discrete-math

[dim magnet]

What is the axiom of specification?

[sleek swallow]

Well, I can state it for you.

Let A be any given set. Let p(x) be an open sentence. Then, the axiom of specification tells us that the following set exists:

$B = {x \in A: p(x) }$

That just says that the B is the set of all elements x that belong to A such that a given sentence p(x) holds for some fixed element x.

[vital dew]

Abhijeet Vats:

[sleek swallow]

Basically, what it says is that subsets of a given set A do exist and you can form them by constructing sentences that become propositions for any specific element.

[dim magnet]

What does a sentence mean in this context? Set builder notation?

[sleek swallow]

Uh okay, I should explain this a bit better.

Suppose I told you something like x>3. Okay?

Just suppose that I threw that at you without any context about what x really is

[dim magnet]

ok

[sleek swallow]

Now, the problem is that x>3 can be true or false depending on the value of x

So if x = 5, then 5>3 means that x>3 is true. If x = 2, then 2>3 is clearly false, so x>3 would be false.

So, that's what I mean by a sentence

The moment I give you context about what the variables are, you should be able to determine if that sentence is true for that variable or false

[dim magnet]

right

so I have a set A = {2,4,6,8}, and I have a sentence x<6, then a set B = {x | x is an element of A and x<6} = {2,4}

[sleek swallow]

There are more rigorous definitions for what a sentence is, though. But I certainly haven't learned them yet. I've seen some books on logic and set theory define those things in a more rigorous fashion.

But in any case, that's the basic idea of what a sentence is.

Yes, exactly

However, the way I'd write set B is:

B = {x is an element of A | x<6}

So, the specification portion comes before the open sentence.

However, some authors do write it like how you did. I, generally, don't prefer that.

[dim magnet]

Ah

I do it like that because the current course I'm taking in year 11 high school teaches it that way

Thanks a lot!

oh sorry

I forgot to ask XD

[sleek swallow]

Yea what you're learning now is naive set theory so they're less inclined to use formal notation

So, it's okay to just grab objects and elements out of thin air. However, you can't really do that cos that leads to contradictions.

[dim magnet]

"The answer is that if you look at the axiom of specification and have a set A, then you can define the empty set by means of a contradiction."

How would you define the empty set by means of a contradiction exactly?

[sleek swallow]

Let B be the empty set. Then, B = {x belongs to A | x does not belong to A}

[dim magnet]

wait

the property of belonging to a set can be a statement?

[sleek swallow]

Indeed

In fact, you should have that as an axiom of set theory. If x is an object and A is a set, then (x belongs to A) is a proposition for some fixed x.

That means that it is either true or false but not both.

[dim magnet]

Right

[sleek swallow]

I know this is a bit difficult to understand cos you haven't been acquainted with some basic logic

The point is that all these terms have very strict definitions. So, the only way to grasp the subject is to get a good grip of those definitions.

[dim magnet]

the little logic I know is from some from proofs and some from this book xd

Converse, contrapositive, negation

[sleek swallow]

Oh okay, so do you know what a contradiction is formally?

$p \land \lnot{p}$ is the contradiction that's used to construct the empty set

[vital dew]

Abhijeet Vats:

[dim magnet]

Proof by contradiciton

Is like you negate the original statement you're trying to prove and then look for some sort of a fallacy?

[sleek swallow]

Because what you have is a set that consists of all the elements that belong to another set but also do not belong to that set. Now, an element cannot belong to a set and not belong to it at the same time. So, there are no elements that satisfy that open sentence. Hence, the set is empty.

Well, suppose you have a statement p and you want to prove q from statement p. What you do is to assume that q is false. Then, use p & not(q) to derive a contradiction. Since contradictions don't exist in math, q must have been true.

^That's the basic idea, but there's a formal way of expressing it as a rule of inference

[dim magnet]

I see

Wait

how does "Since contradictions don't exist in math, q must have been true." tie in to defining the empty set using a contradiction?

oh

Let me see if I get it

[sleek swallow]

mmh go on

[dim magnet]

you're proving a set can exist with no elements

wait

hmm

wait I think I can do this

[sleek swallow]

Yes. We're trying to show that a set with no elements does exist. To do it, we need to construct an open sentence that is guaranteed to be false for any object.

[dim magnet]

oh I see

we assume the statement we're trying to prove is false: a set can't exist with no elements

Let A be a given set

[sleek swallow]

Nope.

We don't have to do a proof by contradiction here

The axioms already give us a way of directly constructing the empty set

[dim magnet]

right

[sleek swallow]

A proof by contradiction would be an example of an indirect proof

Let S be a set. Set Theory guarantees that S exists. Now, let x be some object from S. Then, we define the following subset:

$\varnothing = {x \in S | (x \in S) \text{and} \lnot{(x \in S)} }$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So, keep in mind that we're asserting that x is an object that somehow belongs to a set S but also doesn't belong to the set S. That is NOT true for any object x. So, this subset cannot contain any elements. However, it does still exist because it follows all the rules that we've laid out beforehand. Hence, it does exist as a set and since it has no elements, we define it as an empty set.

