discrete-math

[lethal frost]

Answer:

from where did 1,3,5 come?

[sour arrow]

Yeah those shouldn't be there

[lethal frost]

why this answer is from the book

[sour arrow]

Typo

If you want me to check your answer I can

[lethal frost]

yes please

[sour arrow]

Got a picture of how you drew it?

[lethal frost]

nope

but I will change 1,3,5 to 1,2,3

and arrow from 2 to 3?

[sour arrow]

Haha yeah S is correct, they just used the labels 1,3,5 when they should have used 1,2,3

Same with T just swap the question

[lethal frost]

see

[remote cosmos]

your original typing was off,that's all it is

this diagram is correct

the 1,3,5 are element of B

the sets drawn in order are A, B, A, B

but the first 2 show how the relation S operates

and the last 2 show T

[lethal frost]

ohh

my bad

[frosty musk]

Isn't this a mistake? The disjunction isn't negated in the original statement

@agile bobcat

[agile bobcat]

it's not a mistake

@frosty musk

this is a direct application of de morgan's law

~P or ~Q is equivalent to ~(P and Q)

[frosty musk]

oh in that sense

thanks

[agile bobcat]

👍

[frosty musk]

the answer is apparently B = null set, but isn't it wrong?
Doesn't each set contain itself AND the null set, thus the symmetric of difference of A and B in the case of B being a null set isn't A, but A that doens't have as its members the null set?

I was thinking of B = {null set} or something like this

oh wait no

i think i'm wrong because we are thinking about members and not subsets

am I?

[vale cairn]

Yeah, you're thinking about subsets when it should be about members

[frosty musk]

My textbook didn't define division, how do I do it?

[remote cosmos]

division?

[remote cosmos]

that's like normal division i think?

you're dividing two cardinalities?

[frosty musk]

oh right

forgot that it stands for cardinality

but

[remote cosmos]

i don't know jaccard similarity, maybe someone can correct me if i'm off

[frosty musk]

i checked it with the answers before and it doesn't work

[remote cosmos]

can you give an example?

of when you tried it and it was wrong?

[frosty musk]

b) A = {1, 2, 3, 4}, B = {3, 4, 5, 6}

[remote cosmos]

1/3 is your answer for J(A,B)?

[frosty musk]

nvm i'm retarded

[remote cosmos]

oh no worries

you're not lol

1/3 is right though

this is interesting, what book is it from?

[frosty musk]

yeah my answer was different

click on my nickname

[remote cosmos]

oh Rosen

of course

boss

Stroustrop also boss lel

[frosty musk]

it'd be nice if he explained what jaccard similiarity is

but whatev

[remote cosmos]

i mean that's left up to the reader, in typical fashion

[frosty musk]

he'll probably refer to it later

[remote cosmos]

i've had professors give interesting problems and never refer to their provenance or application

i only realize 2 years later what the problem was about

hahahahha

[frosty musk]

i add everything to anki

if he doesn't refer to it later, then no real harm done

i'll just check it myself one day

oh my fucking god

the answer to part d) is 2 pages long

[misty herald]

does A - B = B - A imply set A is equal to B?

[rough crater]

yes

[ember solstice]

what do you call a graph that is 'almost' regular?
meaning most of the vertices have the same degree but not all of them

something like lattice

but in my case the graph is not planar

well i think it can be called lattice...

it can be embedded to $\mathbb{R}^3$

[vital dew]

271828

[ember solstice]

i found the following definition: a graph is called almost regular if the difference between max and min degree of a graph = 1

[drifting hawk]

$$(p \rightarrow q) \wedge r$$ $$p \rightarrow (q \wedge r)$$ How are these not equivalent

[vital dew]

|| Alex ||

[coral hemlock]

hello guys does anyone have the pdf for the textbook discrete math with graph theory?

[split drum]

Does this look good?

[split drum]

I can try and find a PDF, if you want.

If so, just send me the authors and edition you want.

[coral hemlock]

awesome, thank you. Here's the full title: Discrete Mathematics with Graph Theory, by Goodaire and Parmenter. Published
by Pearson, third edition (2018).

[split drum]

Alright, let me check really quick.

@coral hemlock I found 2nd edition 😕

[coral hemlock]

😦

hmmm

cuz I want to make sure I'm doing the right problems for the assignments

[split drum]

Yeah, I know what you mean. My professor makes us check that the problems are right ourselves. It's a lot of extra hassle.

[coral hemlock]

who are you taking 125 with?

[split drum]

I don't think we go to the same uni 😂

[coral hemlock]

Lmao nvm

I found it bro thanks for your help anyways!

[split drum]

Awesome, np!

[weary tiger]

yo, can i get some help with truth tables, i just overall dont understand them

[rich siren]

I feel like b is true*, but is a?

[brisk fox]

For (a), recall the definition of the empty set and what it means for a set to be non-empty.
For (b), recall the definition of subset: $A \subseteq B$ if for all $a \in A$ we have $a \in B$. The empty set has no elements, which means that in showing $a \in \emptyset \implies a \in \emptyset$ there are no elements for which this statement can be false. Equivalently, you can just use the fact that for an arbitrary set $X$, it's always true that $X \subseteq X$.

[vital dew]

Anomalocaris

[rich siren]

Hmm alright

Well I guess {} isnt an element of {}

so a is false because of that reason

while b is true because {} contains all elements of {}

Ty :)

[brisk fox]

That is correct.

[cerulean wind]

more generally, no set can contain itself (in ZF) due to axiom of regularity

[rose cipher]

Hello, I'm new to induction proofs! Can I have help with the following induction proof ```
Prove that 11! + 22! + n*n! = (n+1)! - 1 whenever n is a positive integer.

Here's what I did so far

Basis Step
P(1) = 1*1! = (1+1)!-1
1=1

Inductive Step
Suppose P(k) is true where k is an arbitrary positive int. Prove that P(k) implies p(k+1).

11! + 22! + nn! = (n+1)! - 1
11! + 22! + nn! + (n+1)(n+1)! = (n+1)! - 1 + (n+1)(n+1)!
... stuck here!

My goal is to try and get (n+1)! - 1 + (n+1)(n+1)! to ((n+1)+1)! - 1 so it is in the form of the induction hypothesis and therefore proved. But how can I move the right hand side to get to that? In class, my math prof would always try to factor something out, and thats what I did on the previous question on this hw. But on this question the -1 gets in the way so idk what to do. Help is appreciated

[faint narwhal]

Yeah try to factor something out

Ignore the -1 for now, what can you factor out of (n+1)! and (n+1)(n+1)!

[snow sleet]

those are not equivalent because if p is false and r is false, then the first statement is false (since r is false), while the second statement would be true (vacuously so since p is false). Clearly, false isn't the same as true and so the two statements aren't equivalent. Notice q could be either true or false and we still see that the statements are not equivalent...this is why I only said "if p is false and r is false"

[drifting hawk]

Ah right, thanks

[snow sleet]

You're welcome!

