are the only way for us to see their size, and in this case we say cardinality and not size, so as not to accidentally make an incorrect conclusion based on intuitions that don't apply
glass condor
for example, $\bN$ and $\bN\setminus {1}$ have the same cardinality. Proof: $f(x) = x+1$ is a bijection between them 😛
vital dewBOT
ConfusedReptile:
glass condor
(it's in fact one simple way to prove that a set isn't finite - constructing a bijection between the entire set and a strict subset of it)
karmic nymph
ah nice
glass condor
cooler example:
$f(x) = \frac{1}{1+e^{-x}}$ is a bijection between $(-\infty,\infty)$ and $(0,1)$
vital dewBOT
ConfusedReptile:
glass condor
therefore, $(0,1)$, a tiny part of $\bR$, is isomorphic to the whole.
vital dewBOT
ConfusedReptile:
karmic nymph
i see
short lantern
What does it mean when a eulerian graph G can only be eulerian if and only if it has atleast one nontrivial component?
sour arrow
What is "it"? The first part of your question is very vague
fiery sage
is this the right room to ask questions about finding coefficients in multinomial equations?
short lantern
@sour arrow It refers to the graph G
limber trench
What conditions needs to be fulfilled for a Eulerian path to go into a complete bipartite graph?
limber trench
Yes
stray reef
when either part has only two vertices or both parts have an even number
limber trench
What about when exactly two vertices has an odd amount of edges, otherwise all vertices has even edges
Ive got K2,5
Which implies that ive got 2 vertices that has an odd amount of edges
But if I had for example K3,3, it would mean that every vertice has an odd amount of edges, which doesnt work for it to be a Eulerian path
stray reef
What about when exactly two vertices has an odd amount of edges, otherwise all vertices has even edges
that's $K_{2, 2n+1}$ and i covered it
vital dewBOT
Ann:
limber trench
Thanks!
winged bramble
whats the definition of a proper subset?
quaint star
A is a proper subset of B if it is a subset of B and not equal to B
mystic moss
@short lantern that's... not true?
hardy viper
@spring cave there's a way to checkerboard the map that will help
spring cave
Anything will be very appreciated
hardy viper
also for clarification krimson has to move each hour right?
spring cave
If the helicopter moves one location, so can Krimson
hardy viper
idk what that means
spring cave
Let's say that helicopter is in A
If it moves to B
then Krimson MUST move to one of the adjacent areas
@hardy viper Could you perhaps tell me how to "checkerboard" the map?
I am not very sure whether I understand you
hardy viper
oh let's call it a 2-coloring instead
color the squares so that the same color doesn't touch
once you have that, krimson will always be on one color every odd hour, and the other every even hour
which makes it easier to catch
mystic moss
this is a neat problem :O i hadn't seen this type of scenario before, but the concept is pretty cool
is this for a class?
spring cave
No it isn't, my teacher gave it to me as a challenge and I have to solve it to earn some "honor", you could say.
Any help would be very very appreciated!
hardy viper
ok I solved it 👀
spring cave
Oh damn a genius is here
mystic moss
haha yeah, plumorant's hint is really helpful
hardy viper
the checkerboard is all you need really
spring cave
Honestly I gave it a look and I didn't come up with ideas
hardy viper
shorter problem: find krimson in 5 hours, assuming he starts on B, D, or F
spring cave
Uh i am not sure where you found that, but that was one of the other problems
The answer was 5
hardy viper
lmao
spring cave
We just had to prove from the other direction
spring cave
Afterwards?
hardy viper
you hid the other problems 
hardy viper
after 5 hours you rule out "half" the starting points
like just take your two answers
and put them together
that pic was right
spring cave
Yes
hardy viper
so the answer for 10 hours is BCFCDBCFCD 😮
hardy viper
lol fun problem
winged bramble
does anyone know what a negation of the intersection of 2 sets would look like or is that something that doesnt exist
glass condor
WDYM by a negation of a set? Complement? Then $(A \cap B)^c = U \setminus (A \cap B) = (U \setminus A) \cup (U \setminus B)$
vital dewBOT
ConfusedReptile:
glass condor
where U is the universal set.
winged bramble
oh i meant like if there was an opposite to union or something like that
im kinda bad at wording things
glass condor
you said "negation of the intersection", now you want "opposite to union"? 🙂
glass condor
do you maybe mean symmetric difference or something? It's defined as $A \triangle B \equiv (A \cup B) \setminus (A \cap B)$.