[dim magnet]

The penny has dropped

Far too late xd

Thank you for persevering with me XD I'm not the smartest tool in the tool shed unfortunately

[sleek swallow]

To be fair, the language that's required for this stuff is not known to you yet

So I don't blame you if you don't understand what I've written above

however, keep working at it and you'll get it eventually.

[dim magnet]

We can use the axiom of specification to define the empty set by a contradiction.

[sleek swallow]

yea correct

[dim magnet]

The axiom of specification states that given any set A, and some sentence, there always exists a set B such that x is an element of A and satisfies the sentence

Sorry

That was phrased wrong

x doesn't have to satisfy the sentence

[sleek swallow]

Indeed. It's just that you can define subsets of A using the open sentence, where the open sentence is something that becomes a proposition the moment you use a specific x.

[dim magnet]

and because that subset always exists, regardless as to whether the conditions contradict

we can have the empty set

[sleek swallow]

yeap

in that situation, the condition (which is informal speak for the open sentence) is not true for any element x in the set A. Hence, it is always false and therefore, the set contains no elements.

[dim magnet]

epic

[sleek swallow]

mmh yea

it's cool stuff

[dim magnet]

so now I can actually consider what this means xd

so if we start with the union collection

sorry

arbitrary union definition

for an empty collection

This is how I'm interpreting it

the union = {x | x is an element of nothing an nothing is an element of the empty set}

or do I have it all wrong

[stray reef]

uh

yeah you do

[dim magnet]

I feel like I'm not quite understanding what it means to be an element of the empty set

[stray reef]

it means nothing

well

[sleek swallow]

Well, it means that it's false

[stray reef]

in the sense of

whatever thing you take

it's NOT an element of the empty set

[sleek swallow]

See, the statement $x \in \varnothing$

[vital dew]

Abhijeet Vats:

[sleek swallow]

is actually false for all objects

There is a definitive meaning that it has. It just means that if you find me an object x and ask me to check if that object belongs to the empty set, I will ALWAYS tell you that that's false. It can never belong to the empty set.

[dim magnet]

what would A be in this case then?

the subset of the empty set

That's what I'm not understanding

[sleek swallow]

Now, if you're talking about the empty union, then let's look at it a bit more simply with just a family of two sets.

If x belongs to the union of the family F = {A,B}, where A & B are sets, then we are saying that x must belong to either A or B or both. That means that there is a set in F such that the statement (x belongs to that set) is true.

Now, let A be the empty family and U A be the empty union. So, we are asserting that this set contains all the elements such that there exists a set in A where (x belongs to that set) is somehow true. However, that cannot be the case because there is no such set in A. A is empty, after all. Hence, the open sentence is false for all objects x. Thus, the empty union is empty.

[stray reef]

why not write ∅ for the empty family lmfao

[sleek swallow]

how do you do that without latex

i didn't want to use latex lol

[stray reef]

oh i'm using autohotkey lol

got it configured to let me type certain math symbols

[dapper rose]

option shift o for mac

[stray reef]

[sleek swallow]

😦

haizzz, i'm just really dumb tbh. I should probably use latex more

also, $\phi$ will always be the empty set for me ❤️

[vital dew]

Abhijeet Vats:

[stray reef]

yikes

no

please at least use $\emptyset$ or $\varnothing$

[vital dew]

Ann:

[sleek swallow]

NO 😄

[stray reef]

38(

[sleek swallow]

no u

[dapper rose]

@dim magnet *element of the empty set
is more accurate

[sleek swallow]

yer mum is more accurate

OH SHIT REKT

❤️

[dapper rose]

disscrete-math

[gilded cove]

Hello guys. First time submitting anythinh to this forum. I have a Hard time to get started with this homework. Can anybody nudge me to th right direktion. PS sorry bad engrish

[reef thistle]

| is divides, as in a | b iff a is a factor of b, don't use it as a division operator @gilded cove

Did you transcribe the question accurately?

[spice lance]

can someone help me understand the language of this question. Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives). a) ¬∀x∀yP(x, y)

[gilded cove]

transcribe = make out or write out. Yes. And i do understand that it is a two part question @reef thistle

[spice lance]

oh sorry didnt see a question was already active in this channel

ill leave

[reef thistle]

do you understand fermat's little theorem?

[gilded cove]

But there seems to be that i have forgoten some kind of therom

there it is

no first time i have heard of fermat's "little" theorem

but i will look in to it

only heard of Fermat's theorem, thanks for the information

[reef thistle]

I think it's likely to be the same one

unless you mean this: https://en.wikipedia.org/wiki/Fermat's_theorem_(stationary_points)

[gilded cove]

i first ment that one yes

but i googled it and got https://en.wikipedia.org/wiki/Fermat's_little_theorem

witch is the first time

gonna look in to it and see if i can worksomthing out

if you can give any more advice, pleas do, and thank you in advance

[zinc pewter]

How do I go about solving this problem

[teal canyon]

not and and are sufficient to express any logical connective

say, a or b is can be defined as ~(~a and ~b).

Do the same for equivalence.