[snow sleet]

Like someone else already said, you can factor $(n+1)!$ from the right hand side. Doing so would turn the right hand side into $(n+1)!(1+(n+1))-1$. This becomes, after simplification, $(n+1)!(n+2)-1$.....then you're almost there, there's just one more simplification you need to make! Got it from here?

[vital dew]

logician

[snow sleet]

also, @rose cipher , you should really be using k's instead of n's there and your assumption should be more clear. Assume P(k) is true FOR SOME integer k>0. For such k, show P(k+1) is also true.

[rose cipher]

Oops you’re right. K. I think I got this from now, thanks for your help!

[snow sleet]

You're welcome!

[red wharf]

How do I do 3 times 92

[cerulean wind]

just do it ☑️

[weary tiger]

in my head id do 9x3 -> 27 so 90x3 -> 270 add 2x3->6 so 276

cause its (90+2)*3

[misty herald]

if A xor B = A xor C does this always imply B = C?

[pale epoch]

just check the truth table

[misty herald]

I meant for sets

[pale epoch]

you can still do a truth table

just have columns for x in A, x in B, x in C, ...

[misty herald]

The conclusion I'm getting is that this is true, but how do i justify it in words and not using truth table?

[pale epoch]

you emulate the truth table basically

start with some element in B, consider the case that it is also in A and the case that it is not in A, conclude that it is in C in both cases

do the same starting with an element in C

[misty herald]

oo that makes a lot more sense, thanks

[grizzled coyote]

Can someone explain to me what this means?

The codomain is R but what is the domain? I didn't understand it.

[faint narwhal]

The real numbers without 1

[grizzled coyote]

Oh

I'm dumb

[prime steeple]

• is kinda not the greatest notation

[rocky blaze]

Yea, usually my teacher writes R \ {1}

[split drum]

Can someone check this for me? I think it's right, but just want to he sure.

[weary tiger]

Everything is fine except the last one

@split drum

[split drum]

OK, I'll take a look

@weary tiger OK, I guess I'm missing something really simple but I'm not sure why it's wrong. You're referring to the interval [5, infty), correct?

[weary tiger]

Yeah, why did you cross out (4, infty)?

[split drum]

Because I was thinking that 4 is in C1 and C2, but not C4.

Ooh, wait.

That's exactly what that means.

[weary tiger]

Haha ok seems like you got it

[split drum]

Lol, thanks. 😂

[remote cape]

so anyway, i realized this works as well for the Gaussian integers, so just using, 0,1,-1,i,i works perfectly

[daring beacon]

S = {/0,{3},{/0},{4}} is the cardinality of this just 4, or is it 3 b/c the empty set /0 has 0 cardinality

[weary tiger]

The cardinality of the elements doesn't matter

That would be like saying {} and {{}} are the same set

[daring beacon]

wait I'm confused so what is it though

[weary tiger]

It's 4

[daring beacon]

so would {} and {{}} both have 0 number elements or would {{}} have 1 element

[weary tiger]

{{}} has one element

[daring beacon]

ok sorry Ik I'm overthinking and making this more complicated for myself but I was afraid of being wrong

[weary tiger]

All good

[oblique cove]

are these correct answers ?

[weary tiger]

No

Do you want to know which or do you want to try again

[shut jacinth]

ignoring the direction of the edges, this would be the longest path correct?

[weary tiger]

Sure

But it is equally long as JGMSP

[shut jacinth]

yeah that's where i was confused

is it ok to say the longest path is DGMSW

even though there are other paths with equal length

[weary tiger]

I guess that's ok

Like there are no longer paths, so it's all syntax at that point

[shut jacinth]

true, thanks

bofa

[weary tiger]

bofa

[daring beacon]

S = {(x, y)|x ∈ A, y ∈ A, x ≤ y}. how do you go about solving the cardinality for something like this when you don't even know what x and y are

[snow sleet]

well it certainly depends on what A is! @daring beacon. If A is finite, so is S.

[weary tiger]

How would I explain if graphs G and H are isomorphic, their complements would also be isomorphic? I just need a hint in the right direction

[faint narwhal]

you can explicitly construct the isomorphism

Something like, let f be the isomorphism between G and H, then we define g to be a function from the complement of G to the complement of H such that .... and then show that this g is an isomorphism

[weary tiger]

oh ok that makes sense, thank you!

[weary tiger]

If A is a finite set of real numbers then the cardinality is just n(n+1)/2 I'm pretty sure

Which is the nth triangular number, since the pairs form a triangle

[daring beacon]

A would be finite in this case

[snow sleet]

@daring beacon okay since A is finite, so is S

[daring beacon]

@snow sleet how would you find the value of x and y though

[snow sleet]

I don’t think find is the right word here. Given A, it should be clear what (x,y) will be given also the assumption that both x and y are in A and x is not greater than y.

Do you mind sharing what A is exactly? Because this is quite hard to go further from here if all you’ve told me is that A is finite.

@daring beacon

[velvet junco]

We got this definition in class but did not talk a bit about like where m comes from? Does anyone have experience applying this definition to find the inverse of a modulo function?

[daring beacon]

{0,1,2,3,4,5}

[pale epoch]

this looks very weird
what is N_n? what is a, b, k, c? what does mod n mean?

[velvet junco]

here ill show you the question on my homework for context as well but im not asking for the answer just showing

So I've found that f^-1 does exist because gcd(5,7) = 1

And that c = 4, f(4) = (5(4)+1)mod 7 = 21 % 7 = 0

that leaves me with an inverse of f^(-1)(x) = (((1-7m) / 5)x + 4) mod 7 for some integer m in Z

But I am not sure this is an adequate answer, and also wouldn't that definition of the inverse imply that there could be multiple inverses

[pale epoch]

🤔

those are some really weird definitions tbh

[velvet junco]

yea this entire textbook is like this

[pale epoch]

but let me see...

[velvet junco]

also the professor is not the best doesnt really explain much so its the perfect storm

[pale epoch]

but in general if an inverse function exists, it is unique

but the main issue with those "definitions" are that it's not clear at all that those are even functions to begin with

[velvet junco]

Oh for more context this is a CS class and a CS textbook so I think that might clarify why mod is just talked about regularly

since its a pretty standard definition in CS

[pale epoch]

this is still very bad

[velvet junco]

yea its not my favorite textbook

[pale epoch]

anyways, for your inverse you need to find k and m such that ak + nm = 1

[velvet junco]

the proofs are pretty bad too it solves like 2 steps of every proof then pulls the leave the rest for exercise

[pale epoch]

and what this does is ask when a linear function x \mapsto ax + b has an inverse modulo n, this boils down to finding an inverse of a modulo n and this exists iff gcd(a, n) = 1

then bezouts lemma gives you k and m such that ak + nm = 1 and this k is the the inverse of a modulo n

you can compute k and m using the (extended) euclidean algorithm

[snow sleet]

Okay thanks. So x can range from 0 up to And including 5. When x is 0, y could be 0 through 5. When x is 1, y could be 1 through 5, etc

[snow sleet]

I’ll give you a hint for when x=0. When x=0, the ordered pairs: 0,0),(0,1),(0,2),(0,3),(0,4),(0,5) are all in S. Now do this for x=1,2,3,4,5 as well

[snow sleet]

@daring beacon

[hidden crystal]

HELP!!! HELP!! I NEED HELP WITH SET ROSTER NOTATION 😦

[worthy mist]

calm down a bit and ask your question

[stray reef]

WHY ARE YOU YELLING??