vital dewBOT
ConfusedReptile:
glass condor
In other words, it's the elements which are in exactly one of the sets A, B.
winged bramble
gotcha
weary tiger
I’m stuck on q1 please can someone help me? I’ve got that it’s false for every one but I have a strong feeling I am wrong
vital dewBOT
ConfusedReptile:
upbeat lodge
i need help w this hw question, ive done definition of set difference but i dont think im doing this right
so by that i did x is an element of a u c and x is not in b
weary tiger
for this question, would the answer be "D"?i chose "b" but my teacher said "b" was partially correct
unique herald
Can someone guide me through this?
p = George has eight legs
q = George is a spider
Argument: ¬p --> ¬q
And we can not conclude that the conclusion is true if the premises are true
Is this correct?
pale epoch
the contraposition of your argument is q -> p
pale epoch
isn't q the given premise and p the given conclusion?
so we should be able to conclude that it is true?
unique herald
oh true
Wait so is it asking me to put the first sentence into argument form?
and then asking me to say whether we are able to conclude if the second statement is true?
pale epoch
i'm not sure, it's kinda weirdly written
but i would guess that what you did is correct
other than the fact that the conclusion is true if the premises are true
like, i'm not sure what "argument form" is supposed to be
pale epoch
but it seems that p, q, ¬p --> ¬q and q are given
and p is the conclusion
sorry, that is weirdly phrased
what i mean is
let p and q be as you stated
then ¬p --> ¬q and q are given
and the conclusion is p
and then it asks whether that is valid reasoning
and it is
unique herald
Right
I put my final answer as:
Argument: ¬p --> ¬q
Therefore, we can conclude that the conclusion is true if the given premises are true.
golden totem
u have to prove that all degree 3 polynomials have at least one real zero
not just one
analog sonnet
correction: not all the degree 3 polynomials
karmic flint
If A is a set, and P(A) is the power set of A, A=A-P(A) right... this is tripping me up
Difference
Aka set difference
Hmm but A exists in P(A), but elements of A exist in P(A) not as elements, but as elements of elements because each element in P(A) is a set?..
Right. So A = A - P(A) seems to be correct because A isn’t a subset of P(A), but rather A exists in P(A)?
tawdry pasture
Prove by contradiction: For all integers k and l, if 7 divides kl, then 7 divides k or 7 divides l.
steady kernel
@tawdry pasture what do you have?
jade pewter
sorry if its too much. i'm basically stuck on everything and need help.
tawdry pasture
@tawdry pasture what do you have?
@steady kernel so my teacher told me to use these to get the contradiction
k = 7r where r is not an integer
l = 7s where s is not an integer
kl = 7n where n is an integer
kl = 7rl
steady kernel
Instead of rl isn't rs?
tawdry pasture
wdym?
idk how she wants me to write the contradiction
she did say that 7 is prime is also a hint
but like no shit
@steady kernel
steady kernel
Do you division's algorithm?
steady kernel
I mean, given to numbers, a, b, you can write a = bq + r
r is the residue
Of the division
tawdry pasture
can you give them values and restate the a = bq + r?
so like 15 = 7 x 2 + 1
meaning 15 / 2
tawdry pasture
so how does that apply to this proof
tawdry pasture
so you get 7sr = n
but that doesn't provide a contradiction all that says is 7 x a non integer x a non integer = an integer
and that is possible
tawdry pasture
@steady kernel my professor just told me that i am correct when i got to r x l = n
steady kernel
You have that 7r and 7s are integers, but not r,s, that means that their denominators are ± 7, then the denominator of sr is ±14
7sr can be an integer knowing that?
@tawdry pasture
tawdry pasture
no k = 7 x r where r is a non integer and l = 7 x s where s is a non integer
the reason that is stated is to have the contradictory assumption that k / 7 and l / 7 does not produce an integer therefore 7 does not divide k or l
@steady kernel
i got it
n = r x l
nope i didn't my mistake
tawdry pasture
yea but how can you say that
steady kernel
A integer
steady kernel
You have that 7r and 7s are integers, but not r,s, that means that their denominators are ± 7, then the denominator of sr is ±14
@steady kernel
Think of this
tawdry pasture
wait explain
i see what you're saying now
7r is an integer because k = 7r and k is an integer
so rs = k/7 x l/7
steady kernel
Yes
steady kernel
The numerators aren't multiples of 7
That's where you use is prime
Because it can't because otherwise it would be integer
steady kernel
Yes
tawdry pasture
this is what i got bro
@steady kernel
do you think i need to explain more in depth of why it won't produce an integer
tame hound
hi
can anyone help me with this question?