[zinc pewter]

so are you saying that since <--> takes an argument of both true or false that we can change the logical expression to be

[teal canyon]

I'm saying that there is a function f written in /, ~, that f(a, b) is equivalent to a<->b

[zinc pewter]

Like this?

then use distributive properties

whoops forgot and

[teal canyon]

say (a <-> b) is equivalent to ((a->b) /\ (b->a))

then you can rewrite -> in and and not

[zinc pewter]

I can't use -->

[teal canyon]

Not in the answer, but it might lead you to the right answer

[zinc pewter]

oh

[teal canyon]

you'll need to express implication through conjunction and negation, so it won't be there

[zinc pewter]

are you talking about biconditional equivalence?

like <--> becomes P --> Q and Q -->

[teal canyon]

yes

[zinc pewter]

that's an idea i had but I don't know how to do it to an argument with 3 variables

I know what P <--> Q is

but how do I apply the law to something complex

do I simplify the left side first?

or can I do it straight out of the gate

[teal canyon]

you apply it to subterms, replace P and Q with ~(p/~r) and q correspondingly

how do you write implication through conjunction and negation?

[zinc pewter]

demorgan's law?

[teal canyon]

Like this: (a->b) is equivalent to ~(a/\ ~b)

[zinc pewter]

Oh I see what you mean

I think I get it

Is this correct?

I used distributive law after

@teal canyon

[teal canyon]

it seems like outer ~ from implications are lost

and q in the second subterm is negated

[zinc pewter]

I used demorgans law

Unless I messed it up somehow

[teal canyon]

nah, that one flips conjunction to discunction, so you can't use it

well, you use it only once to be exact

on the disjunction

[sleek swallow]

$P \iff Q \iff (P \implies Q) \land (Q \implies P)$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Let $P :\iff p \lor \lnot{r}$ and $Q :\iff \lnot{q}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Now simplify

[zinc pewter]

I can turn p V ~r to ~(p and r) right?

wait

[teal canyon]

you lost one negation

[sleek swallow]

$(p \lor \lnot{r}) \iff \lnot{(\lnot{p} \land r)}$

[vital dew]

Abhijeet Vats:

[zinc pewter]

so ~(~p and r) and ~q?

[sleek swallow]

[zinc pewter]

I'm confused

https://cdn.discordapp.com/attachments/621777911808000021/668614819179397120/Screen_Shot_2020-01-19_at_7.36.23_PM.png

[sleek swallow]

Okay wait. Originally, you wanted to find, using only negations and conjunctions, a logical expression equivalent to $(p \lor \lnot{r}) \iff \lnot{q}$.

By the biconditional law, that's just equivalent to:

$[(p \lor \lnot{r}) \iff \lnot{q}] \iff [(p \lor \lnot{r}) \implies \lnot{q}] \land [\lnot{q} \implies (p \lor \lnot{r})]$

[vital dew]

Abhijeet Vats:

[zinc pewter]

Yep

[sleek swallow]

Now, we simplify each one individually:

$[(p \lor \lnot{r}) \implies \lnot{q}] \iff \lnot{(p\lor \lnot{r})} \lor \lnot{q} \iff (\lnot{p} \land r) \lor \lnot{q}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Similarly:

$(\lnot{q} \implies (p \lor \lnot{r})) \iff (q \lor (p \lor \lnot{r}))$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Okay, because this is getting a little bit complicated with 3 primitive propositions, I'm gonna make some simplifications and let $P :\iff p \lor \lnot{r}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So, now, we have:

$[\lnot{P} \lor \lnot{q}] \land [P \lor q]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So, the question is, can that be simplified?

[zinc pewter]

you can factor out the negative for the first term and that would make it and

I think

[sleek swallow]

Hmm

Uh give me a while, i'll do this on paper

[zinc pewter]

I got this answer

[sleek swallow]

Hmm lmao i might just try expressing this in terms of the sheffer stroke

[sleek swallow]

,rotate

[zinc pewter]

that's what I'm getting

[sleek swallow]

That negation at the front of the entire expression oof

That seems to not be the way to properly express it in conjunctive normal form

Uh check if your teacher is okay with that

[teal canyon]

nah, you don't need cnf there

[sleek swallow]

Hmm i wonder if there's a further way to simplify this

[zinc pewter]

wait so is it correct?

[teal canyon]

you've lost one not before p

[zinc pewter]

where

[teal canyon]

last one

[zinc pewter]

yeah

ur right

[sleek swallow]

Nice interesting problem though

[teal canyon]

also, one (~p /\ r) must be with ~ and another (would have ~~, corresponding q is not negated) without

[zinc pewter]

There's another problem?

[teal canyon]

ye

[zinc pewter]

[teal canyon]

first ~p /\ r shouldn't be negated

[zinc pewter]

I'm using two laws

https://cdn.discordapp.com/attachments/496785905474994186/668604290256076800/unknown.png

this

and I'm converting it to make it and

so I flip the signs

so 3 nots

oh

you meant the right side

wait

yeah it should work

[teal canyon]

you see that when you have q negated, the corresponding ~(~p /\ r) isn't negated

but when you have a bare q, it will be ~(~(~p /\ r)) next to it

it follows from the law you posted

the bare q comes from q -> ~(~p /\ r)

and negated one from ~(~p /\ r) -> q

what's the expression for q -> ~(~p /\ r) by your law?