[remote cosmos]

[weary tiger]

😂😂😂😂😂😂

[snow sleet]

I think you mean set builder notation. I have no idea wtf set roster notation is tbh

[stray reef]

roster notation is where you list out all the elements separated by commas.

[snow sleet]

Hmm gotcha

Thanks

[hoary igloo]

The only thing that confuses me here is the line "it is the subgraph obtained from G by deleting the verticies in V' with their incident edges"

Arent those the only nodes and edges we are keeping in an induced subgraph?

[red nest]

The second sentence talks about G[V\V’] not G[V’]

V\V’ is the set of all vertices of G that are not in V’

[hoary igloo]

ah thank you

[fresh venture]

Can anyone help me with this?
I am not able to understand this one

[errant bloom]

did you try to write this out in english rather than math symbols ?

[boreal plume]

is this correct for this, or should it look like this? https://cdn.discordapp.com/attachments/824701621598158848/882149996034674688/unknown.png

@errant bloom also

[sharp ledge]

If P(r,s ) : r= s , is this statement true ? ∃r ∈ N : ∀s ∈ Z , P(r,s) How do i proved it with contradiction it is false ?

[fresh venture]

Yes

[snow sleet]

P(r,s): a=b makes no sense. What are a and b?

I think you meant r=s instead

[sharp ledge]

yeap

[snow sleet]

okay, so the first statement is false because no such natural number r exists.

the first statement says that we can find a fixed natural number r that equals every integer...this is not true

if we had: For all r in N, there exists s in Z such that r=s

then this is true

because we could take s=r

[errant bloom]

what did you get?

[snow sleet]

make sense @sharp ledge ?

[sharp ledge]

Yeah , makes sense , thanks a lot

[snow sleet]

You're welcome!

[errant bloom]

looks a bit strange tbh, but it depends what A and what the other sets are in your picture

[fresh venture]

For any x there exists a y then whenever there is x there is also y
Maybe 😕

[errant bloom]

no

$\forall x \forall y P(x,y)$ means for all x it holds that: for all y P(x,y) is true. So if you chose one x, then it has to hold for every y. Repeat this for each x

[vital dew]

lyinch

[errant bloom]

however you got something else in your question

$\lnot \forall x$

[vital dew]

lyinch

[errant bloom]

not for all. This is equivalent to saying that there exists an x where the statement does not hold

[fresh venture]

So the statement is false?

[errant bloom]

so $\lnot \forall x P(x)$ means: not for all x P(x) holds, which is equivalent to $\exists x \lnot P(x)$

[vital dew]

lyinch

[errant bloom]

therefore they are not the same statements left and right

[fresh venture]

Ohh I got it

[errant bloom]

left: there exists a y, such that for all x, P(x,y) does not hold.
right: For all x, there exists a y such that P(x,y) does not hold

[fresh venture]

Thank you very much

I have some doubts in Other questions.
Can I ask them?

[errant bloom]

yes

[fresh venture]

Fibonacci always starts from 1 right?

So can we assume that f0 here is 1

[errant bloom]

no, I think from 0

[fresh venture]

I have seen 1, 1 , 2 , 3, 5.....

Never seen 0

[errant bloom]

mmh I guess that then just depends

do the way you've seen it in the lecture

[fresh venture]

Nah not lecture

I googled it

[pale epoch]

it starts from 1 here or the theorem is false

[errant bloom]

because there's not much difference between 0 1 1 2 and 1 1 2 for some people 🙂

[fresh venture]

Ohh

[pale epoch]

or actually wait

from 0 lmao

[fresh venture]

It starts from 0?

[pale epoch]

for n=1 you get 0+1 = 1^2

actually both seems to work

so it doesn't matter

[fresh venture]

I see

[fresh venture]

Then how should I solve this?

[fresh venture]

Oh btw @errant bloom
The negation is on y on both sides.
Did u used the reverse method?

[errant bloom]

not sure what reverse method, but I explained the meaning

[errant bloom]

here

[fresh venture]

Ohh ok

I just mean ur statement said 'there exists a y '

But there is a negation in the question on y

So shouldn't it be 'y doesn't exist'

?

[errant bloom]

but the negation is in front of the forall

normally it's forall y it holds that

[fresh venture]

In the right side it isn't

[errant bloom]

not you say not for all, which means that there is at least one for which it doesn't hold

but the negation of for all doesn't mean that nothing holds. It means that at least one doesn't hold

[fresh venture]

Ohh I see

Need help with this 😞

[fresh venture]

Possible?

<@&286206848099549185>

[faint narwhal]

@fresh venture what have you tried?

[fresh venture]

Umm... From well ordering it means we need to show that the number is expressible as product of primes
Right?

[faint narwhal]

Uh no, not really. Do you know what well ordering states?

[fresh venture]

Kind of

[faint narwhal]

Can you tell me what it says?

[fresh venture]

All natural numbers except 1 can be expressed as product of primes

[faint narwhal]

No that's not the correct statement

You might want to check your notes or whatever you're getting this from

[fresh venture]

I read tbh

[faint narwhal]

Huh?

That theorem can be proven using the well ordering principle, but is not what the well ordering principle says

[fresh venture]

It says all non empty subset of positive ing have a least element

[faint narwhal]

Right

[fresh venture]

...

[faint narwhal]

Yes?

[fresh venture]

Uk any lead in solving that question?

@faint narwhal

[faint narwhal]

I mean I know how to solve it, but I think you should think about it too

Here's a hint, consider the set of natural numbers that don't satisfy the condition

Try to use well ordering on this set

[fresh venture]

All integers divisible by 7 are natural numbers

I mean ofcourse 1 can say -7 is divisible by 7 giving -1
But idk if we count those

From my pov I would say answer is uncountably infinite

Cause ig we call countably infinite which are really really too damn many but still isn't out of reach

But no matter how further you go there will always be another which is divisible

[rotund frigate]

N is countably infinite as well

But anyways it's obviously not finite or uncountably infinite

As there are an infinite number of multiples of 7, and it's a subset of a countable set

[robust mango]

solve for x,y using the equations and see what point(s) A contains

[robust mango]

Yeah one point

thats it

A is a set that contains only one point, which is (1,0)

[stray reef]

yes, it is finite

[pallid lintel]

Reading this paper, it's claimed each v_i is simplicial in a perfect vertex elimination scheme. But v_3's neighbourhood isn't a clique.

relevant definitions

v_3 has v_4, v_2, v_1, v_10 as neighbourhood. There is no edge from v_1 to v_4 so it cant be a clique. I must be misunderstanding something here

[trail steppe]

v7 does not connect to v4 either.
A perfect vertex elimination scheme is ordered. You have to remove the nodes from the left. Could be better to code it and draw the resulting graph after every iteration but I am lazy

[pallid lintel]

ok i get it now, i drew it with ordering and seems to work

[hidden crystal]

why isn't {2} an element of {1,2,3} ?