Let a and b be integers. Prove that if 3 ∤ ab, then 3 ∤ a and 3 ∤ b.```That's the reciprocal of BoopBoop question
Prove the equivalent statement if 3 divides a or 3 divides b then 3 divides ab
@tame hound
tame hound
i see
so if i prove that 3 divides ab then i can prove it also means that it proves the inverse
last timber
Let a and b be integers. Prove that if 3 ∤ ab, then 3 ∤ a and 3 ∤ b.```
@tame hound
wait
ah wait
it not divide
then it is true
steady kernel
so if i prove that 3 divides ab then i can prove it also means that it proves the inverse
@tame hound
Yes
$(p \implies q) \iff (\lnot q \implies \lnot p)$
vital dewBOT
Enigsis:
tame hound
Thank you
also how could i do this one?Let x and y be integers such that x + y ≡ 0 (mod 3). Prove that if a, b ∈ Z such that a ≡ b (mod 3), then ax + by ≡ 0 (mod 3).
if you or anyone could help me with this of course
?Now im just comfused
the whole thing to be honest. But ill just try to understand this for now. How did you get to that conclusion for this formula?
yes
uhm i see
but how do i actually prove that based on my question. That's what i truly don't understand. It's laid in front of me and i sorta get the idea now. But how do i do this now?
wait are you asking because you want me to answer or you don't know
im sorry former?
last timber
we want you to answer
tame hound
ah ok, it means m divides (a - b)
well that's the definition of it
not really explaining anything
tame hound
wait...
wouldn't it be something like this: 3|(ax +by) - (ax + ay))?
im not sure, do i have to simplify what is inside of the parenthesis first?
or do i need to use a formula
3|(by - ay)?
i can isolate the y?
ok so now i have 3|(y(b-a))
a = b (mod 3) means 3|(a - b)?
so 3|(b-a) = 3|(a-b)
well you wrote b-a when i wrote a - b so i assumed that that's what you meant
yes
mh....
| means divides
last timber
is it true that if a | b or a|c then a|bc?
sick swallow
yea lol
last timber
i am not asking you
cunning thistle
Euclid's lemma: if m|ab then one of m|a or m|b must hold. Think of it in the reverse order
sick swallow
this is gauss' lemma
tame hound
@worthy palm i still do not get how i can get rid of that y
sick swallow
also it is easy if u consider fundamental thoerem of arithemetic
last timber
@worthy palm i still do not get how i can get rid of that y
@tame hound what we have
Getting back to it, you want to show that 3 |(y(b-a))
@worthy palm we have this
tame hound
it means a divides b
last timber
now read couple of messages above
stray reef
@sick swallow @worthy palm one of y'all has to stop being a default pfp gremlin
sick swallow
it means there exist and integer x such that b/a =x
last timber
KANE YOU ARE NOT CONTRIBUTING
sick swallow
rip he wasnt gonnna say it anyway
also that is not what u said lol @cunning thistle
u said the general version of m and n, which is precisely gauss' lemma
last timber
ok lol kane is second poly as i see
cunning thistle
completely irrelevant
sick swallow
no need
tame hound
yes
sick swallow
but like this is literally definitions at this point, he should do it himself
sick swallow
hint: ive already said it
last timber
shut up kane
sick swallow
lol
bronze bronze
guys, i am in confusion with understanding the difference between statements and propositions , is this sentence right ?(this is from my thoughts): statements are ways to convey propositions
last timber
um..
do you have definitions?
as far as i know there is no much difference. they are just aliases for each other (if you define proposition to be a sentence which is always either true or false)
bronze bronze
i read this and i just wanted to check if my understanding was correct by that sentence i formed from my comprehension
bronze bronze
yeah, in essence it is saying, all propositions are statements but not all statements can be propositions, right?
propositions must be binary, but only statements that are declarative and have binary values lied within are propositions , the rest of statements aren't propositions; did i comprehend correctly?
if i get the distinction right, then i can move on...
last timber
E.g. "snow is white" is a statement that itself doesn't have a truth-value, but instead expresses the proposition that snow is white, which happens to be true. That's pretty much it.
hm i guess you are mainly correct
In philosophy of language (and metaphysics), statements are linguistic objects, like sentences of a natural language. Propositions are (traditionally understood as) the meanings of sentences (of a language) (in a context of utterance).
ah so
2+2=4 is just a string expressing the proposition "two plus two equals four" (or vice versa)
you know it is like
bronze bronze
E.g. "snow is white" is a statement that itself doesn't have a truth-value, but instead expresses the proposition that snow is white, which happens to be true.