[zinc pewter]

it's cause q is negated from the beginning

[teal canyon]

nah

it's not

https://cdn.discordapp.com/attachments/496785905474994186/668600911672770582/unknown.png

[zinc pewter]

^^

That's the question

the one above that is my horrendous answer

[teal canyon]

oh

[zinc pewter]

is this endgame

have I finally cracked the code

[teal canyon]

seems to be okay then

[zinc pewter]

we made it

[teal canyon]

the last ~p is still lost

[zinc pewter]

:>

p

hmm

it's negated

from the beginning isn't it

and due to the --> rule

we negate the whole term

[teal canyon]

p has lost its negation

[zinc pewter]

where

[teal canyon]

last one

~p /\ r must be the same on both sides

[teal canyon]

yes

[zinc pewter]

you're right

now we're talking

,rotate

fuck

[teal canyon]

ye

[zinc pewter]

sweet ty

[teal canyon]

you can remove double negation ~~(~p /\ r) into (~p /\ r)

[zinc pewter]

on the left side or the right side?

cause if i remove the left side

it will make my equation have discontingency

[teal canyon]

you have only one double negation

[zinc pewter]

wouldn't that make "or" q

because i'm multiplying through

[teal canyon]

nah, I mean ~(~x) is equivalent to x

[zinc pewter]

that's a law

[teal canyon]

and there is ~(~(~p /\ r))

do you see it?

[zinc pewter]

yeah but you're multiplying through aren't you

the reason why ~(~p) = p is because it's making the statement false then true

or true then false

~(~(~p and r) and q)

[teal canyon]

that's not a double negation

because there is "and q" thing

there is ~(~(~p /\ r)). exactly that

[zinc pewter]

Oh

you're correct

[glossy relic]

Hey , why is the 12 a combination and the 13 a permutation ? Shouldn’t the 13 be a combination as well since the cards are random so we don’t care about the order

[earnest meadow]

For 12 it's because the order of the cards in your hand doesn't matter. For example A ♠️ 2 ❤️ 3 ♦️ 4 ♠️ 5 ♦️ would be a straight no matter which order you arranged the cards.

[stray reef]

we do care about the order in 13 @glossy relic

we care who gets what card

[earnest meadow]

As for 13 when you're giving cards to different players, that's when order matters, because giving an Ace of spades to player 1 would be different from giving an ace of spades to player 2

[glossy relic]

So the order matters the moment she mentioned the three players ?

[earnest meadow]

Yes

[glossy relic]

I see thanks ! Are there any keywords that would directly lead to knowing if the order is important or not ?

[earnest meadow]

You'll notice the patterns as you do more problems, but in general if there's a unique distinction between different results then use a permutation.

For example electing 3 council members from a class of 10, you'd use a combination, but electing a president, vice president, and treasurer then order would matter.

[glossy relic]

I see , thanks !

[glossy relic]

Why did the teacher use permutation here ? Aren’t we just looking for 3 people out of two groups , there should be no ordering right?

[obtuse lance]

,rccw

[oak yew]

@glossy relic whats original exercice text?

[weary tiger]

Theres no ordering, if you have 6 mathematicians and want to pick any 3,that would be 6C3

Same with physicists

[stone nymph]

Sup, I am having trouble proving the correctness of this algorithm with induction
double expRecursive(double x, int n) {
if (n <= 4) {
return expIterativ(x, n);
}

return expRecursive(x, n/2) *
       expRecursive(x, (n + 1)/2);

}

Have tried it for quite the while but not getting anywhere

[analog sonnet]

What's this algorithm supposed to do?

What's expIterativ?

[tacit willow]

 Let C = D = {−3,−2,−1, 1, 2, 3} and define a relation S from C to D as follows:  
For all  x,y∈C×D, (x,y)∈S (x, y) means that 1x−1y is an integer. 
For all (2,2) is part of CxD, (2,2) is part of S
Is 2 S 2? Is −1 S  −1? Is (3, 3) ∈ S? Is (3,−3) ∈ S? 

Could anyone help me with this one?

[pale epoch]

where is the problem?/what have you tried?

[tacit willow]

well, i don't really understand how to read it

that's my main problem

[pale epoch]

then the first step is to look up all the definitions you don't understand

i.e. what a relation is

[stray reef]

is that the problem statement verbatim

[tacit willow]

i understand the definitions. but i am struggling to understand this particular task

To go through it step by step

the third line is my own sad attempt btw

hm 😦

[stray reef]

can you show the original problem exactly as it was stated just so we're on the same page

[tacit willow]

Sorry, there was some sort of formatting error

[wild flame]

Just to review.

https://i.imgur.com/unjYiAm.png

Based on the order of precedence. is the add logical operator

supposed to have higher precedence than the or?

Or are the same precedence?

I imagine the and/or are on the same precedence, but the numbering on the right confuses me.

[teal canyon]

they aren't

a and b or c and d is parsed as (a and b) or (c and d)

[wild flame]

I see. what about

a or b and c and b

would it be (a or b) and (c and b)?

[sleek swallow]

a or (b and (c and b))

[wild flame]

Ahh.

[sleek swallow]

Where the and is associative so you can change the placement of the paranthesis

[wild flame]

Okay.

so I need to work on committing that to memory exactly. No problemo.