I see the number 2 in the set 1,2,3.

But 2 is different from {2} in builder notation. how?

[pale epoch]

2 is just 2

{2} is the set containing 2

an apple and an apple in a box are two different objects

[hidden crystal]

so {2} isn't an element of {1,2,3} because {2} isn't an element, it's a set?

[pale epoch]

not quite

{2} is not an element of {1, 2, 3}

sets can also be elements

e.g. {2} is an element of {1, {2}, 3}

(this is the set containing 1, 3 and {2})

[hidden crystal]

but in that specific question, {2} is not an element in {1,2,3}. Ok I get it

thanks

[mint bane]

i asked something similar to this before but again - i can prove this algebraically pretty easily but is there a more "combinatoric" way of understanding

[snow sleet]

@mint bane yes. Consider a department with n men and n women. We wish to make a committee with n individuals

Each side of the equation counts the number of possible committees

Choose n from the total (2n) is one way to count the number of committees. We could also choose i of the men and n-i of the women. Let i run from 0 to n and sum up the terms

Each side counts the same thing, so they’re equal

[mint bane]

ok semi related question first, my prof showed us a notation that went like (n r,n-r) [imagine that the r,n-r is under the n, idk the latex for it] what does that mean

because arent the two 'choose's on the right going to evaluate to the same

[snow sleet]

Uh I think that means n!/(r!(n-r)!) though I’m not sure

[mint bane]

hmmm

ok im gonna give it some thought i'll brb

mucho thanks

[snow sleet]

You’re welcome!

[brisk fox]

$\binom{n}{r, n-r} = \binom{n}{r}$ in this case. you can generalize it to $\binom{n}{r_1, \dots, r_t} = \frac{n!}{r_1! \cdots r_t!}$ where $r_1 + \dots + r_p = n$.
it's called the generalized binomial coefficient since it lets you express $(x_1 + \dots + x_t)^n$ as a sum

[vital dew]

Anomalocaris

[mint bane]

ok i understood the first part of this but how does summing each of the terms tie it together

like i get that each element of the summation represents different possibilities for how many men/women in a group and ya add together each one

but how does that end up being 2n choose n

[snow sleet]

Give me a sec @mint bane . I’ll reply once I fire up my laptop

[mint bane]

👉 👈

[snow sleet]

okay @mint bane I'm back

so each term in the sum is the number of different possible combinations of men and women given that there's i men in the committee at that time

so i can only run from 0 to n

and that's ALL the cases

therefore the sum

equals 2n choose n

2n choose n represents choosing n different people from 2n different people

that implicitly takes care of the different number of males we can have in a given committee

make sense @mint bane ?

[mint bane]

im gonna sleep on it it's late where i am

but i think i get it

[snow sleet]

aight

have a good night

[sharp ledge]

How should i explain that for all natural number r and s , if r^2 = s^2 , r=s ?

[sour arrow]

You could go with a difference of squares:
(r + s)(r - s) = 0

If r + s = 0, then r = -s. However, there's no such thing in the naturals.

Leaving us with r - s = 0

[sharp ledge]

What about verbal explanation ? instead of using equations and number ?

[sour arrow]

Is there a format you're trying to match?

[sharp ledge]

Dont really understand this, any easier explanation ? in sentence maybe ? I tried to answer as : Since a is in domain of natural numbers where there is no negative numbers and b is in integer, a will not be equal to b for every b , how is it ?

[spiral pilot]

i can't get the compound propositions

[sour arrow]

I'll use ' for complement, or logical NOT.
Chest A says "it isn't true that A and B both have no treasure"
That is, A says (a'b')'

That one make sense? @spiral pilot

[spiral pilot]

@sour arrow i think i got it

[sour arrow]

The whole thing? Okay good!

[spiral pilot]

yeah i believe so

you wanna check?

[sour arrow]

Go for it

[spiral pilot]

@sour arrow

the 7 looking thing is negation

[sour arrow]

I'm with that first one

But the second one, I'm not following

Trunk B basically just says NOT a

[spiral pilot]

the second prop is if both inscriptions are false

so not not a

[sour arrow]

Oh I see

Yeah okay I'm with it

Just delete the original question

[weary tiger]

bit binary floating-point number representation of 27/128. You must show how
your derived each component of your bit string answer.```
can anyone do this for me im stuck and show working pls

[sharp ledge]

Is it possible to explain that this statement is true using proof by contradiction ?

[sharp ledge]

Would you explain about this sentence ? Why could we only choose r-s = 0 ?

[proven silo]

zero-product property. If a * b=0 then a=0 or b=0. So r+s=0 or r-s=0

[worldly flame]

How do I find the number of integer solutions to a linear inequality?

[vital dew]

Neil_P

Neil_P

[robust pasture]

yo i need help

with math

[meager prairie]

[tepid bay]

So.. i dont quite understand why (n!)! > (n!)^(n-1)!

Looking for an argument as to why

[stray reef]

@robust pasture http://dontasktoask.com/

[robust pasture]

sry i will be mindful next time

[stray reef]

post the thing you need help with

right away

[robust pasture]

dont need help no more its ighT

ight*

[stray reef]

instead of just going "i need help with math" because like. no shit sherlock you've come to a server dedicated to math help

[robust pasture]

mb

[proven silo]

x_1 can be in [0,100] for there to be a solution to 5x_1+x_2<=500.
If x_1=0 x_2 can be in [0,500].
If x_1=1 x_2 can be in [0,495] and so on. You should be able to find a formula and find the amount of solutions from here

[outer hinge]

Whenever we decide to say something like “ForAll a”, or if we say “A = { a such that..”, or we say “aFb” are we treating these lower case letters like a constant or a variable ?

I was under the impression we were treating them like a variable but then I got confused when we went over the formal definition of a function

and when we went into relations

I’m a little confused because we defined F as being a subset of A x B, and then we also can say aFb which is the same as F(a) = b

But then also a cannot be paired with more than one b

[weary tiger]

ye

so what r u confused about?

[tepid bay]

still looking for some assistance on this

[weary tiger]

do u know induction?