@last timber so from this quote we can get that statements in essence dont contain any truth-value, the meaning of them contains ...... RIGHT?
bronze bronze
ok
last timber
physically speaking i just wrote a sequence of symbols
i.e string
3+5=8 does not have any meaning itself
if you are a newborn you just see that i wrote some funny symbols
bronze bronze
yea
last timber
but you are not a newborn so you know what 3+5=8 expresses
and that underlying meaning is a proposition
while 3+5=8 is a statement containing no information if we remove meaning
bronze bronze
aha !
so, the fact that something is proposition is independent from the person who wants to research it's true-false value... right?
last timber
if i understood you correctly then yes
we are assuming some meaning behind statement which is a proposition
but proposition can be written by different statements
i can say P is statement meaning 3+5=8 and another can say X is a statement meaning it. We wrote diff symbols but the proposition is the same
bronze bronze
but proposition can be written by different statements
@last timber proposition is the end result, but we have lots of ways(statements) to reach it, just like programming stuff, where you have many ways(algorithms here i.e statements) to reach the end result(output here i.e proposition), right?
last timber
ye also way to see it
last timber
yw
fiery sage
can anyone help with inequations with conditions?
stray reef
can you be more specific and/or post your problem
fiery sage
how many full number results this has
im not native sorry for bad english
I need also how not only what
Ive been stuck on this for the whole day
but i guess its fricking long and I have to post the whole thing till 18:00 cest so i guess ill just skip it
stray reef
hmm
limber trench
I need help with deciding how many part-graphs there is to a complete graph with 4 vertices
I know that for one edge, I need to use 4choose2 since two vertices makes one part-graph in the complete graph
Am I speculating in the right direction?
signal basin
If u: R -> R is injective, then it must be bijective, since the domain and the codomian is the same ? right?
signal basin
okay, I see. I rest my case.
fallow raft
its valid to just state that if we know x <= y-1 then x <= y right? Assuming we're working with values > 100
stray reef
this assumption is unnecessary
x ≤ y-1 < y no matter if x and y are greater than 100 or not
fallow raft
true, i was just thinking about vals greater than 100 since thats the constraint i saw on a problem earlier. thought this was given but wasnt 100% cause idk maybe sme niche rule mde by someone w/ a biggerbrain
near root
How do I prove that if a vertex has an uneven degree in the complementary graph she has an even degree?
Is there a sentence or proof anywhere I can use ?
ocean turret
But that is not a true statement?
near root
Lemme correct myself, if exactly 2 vertexes in a graph have an even degree, how do I prove that exactly 2 vertexes in the complementary have an uneven degree
The graph is on n>=2 vertexes
The graph is simple and NOT a connected graph
glass condor
The total degree (sum of all degrees) has to be even. So if you have exactly two vertexes with even degrees, the there must be an even number of other vertexes (all with uneven degrees), so n is even.
Then when you move to the complement, each vertex's degree goes from k to n-k, which is even if k is even and odd if k is odd.
And then you need to somehow use that:
The graph is simple and NOT a connected graph
to finish the proof, not sure how.
has exactly two vertices with an even degree
uhh, it has degrees 2, 4, 2 though?
oh, I see what you mean
6 total vertices in 2 connected components, that's 1,2,1; 1,2,1 degrees
analog sonnet
K_{3,3} with two additional edges has 4 vertices of degree 4 and 2 vertices of degree 3...
proven shard
are these dynkin diagrams ? 
glass condor
ah, that's right
so the parity is flipped.
wait, that's the entire proof, isn't it? 😅
near root
So you're saying that this isn't true?
glass condor
There were 2 vertices with even degrees. Then the remaining (n-2) vertices, all with odd degree, sum to an even number (since the total degree must be even)
therefore n is even
therefore when you go to the complement, k -> n-1-k, parity is flipped, and you get 2 vertexes with odd degrees (the ones that were even before)
so, uhh, now the real question is why the hell did we get provided these conditions:
The graph is simple and NOT a connected graph
since they don't seem necessary
analog sonnet
Just prove a stronger statement 😎
near root
Well it's given in the question... Also it's needed to prove that in the graph exists Euler's path
And to also prove that the graph does not have Euler's circle
The complementary***
@worthy palm what does k represent in the n - 1 - k in the complementary formula?