[sleek swallow]

It's not a very difficult thing though. For me, I just use the paranthesis cos that's easier. You can also learn it subconsciously just by doing lots of problems in logic

[wild flame]

yeah, its pretty simple. Right now Im just reviewing to reinforce my knowledge.

[tacit willow]

So back to my problem? 🙂

[pale epoch]

honestly its just plugging in the numbers

[sleek swallow]

Anyways, going back to your question Joakim:

$S = { (x,y) \in C \times D: \frac{1}{x}-\frac{1}{y} \in \bN }$

[vital dew]

Abhijeet Vats:

[pale epoch]

if you understand what a relation is, this should be no problem

[sleek swallow]

So, first thing to do is to write out C x D explicitly

Then, think about what values of x and y allow 1/x - 1/y to be in the integers

After that, just pick the relevant ordered pairs that satisfy that property. Then, you're done.

So, for example, (3,3) satisfies it since 1/3 - 1/3 = 0 and 0 is an integer.

[tacit willow]

yea, so the next question was:

a. Is 2 S 2? Is −1 S −1? Is (3, 3) ∈ S? Is (3,−3) ∈ S?

I just answered yes, yes, yes and yes since theyre all = 0

but that didn't seem right?

[sleek swallow]

Okay, so if (2,2) in S is true, then it should satisfy the open sentence that allows it to be in S, no?

Now, apply that to the others. Is it true for each one of them? Look very closely at each of them, yea?

[tacit willow]

aha, i see the last one is 3, -3

[sleek swallow]

Mmh understand now?

[tacit willow]

i think so.

[sleek swallow]

Uh i'd say that since you're learning this stuff, you should try and write out C x D explicitly

Just to get some practise in forming the cartesian product of two sets

[tacit willow]

So in order to write S as an ordered pair i would do S = {(2.2), (-1-1), (3,3)}?

and the CxD would be C = {−3,−2,−1, 1, 2, 3}, D = {−3,−2,−1, 1, 2, 3}?

[sleek swallow]

'In order to write S as an ordered pair'

No

That is just wrong

[tacit willow]

im sorry. a set of ordered pairs

[sleek swallow]

S is NOT an ordered pair. It's a subset of C x D. C x D is a set.

What you have written for C x D is also wrong. You literally just put the sets C and D side-by-side and called that the cartesian product

You have to form the set of all ordered pairs from C and D

So, for -3, you have the following ordered pairs:

(-3,-3),(-3,-2),(-3,-1),(-3,1),(-3,2),(-3,3)

That would be the ordered pairs where the first number is fixed at -3.

[tacit willow]

aha, i see.

[sleek swallow]

Now, repeat the same process with the other numbers in C. This is the sort of thing that should also have been covered in your notes/textbook, so do review that as well

{1,2} x {1,2} = {(1,1) , (1,2) , (2,1) , (2,2)}

^That's an example

[tacit willow]

aha

this helped alot

[tacit marlin]

Can somebody explain where I messed up? I am not sure what I did wrong

[faint narwhal]

You substituted k+1 wrong on the right side

In the line after "true for n = k+1"

[tacit marlin]

What would it be? I don't see how it is wrong

dont i just replace k with k+1?

[faint narwhal]

Yes you do

What's 3k when you replace k with k+1

[tacit marlin]

isnt it 3k+1?

or is it 3(k+1)?

[faint narwhal]

the latter

and you should think about why that is

[tacit marlin]

I think I got it im stupid. Thank you

[shut fjord]

How would I approach solving this type of question?

To check if its consistent, wouldn't each proposition have to have the same truth value in the same column

[pale epoch]

this?

those are some dumb questions tbh

[shut fjord]

that looks very familiar

I haven't did discrete math in sometime, but now I'm actually taking a course haha

[weary tiger]

https://cdn.discordapp.com/attachments/624693291014946829/668948076064407583/image0.jpg

,rotate

[weary tiger]

Can someone pls help me solve B

I cannot do it for the life of me

idk how to handle so many implications

[sour arrow]

A → B = ~A U B

[sleek swallow]

What have you tried?

One way to do it is to assume that there exist a set of truth values for (p,q,r) that cause the given proposition to not be satisfiable.

In that case, r has to be false and the entire antecedent has to be true. However, for that to happen, all the individual compound propositions making up the conjunction have to be true.

In particular, if r is false, then p and q must be false so that p->r and q->r are both true by ex falso quodlibet. However, that means that (p or q) is false and that causes one of the compound statements in the antecedent to be false. Thus, the antecedent must be false, which is a contradiction.

@weary tiger

If you want to manipulate them, then use the fact that $p \implies q \iff \lnot{p} \lor q$

[vital dew]

Abhijeet Vats:

[weary tiger]

thats all ive got and i dont even think its right

@sleek swallow

[sleek swallow]

Your negations are placed wrongly

[weary tiger]

when i do implication i dont put negations on the inside of the first term?

[sleek swallow]

Also $(p \implies r) \land (q \implies r) \iff (\lnot{p} \lor r) \land (\lnot{q} \lor r)$

[vital dew]

Abhijeet Vats:

[sleek swallow]

What do you notice about that immediately?

See $\lnot{(p \lor r)}$ is not the same thing as $\lnot{p} \lor r$

[vital dew]

Abhijeet Vats:

[weary tiger]

ok let me retry this

[sleek swallow]

Answer my question above

What do you notice about the conjunction on the right of what I wrote above?