[tepid bay]

yes

i have this breakdown of (n!)! but i dont quite understand it

this is in the memo

[weary tiger]

what do you not undersnad

[tepid bay]

say for instance that 2nd factor

(n+1)...(2n)

[weary tiger]

ye

[tepid bay]

i dont understand how they get to that

the first factor makes sense

[weary tiger]

so they list the nunmbers up to n!

and then multiply them

[tepid bay]

by?

[weary tiger]

themsevles

thats how factorial works

like n! is 1*2... n

[tepid bay]

yes i understand how the factorial works

[weary tiger]

so n! ! would be 1 2 ... * n!

and every n terms they group them

first n terms is just n!

2nd n terms is bigger than n!

like myabe if you see n=3 you might understand it better

[tepid bay]

ah i see

thanks

i was trying to think of it as n!! = n! x (n! -1) x (n! -2)..

[weary tiger]

same thing though

but from the other side

[tepid bay]

can you explain this?

[weary tiger]

ok see case for n=3

[tepid bay]

yes

[weary tiger]

(3!)! is something like 720

I think

,w 6!

[vital dew]

Results provided by WolframAlpha

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

Requested by Ledog#7947

[tepid bay]

nice

[weary tiger]

actually that doesnt matter

[tepid bay]

yes

anyways

[weary tiger]

but (3!)! would be multiplying all integers up to 3!

so just 1 x 2 x...x6

you group them by 3

(1x2x3)x(4x5x6)

first term is 3! 2nd is bigger than 3!

[tepid bay]

ok so now its the same concept but for n

[weary tiger]

ye

[tepid bay]

why group them by n though?

just for sake of the argument ?

[weary tiger]

because if you group them by less you might not be sure if its bigger than n!

[tepid bay]

ok yes, n!^(n-1)! you mean?

so we are trying to get the same number of terms, and see how each term size compares

[outer hinge]

I guess I’m confused as to wether or not we treat something as a constant or variable and I’m a little confused at relations

So, when we denote an element a, or an element b, within the set A or B, is this a constant or variable ?

Also, I know a relation is a subset of a Cartesian product, but I’m confused about what something like aRb would be

Is that stating a specific ordered pair in set R ?

Which could be considered R(a) = b

?

[pale epoch]

aRb means (a, b) is an element of R

R(a) = b is bad notation, since relations are in general not functions

and there might be multiple different b such that aRb

(think of the relation <)

then yeah a function F is a special kind of relation, where for each a there is exactly one b such that aFb and in this case we can write F(a) = b

[pale epoch]

here the things are variables

everything is a variable unless it is explicitly given

or sometimes we "fix" certain objects and give them a name, but that is stated

i think it just takes a while to get used to that

@outer hinge

[outer hinge]

Well thanks. I just went by the math lab and they helped me out with this

I was mainly confused about how we were defining a function

I have to clean up my notation though

I also forgot to put a “such that” symbol

[pale epoch]

👍

this is mainly just formalising of what our intuition already is

a function is a "machine"; you put a thing in and get some output

same input always means same output (i.e. one input cannot produce multiple different outputs if you run the machine at say different times)

[drifting hawk]

Express $$\xor x \in A, P(x)$$ with other standard logical operators. Does $$(x \in A) \oplus (y \in A), P(x), x\neq y$$ work

[vital dew]

|| Alex ||
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[north dust]

My combinatorics class is going well so far.

[sharp ledge]

If P(r,s ) : r= s , ∃r ∈ N : ∀s ∈ Z , P(r,s) How do i proved it with contradiction it is false ?

[snow sleet]

well suppose such a natural number r exists.

Then s=r+1 is an integer and so we're guaranteed r=r+1. This is a contradiction

therefore, the statement is false

make sense @sharp ledge ?

[sharp ledge]

Why is it adding by 1? for r

[warm zinc]

halp

[snow sleet]

I just did that to keep it simple. You could have taken s=r+2 or s=r+3 or so on

[snow sleet]

x is in A U B if and only if x is in A or x is in B. y is in A intersect B if and only if y is in A and y is in B.

[sharp ledge]

Why are we adding it with some number ?

[warm zinc]

im just 13 can you like explain it more clearly

[snow sleet]

One proof that proves this statement is false would look like this: Suppose there is a natural number r such that r=s for all integers s. For such r, note that r+1 is an integer due to the closure of the integers under addition. Then our assumption guarantees r=r+1, a contradiction.

[snow sleet]

what do you mean by "it"?

[sharp ledge]

The r

[snow sleet]

okay

[snow sleet]

read this^

that might clear it up^

[snow sleet]

sure. Look at the things (called elements) that A and B share. Those two things are 3 and 4. Therefore, A intersect B is just {3,4}. For A union B, write down all of A and then add in the elements from B that are not in A. So A union B is {1,2,3,4,5,6}

[snow sleet]

does this make sense @sharp ledge ?

[warm zinc]

@snow sleet can you solve Find the HCF of 45, 75 and 120 by division method.

[snow sleet]

You mean GCD?

[sharp ledge]

For such r, note that r+1 is an integer due to the closure of the integers under addition. Then our assumption guarantees r=r+1, a contradiction. What does this means ? Closure of integer ?

[snow sleet]

since r is an integer, and since 1 is also an integer, r+1 must also be an integer @sharp ledge. That is closure of the integers under addition. Given two integers x and y, x+y is an integer

[warm zinc]

mabye lel

[sharp ledge]

Yup

[snow sleet]

wtf you mean maybe lmaoo. What does HCF stand for? Highest Common Factor?

[warm zinc]

yus

[snow sleet]

okay cool so does my proof that the statement is false make sense to u?

[snow sleet]

okay that's also known as the GCD just so you know lol

[warm zinc]

ah

[snow sleet]

anyway, I'd solve that using prime factorization @warm zinc

"division method" makes no sense to me as that's very ambiguous

[warm zinc]

[snow sleet]

well, write each of the three numbers you stated as a product of primes

[vital dew]

logician

[sharp ledge]

What about the statement that r= r+ 1?

[snow sleet]

r=r+1 is a contradiction

because 0 doesn't equal 1

do you know why our assumption guaranteed r=r+1? @sharp ledge

it's because they said r=s for every integer s. So since that's true for every integer s, it's also true for r+1. Hence we have r=r+1.

and then we see r=r+1 can't possibly happen. So it is a contradiction

make sense @sharp ledge ?

[sharp ledge]

Should s+ 1 also when r+ 1? since r=s we assume ?

[snow sleet]

this is actually quite easy to check that the GCD is 15 because each is divisible by 5 and each is divisible by 3 (note that 75 isn't divisible by 9) and after viewing the prime factorization of 45, it can't be any higher than 5 times 3, which is 15. Thus the gcd of 45, 75, and 120 is 15

[snow sleet]

s was just a place holder in the statement "r=s for any integer s". In my proof, I took r+1 to be s. And I'm allowed to do so because r+1 is an integer. So talking about s+1 doesn't really make sense here.