merry shard
my professor docked me points for having a proof that is "more extensive than it needs to be"
😐
drifting sail
well, nobody wants to read two pages for something that could've been proven in two paragraphs
a proof is a proof, but a short proof shows understanding and clarity in your ideas
then again I probably wouldn't deduct points unless it was a ridiculously long and/or unclear proof
pallid lintel
in my view longer proof can also be better if it contains ideas that can be used to tell us about results in other proofs or areas of math. In any case if you got a reasonable case to be made take it up with your proff, im sure you can both listen to reason.
bronze bronze
i dont understand why the conclusion is True when our based condition(p) is false, my mind doesn't get around this... can someone help me with understanding this? i just dont get that part.
pallid lintel
It doesn't need to make sense, the rule is just defined to be true. Note that 0=1 therefore im the pope is also a perfectly valid argument in propositional logic. There are alternative systems of logic that work around those rules that just arn't very intuitive. which a lot of philosophers and mathematicians work togetherr on. For example, intuitionist logic. Graham Priest is a philospher who has a textbook on nonclassical logics if you want to look into it.
bronze bronze
so.... you mean, this doesn't work with classical logic that we are familiar with?
It doesn't need to make sense, the rule is just defined to be true. Note that 0=1 therefore im the pope is also a perfectly valid argument in propositional logic. There are alternative systems of logic that work around those rules that just arn't very intuitive.
@pallid lintel
pallid lintel
here's another one. (p->q)->(p->q)...do this a million times. If i take one hair from my head then im still not bald. but if you keep taking hairs off your head and you join all those propositions togethor, at some point people would start saying your bald!
thats a criticism of logic that greeks came up with
you gotta make choices in math about how your logical system is going to be, some of it won't make sense. And there are criticisms of math, which are perfectly reasonable and perhaps something we should keep in the back of our minds.
here's another one: The liar's paradox. You can google that 🙂
bronze bronze
i'll get back to you :))))))
pallid lintel
what im saying is you have some good reasons to be suspicious of this logical system your being taught. It has been discussed a lot before. In my view it just comes down to personal choice.
bronze bronze
if i get the liar's paradox, then i can get this :), huh?
pallid lintel
nono
nvm about that. its something different. I'm just feeding you things which your curious mind might want to look into later on 🙂
bronze bronze
ohum, i have class now, so i'll come back to this lovely chat after i did a study on those
thanks, fam, i'll get back to u
pallid lintel
well pm me because theres no way i keep up with this chat here
bronze bronze
that's great, i'll
wild flame
@bronze bronze
Simple.
Here is an example. I'll list these two variables with each possible outcome.
P: I studied
Q: I pass the class
T ->T
P -> Q
If I studied, then I pass the class.
F -> T
~P -> Q
If I don't study, then I passed the class.
T->F
P ->~Q
If I studied, then I did not pass the class
F -> F
~P -> ~Q
I did not study, then I did not pass the class
The conditional logical statement is saying that if P is true, then Q is true.
All in all, this means that in order for Q to be true, P HAS to be true, otherwise the statement makes no logical sense.
So with the class example.
If you study, that means you'll pass (T ->T). Likewise, if you don't study, chances are that you won't pass (F ->F).
at the same time, its logical that some people are just geniuses and don't really need to study. Or mayhaps they already the information an are retaking the class.
In the scenario they won't study but they'll still pass.
However, in any scenario. If a person studies enough, there's no way they won't pass the class.
wild flame
Yeah. That's essentially the logic behind it. Let me know if you run into anymore confusion or need some clarification.
Try not to think too much on why it makes sense and remember the rules for now.
@bronze bronze
Also I mean conditional logical statement, not AND.
weary tiger
big brain
bronze bronze
Yeah. That's essentially the logic behind it. Let me know if you run into anymore confusion or need some clarification.
Try not to think too much on why it makes sense and remember the rules for now.
@wild flame thanks fam, that was a big help, i'll let you know if i ran into any problem
tawdry turtle
naive saffron
yes
livid shadow
steady kernel
is gcd always positive?