[weary tiger]

um

i am not sure, thats just the implication definition isnt it?

[sleek swallow]

Well, see how r is the common proposition?

[weary tiger]

yes

[sleek swallow]

Could you remove it using the distributive laws and pull it outside?

[weary tiger]

oh wait

r v (!p ^ !q)

So like thus

@sleek swallow

Do i do demorgans next?

[sleek swallow]

What do you think

Just try it

See where it leads ya

De Morgan does look like the way to go

[marsh mountain]

hey I am new to the server I was wondering where I can have someone teach me a few things

[sleek swallow]

Uh mr k, can you use an unoccupied questions channel

[weary tiger]

I think i did something wrong

but I got part of it T atleast 😂

[marsh mountain]

ty @sleek swallow

[sleek swallow]

Lol

Let me get on the bus first and i’ll try typing it out

[weary tiger]

Thank you for helping me

I appreciate it a lot

[sleek swallow]

Gosh typing in latex on phone will be painful

Here goes

So we’ll consider the antecedent first, yea?

$[(p \lor q) \land (p \to r) \land (q \to r)] \iff [(p \lor q) \land ( \lnot{p} \lor r) \land (\lnot{q} \lor r)]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Understand that? @weary tiger

[weary tiger]

yes

[sleek swallow]

Okay

So that’s equivalent to:

$(p \lor q) \land [(\lnot{p} \land \lnot{q}) \lor r]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Understand?

[weary tiger]

yes

[sleek swallow]

Now, let me define the following:

$P :\iff p\lor q$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So now we have something of the form $P \land (\lnot{P} \lor r)$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Understand? All I did was use De Morgan’s

[weary tiger]

why is
$P :\iff p\lor q$

[vital dew]

Skwiid:

[sleek swallow]

I just defined a new proposition

That’s all

[weary tiger]

Oh ok

[sleek swallow]

Just to simplify things

Now, I can distribute:

$(P \land \lnot{P}) \lor (P \land r)$

[vital dew]

Abhijeet Vats:

[sleek swallow]

See an absurdity in there?

[weary tiger]

p and not p is false

[sleek swallow]

Indeed

So we can throw that away

It’s not gonna be relevant to the truth value of the disjunction

Hence, we actually have something of the form:

$(P \land r) \implies r$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Keep in mind that we were only manipulating the antecedent just now

[weary tiger]

yes

[sleek swallow]

$[(P \land r) \implies r] \iff [\lnot{(P \land r)} \lor r]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Understand?

[weary tiger]

yea implication definition

[sleek swallow]

That’s not the definition of the implication lol

But it’s a tautology you can use

Okay anyways

[weary tiger]

oh

[sleek swallow]

By De Morgan’s

$[\lnot{P} \lor \lnot{r}] \lor r$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Now, use associativity of the disjunction:

$\lnot{P} \lor (\lnot{r} \lor r)$

[vital dew]

Abhijeet Vats:

[sleek swallow]

See a tautology in there?

[weary tiger]

Yes

!P or True

then domination Law

[sleek swallow]

Domination law?

Gosh such kinky names

[weary tiger]

LOL

[sleek swallow]

Anyways yea

[weary tiger]

my professor taught me this shit man 😂

but I understand now, thanks so much

[sleek swallow]

The disjunction is true because one of the propositions is a tautology

No problem

Did you understand the verbal argument I gave initially, though?

Generally, when you’re trying to prove that something is a tautology, you just need to show that there are no propositional truth values that could make it false

@weary tiger

[weary tiger]

ahh

wait question

https://cdn.discordapp.com/attachments/496785905474994186/668964380368437248/340191992636243968.png

https://cdn.discordapp.com/attachments/496785905474994186/668964881302552596/340191992636243968.png

!P isnt !p or !q

its !p and !q

is that a mistake?

@sleek swallow

[sleek swallow]

De Morgan bruh

$\lnot{p} \land \lnot{q} \iff \lnot{(p \lor q)}$

[vital dew]

Abhijeet Vats:

[weary tiger]

oh my god

i forgot demorgan can go backwars

-0eeofjowpjf;elsdf

why is this harder than calculus 3

[sleek swallow]

What’s calculus 3?

But yea, this stuff takes a bit of time to get used to

[weary tiger]

Series

calc 3 was just the leftovers of the calculus book

[sleek swallow]

Oh

What I suggest for propositional logic is to prove all of your basic tautologies from scratch

You’ll remember them better

[weary tiger]

alright, thanks for the tip

our professor just taught us these laws and thats what were relying on

[sleek swallow]

Did he show you the proofs?

[weary tiger]

nope

[sleek swallow]

Wow what a dumbo

He’s such a cock

[night frost]

How do I go about starting to disprove statements with like 3+ quantifiers

[sleek swallow]

Well, it depends

Give an example

[night frost]

the statement I'm working on is this

$\exists M \in \mathbb{R}^+, \exists n_0 \in \mathbb{Z}^+, \forall n \in \mathbb{Z}^+, [(n>n_0) \Rightarrow (n^4 \leq Mn^2)]$

[vital dew]

darkninja175:

[night frost]

It's just an exercise in doing these kind of proofs, nothing special

but it just gets a little confusing with these quantifiers

I was thinking it might be easier to negate the equation and prove that but im not sure

like can I negate it and apply the implication identity to get this?