[brittle dove]

A perfect number is a positive integer n such that the sum of its factors equals 2n. For instance, 6 is a perfect number since 1+2+3+6 = 12 = 2×6. Prove that a prime number can’t be a perfect number.

[brittle dove]

Can anyone help with this?

[snow sleet]

does that make sense @sharp ledge ?

[stray reef]

sammee_here, i think you're interrupting a conversation here.

[sharp ledge]

Yup

[stray reef]

either wait or move to a free questions channel

[snow sleet]

okay great

@brittle dove you're free to ask now

next time don't just barge in

[brittle dove]

A perfect number is a positive integer n such that the sum of its factors equals 2n. For instance, 6 is a perfect number since 1+2+3+6 = 12 = 2×6. Prove that a prime number can’t be a perfect number.

Can you help with this?

[stray reef]

maybe

have you made any progress on this so far?

any thoughts at all on how this problem may be approached?

[brittle dove]

nah I was like blank

[stray reef]

okay

do you know what a prime number is?

[brittle dove]

Yeah

[stray reef]

okay

[brittle dove]

a number that can be divided exactly only by itself and 1,

[stray reef]

exactly

so what are the factors of a prime number p?

[brittle dove]

1 and itself

[stray reef]

and what is their sum?

[brittle dove]

1+p

[stray reef]

and can it ever happen that 1+p equals 2p when p is a prime?

(if yes, for what p? if not, why not?)

[brittle dove]

this is where i am struck

[stray reef]

1 + p = 2p

[brittle dove]

if p is 1

it can be equal

[stray reef]

great

does this equation have any other solutions besides p=1?

[brittle dove]

i guess no

if p = 2

Then 1+2 != 4

[snow sleet]

can I suggest an algebraic approach to answering Ann's question @brittle dove ?

[stray reef]

you "guess"?

sounds like you're massively overthinking it, sammee_here

you are familiar with algebra, yes?

[brittle dove]

it cant be Yes other than p =1

[brittle dove]

Yeah but not good

[stray reef]

correct.

the equation 1+p=2p has only one solution, p=1.

but 1 won't do, because 1 is not a prime.

[brittle dove]

yeah

[stray reef]

therefore no prime satisfies 1+p=2p

therefore no prime is perfect

QED

[snow sleet]

https://tenor.com/view/clapping-leonardo-dicaprio-leo-dicaprio-gif-10584134

[brittle dove]

Thanks a Lot @stray reef !!

[snow sleet]

This channel is open for questions

[brittle dove]

Will Come here with more doubts then😀

[warm zinc]

B={x: x=3y+2, y<6; y N} wha dis means

@snow sleet

[stray reef]

have you ever seen set-builder notation before?

also, i presume that you meant $y \in \bN$ at the very end?

[vital dew]

Ann

[stray reef]

@warm zinc

[snow sleet]

this depends on how you have defined N. Some profs include 0 in N and some don't

[warm zinc]

is this correct? {(1,5)(2,8)(3,11)(4,14)(5,17)}

[snow sleet]

no

elements of that set are not ordered pairs

[sharp ledge]

What about proving with contradiction for :: For all r in N, there exists s in Z such that r=s , which should be true. How different is the steps in proving in compared to previous one ?

[snow sleet]

there's no need for a proof by contradiction for this statement. It is quite easy to prove it directly like so: Given a natural number r, clearly r is an integer and r=r. Therefore, the statement is true.

make sense @sharp ledge ?

[snow sleet]

you can write B equivalently as ${3y+2:\text{$y<6$, $y\in\mathbb N$}}$

[vital dew]

logician

[snow sleet]

again @warm zinc , B depends on how you have defined N as I said before here^

[plush dock]

Hey, I'm struggling with this problem a bit:

Write a generating function for the following equation and formulate it: x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7=n
Where: x_1, x_2 & x_3 are divisible by 3, and x_4, x_5, x_6 & x_7 are not divisible by 3.

I think I understand that, the generating function itself would be:
f(x)=(1+x^3+x^6 ...) ^ 3 * (1+x+x^2+x^4+x^5+x^7 ... ) ^ 4

I managed to formulate the left part of the equation: 1/(1-x^3)^3. However, I'm not sure how to formulate the right part (1+x+x^2+x^4+x^5)^4.
Would super appreciate pointers, since I feel like I'm missing something very fundamental.

Thank you very much!

[faint narwhal]

Try splitting it into two sums

@plush dock

Alternatively, you know what 1 + x + x^2 + x^3 + x^4 + ... is, and in some sense, the left and the right parts make up that

[plush dock]

That makes sense, thank you

[plush dock]

I think I get it! @faint narwhal
Would it be correct to do this:
[(x+x^2)^x^3 + (x+x^2)^x^6 ...]^4
[1/1-x^3 * (x+x^2)]^4

[faint narwhal]

Yes that works

[plush dock]

Thank you!

[brittle dove]

Hii

Can anyone help me to prove this?

[buoyant bloom]

help

[brittle dove]

@stray reef ?

[stray reef]

don't ping me like this please

[cerulean wind]

factor it. n^3-n = n(n^2-1) = n(n+1)(n-1)

then just note that this is a product of three constructive integers.

[sharp ledge]

how do i prove that if n is even integer, then there exist an integer p such that n is sum of (p-1) and (p-3)?

[pale epoch]

you use the definition of even

[sharp ledge]

Yeah, i mean i first assume n = 2k , then summing up the given p-1 and p-3 , i got 2p - 4,which is then factorised to get 2 (m-2) , and how should i proceed ?

[last timber]

solve for p

@sharp ledge solve 2p-4 = 2k for p

[sharp ledge]

ok, so p = k+ 2
Then ?

[last timber]

then qed

you have constructed desired p

[sharp ledge]

Why is it that p in terms of k instead of k in terms of p ?

[sharp ledge]

Anyone ?

[brittle dove]

Anyone?

[pale epoch]

because you start with n = 2k and then have to find a p such that your statement is true

[pale epoch]

what have you tried?

[plush dock]

Hi again!
We have the following statement: "Between every 2 numbers, there exists another number that is different from them".
We have 2 options (that I will post down below).
I picked the 1st one since it made more sense to me, but it was incorrect, and I'm trying to wrap my head around why the 2nd option is the correct one.
The fact that z can be equal to x or y really, really confuses me, and I can't wrap my head around it.
In addition, I don't quite understand why the 2nd one is incorrect, though I think I have an idea (probably because it has no conditional statement, and if we were to write it within a conditional statement, it would be different?)

[pale epoch]

pull the negation into the quantifier, it becomes "there exists a z such that z > x and y > z"

i.e. "there exists a z such that x < z < y"

the problem with the first statement is that x < y need not be true and then the statement is false

[plush dock]

Ah I'm such a goofball. I handled the negation into the quantifier extremely poorly.
As for the first statement, what you mean is that if x<y is false, then the entire statement is false?