@tawdry turtle
A common divisor of every integer is one, so gcd ≥ 1
weary tiger
ello
how many different (additive) combinations of 1's and 4's make up n? Order is important.
for example 5 = 1+1+1+1+1, 4+1, 1+4 = 3 combinations. I know the recurranceequation for this but I have no clue how I get to it
near root
If I have (1+x)^5 can I transform it to:
(1-x^2) \ (1-x) or is that not true?
pale epoch
what
latent viper
Can someone help me with a product notation question
near root
Can I transform the binom to this form ? @pale epoch
pale epoch
transform?
shrewd spindle
Could someone help me with this problem:
A tv show has 26 episodes, you can only watch 5, but no two can be consecutive episodes
how many possibilities are possible?
near root
If x1,2,3,4,5 e {0, 1}
Then the binom looks like (1+x)^5 (not opening the exponent)
Then can I say that is equal to (1-x^2) \ 1-x ?
robust mango
@unique herald yes but be sure to write that n and q cannot be 0
weary tiger
I dont get how (2^-1)mod7 is 4
can someone pls explaion
how would I find that using euclidean algorithm
or is that not possible
unique herald
@robust mango right. got it, thx man
unique herald
Could somebody guide me through solving this question.
I dont even understand how i would go about prove this
burnt herald
can someone explain what is antisymmetric?
unreal stump
consider $\leq$ if x$\leq$y and y$\leq$x this implies x=y
vital dewBOT
DrunkenDrake:
burnt herald
i don't understand that definition
from what i read, if there is (a,b), there must not be (b,a)
but i can't grasp the part x=y
if x =y isn't that symmetric?
unreal stump
There must not be (b,a) if a and b are distinct
burnt herald
oh..
hmmm
Let A be the set of all positive integers, and x R y if and only if x | y, i.e., x divides y (equivalently, y is multiple of x). R is anti-symmetric.
Is this true or false?
vital dewBOT
DrunkenDrake:
unreal stump
Because a larger number cannot divide a smaller number
Because mk(Any multiple of m)>=m >n
burnt herald
it says all positive numbers so if i pick x as 3, y as 6,
3 divides 6, but 6 doesn't divide 3
hence it's assymetric?
unreal stump
Yes
unreal stump
That counts as reflexive
For symmetric,if there's a pair (a,b),there SHOULD always be pair (b,a)
burnt herald
a and b must not be the same right?
but in asymmetric is a == b then it's asymmetric
unreal stump
For symmetric case,a and b may or may not be distinct
unreal stump
You could call antisymmetric asymmetric except in a few cases when a=b
burnt herald
sorry i meant antiasymmetric
thank you @unreal stump
Let A be the set of all integers, and x R y if and only if x - y is even.
This is not anti asymmetric because x,y exists (x-y = 2k), but y,x also exists (y-k=-2k)
is this a correct understanding?
unreal stump
Yes
unreal stump
You mean applications?
burnt herald
You mean applications?
@unreal stump yup
i'm studying computer science, but i don't see why i'm learning discrete math (yet)
@weary tiger that's beyond my current imagination haha
near root
x1+x2+...+x10 = n
x1+x2 != 3
x4,5 = 2i(even numbers)
x6,7,8,8,9,10 e {0,1}
Find the number of solution when n=7.
In the previous exercise I did the same thing without the condition on x1+x2.
When I solve this one can I just say that x1 + x2 = 3 --> x3+...+x10 = 4
(With all the conditions as listed above)
Or do I use the (n-1+kCk) formula on x1+x2 and calculate x3+x4+...+x10 = x^3?
(Since D(2,3) gives out 4 and I just divided from x^7 x^4)
drifting halo
@unreal stump It's actually not true. If x | y then |x|<=|y|. For example 4 divides -8 but -8 is less than 4.
unreal stump
I mean,I know the proof
drifting halo
Okay 👍
unreal stump
I mean,You don't have to write (1) . You could just say because g is surjective there is a y such that g(y)=z
(1) is redundant
stray reef
never prove by contradiction what you can prove directly
your goal is to establish the existence for every z ∈ Z of an x ∈ X such that g(f(x)) = z
from z use the surjectivity of g to obtain y satisfying g(y) = z
from y use the surjectivity of f to obtain x satisfying f(x) = y
done
steady kernel
Yes, when you use the facts directly like this proof, means that you can do it without contradiction
gritty crescent
Thanks! @stray reef @unreal stump
night scroll
hi can anyone help me with solving this problem please?