$\forall M \in \mathbb{R}^+, \forall n_0\in\mathbb{Z}^+, \exists n \in \mathbb{Z}^+, [(n>n_0)\lor(n^4>Mn^2)]$

[vital dew]

darkninja175:

[night frost]

then prove that?

that new statement seems trivial

[sleek swallow]

You didn’t negate it properly. That or should be an and

[night frost]

Ah oops, forgot about that when distributing the negation.

makes more sense now

[stray reef]

@iron crescent do you still need help with this

[sleek swallow]

looks okay

[real aurora]

What are some decent root-solver algorithms for finding all real roots for a given polynomial? I'm working with 6-degree polynomials, so I don't need algorithms that work for insanely large degrees.

[faint narwhal]

Do you want exact solutions or what

[real aurora]

I can't solve this analytically, so no. I just need a good enough numerical method

[wild flame]

That you get the job implies that you had the best
credentials.

THis one I can't make heads or tails over. The goal of the problem is to put this in the form of "if, then".

The right answer is "If you get the job, then you had the best credentials."

But that confuses me, since it means that "if you get the job, then you did not have the best credential" is false - when logically, wouldn't that be possible?

[pale epoch]

this is a logic exercise, not a statement about real life

[wild flame]

but they use that same logic for other questions that I got right.

To be a citizen of this country, it is sufficient that you
were born in the United States.

The answer is: if you were born in the United States, then you are a citizen of this country.

The logic is as followed. It is not logically possible to be born in the US, and not be a citizen.

IE, the first statement cannot be true, while the second is false.

[stray reef]

uhh no

[wild flame]

Oh bother

[stray reef]

The logic is as followed. It is not logically possible to be born in the US, and not be a citizen.
you're trying to say that being born in the US is a NECESSARY condition for US citizenship when the sentence says it is SUFFICIENT

two very different things

[wild flame]

I don't understand that. For example, the logic stills tands. The first statement can be false, while the second is true. You could not be born in the US and still be a citizen. But the opposite is not true.

[stray reef]

the sentence "To be a citizen of this country, it is sufficient that you were born in the US." DOES NOT PRECLUDE THE POSSIBILITY of a US citizen who WASN'T born in the US.

[wild flame]

Yeah. But that's not that I said.

[stray reef]

you're saying immigrants can't be US citizens basically. so you've managed to badlogic yourself into racism.

[wild flame]

False.

[stray reef]

and yes it is.

...nvm

[wild flame]

Its not possible to be born in the US and NOT be a citizen.

I never said you had to be born in the US.

What confuses me is why that logic is valid, but its not valid for the job/credential questions.

[stray reef]

ok nvm my bad i misread what you said can you please stop flinging feces at me

[wild flame]

Its al good brehren, I still need help wrapping my head around that initial question.

[stray reef]

But that confuses me, since it means that "if you get the job, then you did not have the best credential" is false - when logically, wouldn't that be possible?
no it wouldn't. you can't have gotten the job and not had the best credentials, as per the sentence

having the job implies having the best creds

[wild flame]

Ahh. I guess that makes sense.

Was going to say. With that arrangement, it would also be possible to not have the job, while having the best credential. just felt weird.

I hate logic questions.

But i'll get there.

[stray reef]

Was going to say. With that arrangement, it would also be possible to not have the job, while having the best credential. just felt weird.
of course

having the best creds is a necessary but not sufficient condition

for having the job

not weird at all

[wild flame]

Explain the difference between necessary and sufficient if you don't mind?

[stray reef]

uh they're dual to each other lol

X is necessary for Y means that you can't have Y without X

X is sufficient for Y means that if you have X you also have Y

very simple example: being over 6' tall is sufficient, but not necessary, for being over 5' tall.

[wild flame]

that makes sense. Alright, going to do a few more problems. Ty, matey.

[dense pier]

Hello

Does anyone know if

!p<->!q the same as !(p<->q)

just wondering

[stray reef]

no

not only are these not the same, they are negations of one another

[dense pier]

Shit my bad

I mean

is !p<->!q the same as (!p<->!q)

can you distribute basically?

i guess not

[stray reef]

wha

[dense pier]

but I hope so

[stray reef]

is !p<->!q the same as (!p<->!q)
those parens are redundant as written

[dense pier]

Sorry

Gimme a sec

I mean

we have p <-> q

right

what if we have !(p<->q) is that the same as !p<->!q

That is what I wanted to ask sorry I confused myself and you

[stray reef]

that's your original question

!p<->!q the same as !(p<->q)

[dense pier]

Yeah

You said they are negations of one another

But how is that

p<->q and !(p<->q) are

[stray reef]

!p<->!q is equivalent to p<->q that's how

[dense pier]

so

they are the same?

[stray reef]

who are "they"

[dense pier]

!p<->!q and !(p<->q)

nevermind

i see my mistake

Thank you so much 🙂

Sorry for the confuson

confusion*

[weary tiger]

16 pick 6. Then count the orders of the group of 6 and the 10.