Sorry, I'm just learning the subject, I may be bit slow

[pale epoch]

the statement after the universal quantifiers that is

so as long as numbers x < y exist, the whole thing is false

which we do not want

to be more precise, 3 and 5 are a counterexample to that statement: set x=5, y=3

(also no reason to apologize)

[plush dock]

Thank you very much, this is an excellent explanation and makes tons of sense to me

I honestly never thought of coming up with a counter example

But this is a really nice trick

[brittle dove]

I just done this

Can you check whether its correct/not @pale epoch

Express the following sentences using logical expression:

No students like Ram

[pale epoch]

in a) it is not clear what you are quantifying over, b) makes no sense since p(x) and R(x) depend on x and x is a variable here (you are missing some kind of quantifier)

what you wrote for c) is "every student takes analysis and geometry

[brittle dove]

Is this correct?

[pale epoch]

this looks better

the notation {S-x} is weird, other than that it is probably fine

[stray reef]

when you write {S - x} do you mean the set S with the element x removed?

because that's S \ {x} (or S - {x} if your class uses normal minus for set difference)

[brittle dove]

Yeah we remove

[slate burrow]

Hello, i would like some guidance if possible. What does it mean "Find two sets A and B" ?

Do i need to show two examples of like A-B {numbers} and A-B {numbers} ?
or like show the numbers that these properties have in common

[vale cairn]

Give an example of a pair of sets A and B such that A union B = {1,2,4,5,6,7} and ... etc

I'm not sure what you mean by 'A-B {numbers}'

[slate burrow]

oh i am suppose to do for each set? Like one for A union B and one for A /\ B?

etc

[vale cairn]

No, you have to give two sets A and B that simultaneously have all of those properties

'that satisfy all of the above properties at the same time'

[slate burrow]

I can't really understand the question, having a hard time to grasp my mind around "two sets a and b that satisfy all of the properties" ,

oh like

[vale cairn]

I think i'd rephrase it as 'Find a pair of sets A,B that satisfies all of the above properties'

[slate burrow]

A - B = {1, 4, 7}?

it satisfies the first and the third one

but not the second one

[vale cairn]

you haven't given any sets there

[slate burrow]

Sorry if i'm slow

[vale cairn]

so if the properties were A union B = {1,2,3} and A intersect B = {2} then an example solution would be 'A = {1,2}, B = {2,3}'

[slate burrow]

A \ B

aha

[vale cairn]

because A union B = {1,2,3} and A intersect B = {2}

[slate burrow]

yeah so like kinda of what they have in "common"

I don't need to use any sign for the answer? Can you just write like A= {1,2}, B={3,4}

or nvm that's not what they are asking for

Thank you, i believe i understand the question!

[vale cairn]

wdym by sign?

[slate burrow]

i meant like "union" or "intersect" but that's not what they are asking for

[vale cairn]

they want two sets so i'm not sure why that'd be relevant?

[slate burrow]

yeah, it's not relevant at all. I just assumed that's what they are asking for.

[simple nova]

https://i.imgur.com/vHZoIZp.png\

Can anyone help me with this?

[weary tiger]

left side will be $(P\land{\neg{Q}})\lor\neg{R}$

[vital dew]

jswatj

[weary tiger]

which u can use distribution laws to reach the right side with

[simple nova]

Conditonal law right

[weary tiger]

uhhhh

i dont know the names

p->q is not(p) or q

[simple nova]

How did you get the not(r)

distrubute the not?

[weary tiger]

demorgans law

[simple nova]

Okay so first I remove the p->q

by switching it to not(p) or q

[weary tiger]

yes

then demorgans

[simple nova]

Okay then I use DeMorgans

Okay

[weary tiger]

then distribution laws

$p\lor{(q\land{r})}\equiv (p\lor{q})\land{(p\lor{r})}$

[vital dew]

jswatj

[simple nova]

Okay I see now

Thank you, I was stuck on the first step but you got me going! I appreciate your help!

[fleet hare]

this rite?

[snow sleet]

sounds like you've already got your question answered here, but perhaps maybe another helpful observation would be that the right hand side (r -> p) ^ (q -> ~r) is equivalent to r -> (p ^ ~q).

[snow sleet]

no

T xor T is false (I'm referring to your second row of T's and F's)

[fleet hare]

?

[snow sleet]

the first row you have F

which is good

the second row

should also be the same

[fleet hare]

y

nvm

lol

i get it

[snow sleet]

dope

the rest looks good once you fix that

[fleet hare]

i think my others look good

ya

[simple nova]

@weary tiger

you still there?

$(P\land{\neg{Q}})\lor\neg{R}$

[vital dew]

purga

[simple nova]

When you get this, how do you distribute it?

to make it (r->p) | (q -> !r)

[snow sleet]

this is distributed like so $(P\lor\neg R)\land(\neg Q\lor\neg R)$

[vital dew]

logician

[snow sleet]

@simple nova see above^

[simple nova]

ahhhhh okay I see now

when you switch it to conditional it gets rid of the not in q

just leaves the not r

[snow sleet]

yea ~Q or ~R is equivalent to Q implies ~R

[snow sleet]

I think you mean & instead of | here

[weary tiger]

everything that logician said

[simple nova]

Yeah typo! Thanks so much @snow sleet

[fleet hare]

sorry.

[snow sleet]

now it's not busy

lol

[fleet hare]

,

its ok.

[weary tiger]

what are you struggling with @fleet hare

[fleet hare]

i did p ^ q

[weary tiger]

okay

[fleet hare]

would (p^q) -> p be TTTT

[weary tiger]

No

[fleet hare]

• FTTTT

sorry

first F

[weary tiger]

No

[fleet hare]

oh then idk

[weary tiger]

T->T is true

when is an implication false?

[snow sleet]

yes

[fleet hare]

?

[snow sleet]

(p and q) implies p is always true this is because p was assumed to be true (if you think about it without truth tables)

[weary tiger]

oh actually

yeah im mistaken

you are right

[fleet hare]

..

[weary tiger]

apologies

[fleet hare]

ty.

its a tautology

.

[snow sleet]

yes

[weary tiger]

yup

[snow sleet]

p implies p is a tautology so (p and q) implies p is also a tautology this is because (p and q) being assumed to be true necessarily entails p is true.

[fleet hare]

so i gues p -> ( p v q) is also a taut

[snow sleet]

yes

because if p is true, then (p or q) is true.

[fleet hare]

right

because p is true

[snow sleet]

it's even easier to see this if you realize p -> ( p v q) is equivalent to [ (~p or p) or q]

[fleet hare]

is a conteignecy when like

(p^q) -> (p->q)

[snow sleet]

no

because that is a tautology

[fleet hare]

or is this still a taugtology

ok

[snow sleet]

think about the fact that we're already assuming q to be true and due to the following equivalence

[snow sleet]

this is equivalent to [(p and p and q) implies q]

[fleet hare]

jeesh

gotcha

[snow sleet]

a contingent proposition is a proposition that is neither a tautology nor a contradiction. One example of a contingent proposition would be p->q

[lethal frost]

hello

does anybody know why this is wrong?

are you sure?