the following stream represents an infinite series:
(define fact (mul-streams integers (stream-cons 1 fact)))
(define ones (stream-cons 1 ones))
(define integers (stream-cons 1 (add-streams ones integers)))
(define mul-streams
(lambda ((a <stream>) (b <stream>))
(cond ((stream-empty? a) b)
((stream-empty? b) a)
(else (stream-cons (* (stream-first a) (stream-first b))
(mul-streams (stream-rest a) (stream-rest b)))))))
can someone walk me through this code and explain what is the value of (last (stream ->listn fact 20))
weary tiger
Hello. Short and sweet question; In set theory, I am proving two sets equal by double inclusion. I have managed to rewrite one side as $A \cap B \setminus C$. At this point is it considered typically to be sufficient in a proof to say $A \cap B \setminus (A \cap C)$?
vital dewBOT
Danieldrd:
honest barn
It's probably way past this point
But those are the same set, and pretty clearly the same set so I'd say yeah or you can just prove they're the same thing
weary tiger
We're doing natural deduction in class right now
And I'm kinda stuck on this question. If we have no gives, what are we allowed to assume?
safe scroll
oblique bay
shell jewel
f(n) is a function right?
@safe scroll looks like a function to me. Do you follow a specific defintion of function?
And I'm kinda stuck on this question. If we have no gives, what are we allowed to assume?
@weary tiger is this only the question? If so does the symbol here denotes implies in your course? In other words, asking if this statement is a tautology?
weary tiger
yeah
unique herald
Could somebody please guide me through this question
im rlly confused on how to prove it
shell jewel
yeah
@weary tiger then i don't see why u need natural deduction
u can just show this by changing implication to a disjunction
(~p or q) or (~q or p)
then u can simplify this to be T
unique herald
<@&286206848099549185>
shell jewel
just use the fact that equivalence of two statements A and B means that for any truth assignment A A(A) = A(B). Which is equivalent to saying A <--> B
So show that for some truth Assignmetn function A it will give the same value for all propositons p1,p2,p3,p4
thus they are all equivalent
latent venture
shell jewel
@unique herald tell me if it wasn't clear
@latent venture it's okay i am done
good luck with ur problem
weary tiger
@shell jewel see I know, but my professor wants us to use natural deduction for some reason 
I figured it out though
shell jewel
Natural deduction is used to deduce a statement. If the question is show that this statement is valid like I understood then I have no clue why natural deduction is suggested O.o
weary tiger
Just kidding I'm still lost
@shell jewel i guess because we're supposed to learn it? I don't really get it either
i asked the professor and he said we can start with like
P or not P, Q or not Q
but I still don't really know what to do with that =/
untold cliff
how do you find the chromatic polynomial of these?
i did deletion contratcion and got k(k-1)(k-2)^4-k(k-1)(k-2)^2(k-3) but i think thats very wrong
void maple
not allowed to
stray reef
what do you mean not allowed
how are we supposed to help you if you don't give us the problems you need help with?? 
do you expect people to just magically know what problems you're doing, and which ones you're struggling on and how?
@void maple?
void maple
yes
stray reef
well we are not telepaths here, so either you post your problems or you cannot get help.
👻
pallid lintel
why is ackermann function not primitive recursive?
stray reef
pallid lintel
we didn't prove it in class but that does make sense
honest barn
I absolutely cannot help you with this since I have no clue what this is, but wtf is the master theorem bruh
That’s the most meme name for a theorem I’ve ever heard
shut fjord
master theorem is a pretty nice way of finding the recurrence relation of a function
if its given in a certain format, you can do plug and chug based on some certain conditions
I wanna know how to determine the c or k value in part c on the right side of the addition sign
lucid marlin
i am trying to prove (-1)a = -a using the axioms of Z. the proof is a bit long so posted it here: https://pastebin.com/ttPLtvxH can someone check it?
stray reef
line 13 has a typo
lucid marlin
thanks, that should be (1)a + (-1)a =
line 25 should read "therefore (-1)a = -a since .." vs "therefore -(1)a = -a since .." -- other than that how does it look?
stray reef
seems ok even if a bit excessive
lucid marlin
thanks for checking
misty merlin
Could someone please explain to me how my teacher got the +1? Thanks
This is the page before that
plucky lake
He means 2^1
or 2^(k + 1)
it's a handwriting issue
Sorry nevermind
It's even simpler
You start with k < 2^k
Then you add 1 to both sides to get k + 1 < 2^k + 1
Then since 1 <= 2^k, 2^k + 1 < 2^k + 2^k
2^k + 1 < 2(2^k)
2^k + 1 < 2^(k + 1)
now since we have k + 1 < 2^k + 1 and 2^k + 1 < 2^(k + 1),
k + 1 < 2^(k + 1)
and the hypothesis is shown
misty merlin
@plucky lake Right I get the +1 because if u add it to one side, u have to add to the otherside. But then why did he replace the 1 with 2^k? Did he sub it in from the hypothesis?
plucky lake
No, he's just saying that 1 < 2^k
That's literally it
So if you add 1 to something it'll always be less than if you add 2^k to it
misty merlin
He's got the <= because 2^0 is 1 right?
It's not always less than, could be equal
Am i thinking correctly?*
plucky lake
You're right, I just assumed it since we're working with k
misty merlin
Awesome I think I got it ty ❤️
plucky lake
When we work with k and we add 1 that sort of implies that 2^k is at least 2
i dont understand the second step of my professor's proof
@solemn dome Think this is a logic question
Someone there might be able to help you
lucid marlin
How does a = d * floor(a/d) + (a - floor(a/d)) in the below image? i'm trying to understand where the d * floor(a/d) and (a - floor(a/d) ) comes from when you start with a =dq +r
This is part of the proof to prove "Show that if a is an integer and d is an integer greater than 1, then the quotient and remainder obtained when a is divided by d are ⌊a/d⌋ and a − d⌊a/d⌋, respectively."
lucid marlin
Ah, I see. so if I am doing my algebra right then d* floor(a/d)+(a-d* floor(a/d)) = a, yeah? so it's saying a = <long expression which reduces down to a> ?
if this is the case, how can we say a = <expression> because a=<expression> is what we want to prove, so isn't that assuming the proof?
unreal stump
Ah, I see. so if I am doing my algebra right then d* floor(a/d)+(a-d* floor(a/d)) = a, yeah? so it's saying a = <long expression which reduces down to a> ?
Yes
No,We are just rewriting it as the long expression which is also a
The long expression is conveniently exactly what we need
lucid marlin
ah, i see. so we haven't proven what we wanted to show yet (that remains to be seen) but we were able to write it in a very convenient way so that we can prove it
weary tiger
@delicate lark PLEASE
weary tiger
@delicate lark pls...unblock its imp
limpid fossil
wat
weary tiger
Ask @delicate lark
limpid fossil
lazy
spring aspen
$ \forall a \forall b (P(a,b) \implies (R(a) \vee R(b))) $
vital dewBOT
carlome:
spring aspen
can i universally instantiate quantifiers with multiple variables like this one?
do I instantiate twice?
drifting sail
sure why not
There is no computer-science room?
@gilded wraith there's a comp-sci server in the #old-network
spring aspen
@drifting sail thanks
spring aspen
i need some help with proofs
i want to prove it by contradiction
then i assume that x + y is rational
sour arrow
Sure. Let's say x and y are irrational, and assume x + y is rational. Then what?
spring aspen
i dont really know how to proceed with this
if x + y is irrational, then it is done. but if x + y is rational, then it must be that $ x + y = (ad + cb) / db $
meaning that x and y must be also rational
is this right?
vital dewBOT
carlome:
spring aspen
so then they dont have to be? this is the contradiction?
drifting sail
the proposition you posted is false in general
sour arrow
Can't prove something that isn't true haha
drifting sail
and for some pairs of irrationals, such as e+π it's not known whether the sum is irrational or not
sour arrow
Ooh good example
spring aspen
i still dont know how to prove it though
proving by contradiction is really difficult
tropic coyote
Ok so this might be a really dumb question, but if I'm listing the edges of a non-directional graph
and say my nodes are labeled 1,2,3,4 etc
and 1 is connected to 2
should I list 1,2 separately from 2,1
they are the same edge since it is undirected
weary tiger
if it is "non-directional", then 1,2 is the same as 2,1
tropic coyote
so it'd be redundant right
weary tiger
yeah, u can list it if u want, but unnecessary
i personally would write 1,2 since u kno numbers in order
tropic coyote
My prof is having us list them in lex order, just wanted to make sure I wasnt being a dummy lol
weary tiger
lucid marlin
I want to prove if a,b,c are in Z, where a=/=0 and c=/=0 such that ac | bc then a | b: I have bc = (ac)k = a(ck), or bc = a(ck). if I could cancel out the c's I'd have what I want to prove, but how can I do that?
sour arrow
ac | bc means ack = bc. Since c is non-zero just divide both sides. ak = b. That's a | b
lucid marlin
is there a reason you wrote ac | bc as ack = bc vs the way I did bc = (ac)k = a(ck) ?