[formal flicker]

agreed

[lethal frost]

yup it is

I checked it

thanks @tough plover '

[waxen nest]

hi anyone can help?

[snow sleet]

sure I can help.

[waxen nest]

am i supposed to prove that the given statement is equals to this?

[snow sleet]

Let's first negate the left hand side of ^

[snow sleet]

no. The given statement has S and T's anyway...not P, Q, and R. That given statement is just for you to note that equivalence and make use of that equivalence in this giant mess they handed you

[waxen nest]

we will use the given statement for later part?

[waxen nest]

using de morganz law?

[snow sleet]

I would say yes

[snow sleet]

first rewrite $[(\neg P\land R)\implies (Q\implies R)]$ before negating

[vital dew]

logician

[snow sleet]

do you know what this^ is equivalent to @waxen nest ?

[waxen nest]

nope

looks like i need to apply implication law?

[snow sleet]

I have no idea what implication law is ( I don't care to remember those things by name), I just know them from experience

[snow sleet]

this statement is equivalent to $[(\neg P\land R\land Q)\implies R)]$

[vital dew]

logician

[waxen nest]

~(~P ^ R) or (Q -> R)

[snow sleet]

yes, but I think my approach is a little simpler it'll likely help if we ever use deMorgan's law

[waxen nest]

hmm what did you just do here

[snow sleet]

I literally pulled that Q out since it's really in the antecedent anyway

because the statement is kinda like: Let t be true. Then if y is true, then q is true. So really, if t and y are true, then q is true

[waxen nest]

okay then what do you do next?

[snow sleet]

you mean after this?

[waxen nest]

yes

[snow sleet]

okay, then we negate this

[waxen nest]

we still have a negation outside right

[snow sleet]

yes

[snow sleet]

negate the whole statement here

[waxen nest]

so it becomes (P V ~R V Q -> ~R)?

[snow sleet]

no

[waxen nest]

i think im wrong haha

how do you do that

[snow sleet]

when you negate an implication a->b, what do you get?

a ^ ~b right?

[waxen nest]

oh yes

[snow sleet]

okay, so what does this become after we negate the statement here?

[waxen nest]

(P V ~R V Q ^ ~R)

[snow sleet]

ahhhhh!!!!

make it as simple as possible!!!!

[waxen nest]

(P V ~R V Q)

[snow sleet]

this negated becomes $\neg P\land R\land Q\land\neg R$

[vital dew]

logician

[snow sleet]

I have no idea what you're doing here.

[snow sleet]

read this^

[waxen nest]

why ~P ^ R for the first part?

wouldnt ^ change to V when you negate it

[snow sleet]

a implies b negated becomes a and ~b

[waxen nest]

i thought ~~P will become P 🤔

[snow sleet]

you literally take the antecedent and negate the consequent

[snow sleet]

because that was literally part of the antecedent

[snow sleet]

we're NOT negating the antecedent!!!

[snow sleet]

read this again^

[waxen nest]

why? you said negate the whole thing but why leave out the antecedent

[snow sleet]

because that's how negating implications work

Again negating (a implies b) yields (a and ~b)

[waxen nest]

a ^ ~b

[snow sleet]

notice how a wasn't negated there

[snow sleet]

with that in mind, negate this^

[snow sleet]

you should get this after doing so^

are you following @waxen nest ?

[waxen nest]

yeah

[snow sleet]

okay

[waxen nest]

its already done here

[snow sleet]

no it's not...well it depends on what you meant by "its"

[waxen nest]

ok from here we can simplify?

[snow sleet]

the negation is done, but we're not done with the problem

[waxen nest]

associative law

[snow sleet]

remember we have an extra "and P" at the end right?

[snow sleet]

yes and commutative

[waxen nest]

R ^ ~R becomes F?

[snow sleet]

the extra P comes from here on the right of the ^

so really we just need to simplify $\neg P\land R\land Q\land\neg R\land P$

[vital dew]

logician

[waxen nest]

ok

so P ^ ~P becomes F as well

and F and F is F

F ^ R ^ Q

[snow sleet]

no

[waxen nest]

oh right

[snow sleet]

look at what you said here

[waxen nest]

R is gone

[snow sleet]

right

[snow sleet]

so this becomes?

[waxen nest]

F ^ Q

[snow sleet]

it can be even more simplified than that

[waxen nest]

yes its F

[snow sleet]

so this whole thing reduces down to F

[waxen nest]

yeah

how can we use the given statement?

[snow sleet]

one way to use it would be to rewrite $[(\neg P\land R)\implies (Q\implies R)]$

[vital dew]

logician

[snow sleet]

but I think that way would be more tedious

that way would require deMorgan's law

[snow sleet]

try using the equivalence they gave to rewrite this^

@waxen nest

[waxen nest]

isnt it what we did earlier?

rewriting that part

[snow sleet]

we rewrote that part, but we didn't not use that equivalence

I used the equivalence $[s\implies (t\implies u)]\iff [(s\land t)\implies u]$

[vital dew]

logician

[snow sleet]

@waxen nest

[waxen nest]

how you get that

i did this

[snow sleet]

this makes no sense. you brought in S and T. The statement you were asked to simplify had neither an S nor a T in it

[waxen nest]

why

yea i assume that S is ~P ^ R

and T is Q -> R

[snow sleet]

okay, you should have stated that in your paper so your teacher doesn't mark you off

[waxen nest]

okay

so i can use that?

[snow sleet]

you can simplify this further

[snow sleet]

your approach? or mine?

[waxen nest]

mine

[snow sleet]

yea, but I'd show more steps

I mean

I'd simplify further

after stating what S and T are

[snow sleet]

you can use deMorgan's law to simplify this further

[snow sleet]

that should be an F not a T in the last line

[waxen nest]

why F

i was trying to sub T, the simple statement inside

[snow sleet]

hang on a sec

I've forgotten how you have defined T

T does not mean true in your case

[waxen nest]

from the given

i used T

[waxen nest]

yea it doesnt

[snow sleet]

right, yea that's awkward notation

most people would think T means true

[snow sleet]

this should be (S ^ ~T ^ P) on the last line

[waxen nest]

right okay

[snow sleet]

now substitute things back in

you really only need to substitute S back in

@waxen nest

[waxen nest]

oh okay

let me try that

ok heres what i got so far

@snow sleet

so after this i can just show that its = F?

since F and anything will just be F right

[snow sleet]

yes

looks good! although your second to last line is exactly the same as the one above it

[waxen nest]

second line?

[snow sleet]

I said second TO LAST line and the line right above that second to last line

@waxen nest