[dense pier]

Hello. Anyone knows the logic behind truth tables? What I mean is I know how to build let's say: a truth table with initially 1,2,3 propositions but if you ask me let's say p,q,r,s I mess up. So what I thought was I don't get what's the pattern. Is there one?

[weary tiger]

@iron crescent you start by picking 2 teams

You have 16 people and you want to pick 2 teams

One with 6 and one with 10

Once you have the two teams account the orders

[magic osprey]

(2-3+5+6-11+3+2)^10 ig hm

[weary tiger]

Can someone help me approach this

[orchid saddle]

How does this become this

[weary tiger]

Implication Definition, then DeMorgan's Law

[orchid saddle]

How did p -> r ^ q -> r become the stuff on the right hand

@weary tiger

[weary tiger]

Distribute

I think lmao idk man im in the same boat as u

im strugglin

[sleek swallow]

No

$p \implies q \iff \lnot{p} \lor q$

[vital dew]

Abhijeet Vats:

[sleek swallow]

That’s an equivalence that you can establish rather easily

@orchid saddle

So they literally just used that and the distributive property of the conjunction and disjunction

[weary tiger]

$\lnot{f} \implies \lnot{d} $

[vital dew]

Skwiid:

[weary tiger]

What can this become

[sleek swallow]

Skwiid, you still good with your problem?

[weary tiger]

im on a diff problem lol

[sleek swallow]

That’s the contrapositive of d implies f

[weary tiger]

The nots confuse me so much

ohh d implies f

ok

[sleek swallow]

Indeed

[weary tiger]

I think i got it actually

[sleek swallow]

So write the full argument

[weary tiger]

ima write it all down now lol

[sleek swallow]

There’s a nice way to do it btw, judging by my sleepy eyes

[weary tiger]

one more question $c \and \lnot{e} $

whats and in latex

c ^ !e

that can simplify to just c

right?

[sleek swallow]

Indeed

[weary tiger]

ok yeah i think i got it then

[sleek swallow]

Law of Simplification

Then, modus ponens

[weary tiger]

do u mind checking my work? @sleek swallow

[sleek swallow]

YES I MIND

prank

Sure

[weary tiger]

sorry bout the glare

those are all hypotheses

[sleek swallow]

Lmao are you doing an entirely formal proof

[weary tiger]

yes man

😂

is it valid

[sleek swallow]

Bruh, you can’t just use a law and not write down the actual implication for it

[weary tiger]

wdym

this is how we were taught 😦

but did i do it right? 😂

[sleek swallow]

Eh give me a while, i’ll look at it

[weary tiger]

ok ❤️

[sleek swallow]

I’m sort of getting ready for work now so give me a few minutes or so

[weary tiger]

no problem, sorry to disturb

ur the only useful resource i have 😂

[sleek swallow]

WOW I FEEL OBJECTIFIED

😄

Bruh how the fuck did ya get not(a) by modus ponens

Lmao

@weary tiger

[weary tiger]

Wat

c implies not a

[sleek swallow]

Oh wait nvm nvm

I misread stuff

[weary tiger]

ok

[sleek swallow]

Lmao dude i thought that first line was numbered 3

[weary tiger]

yeah sorry for the slopiness ive erased it over 8 times probably

[sleek swallow]

Yea looks correct

Uh idk, I don’t like the formatting cos the rules used are not explicit enough

But if your prof is fine with it, then it should be okay

[weary tiger]

Thanks man

[sleek swallow]

You’re welcome.

[half tapir]

Dw ignore that lol

[weary tiger]

y'all recommend any good resources for discrete math?

something that makes it easier to learn 😂

[sleek swallow]

I’m liking the exposition by Conradie in his Logic and Discrete Mathematics book

[weary tiger]

👍

[weary tiger]

Everyone either does not like winter or likes summer or both

How would i write this as a predicate

the grammar confuses me

[sleek swallow]

Let P be the set of people. Then, let us define the following predicates:

$A(x) :\iff \text{x does not like winter}$

$B(x) :\iff \text{x likes summer}$

Then,

$\forall x \in P: A(x) \lor B(x)$

[vital dew]

Abhijeet Vats:

[weary tiger]

i got that but the part that says "or both"

[sleek swallow]

Logical or is inclusive

[weary tiger]

so if something says and/or it implies just or

[sleek swallow]

$(p \land q) \implies p$

$p \implies (p \lor q)$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Those are tautologies

[weary tiger]

Everyone likes spring and/or winter

[sleek swallow]

Everyone likes spring or winter means that everyone likes either spring or winter or both.

In the english language, or is not necessarily inclusive. Hence, you have to build in the extra ‘and’

[weary tiger]

but inclusive or means both can be true

[sour arrow]

We have something called the "exclusive or" which is denoted XOR.
A XOR B means A or B not not both is true.

[sleek swallow]

^

In the English Language, ‘or’ is usually seen as being exclusive. That’s why you need that ‘and’ there to make it explicit that there is the third option such that the entire statement is true

[weary tiger]

but in discrete math $\lor$ is inclusive so it would be correct for and/or

[vital dew]

Skwiid:

[weary tiger]

thats what im asking lol

[sleek swallow]

Yea lol

[weary tiger]

thanks again master

[sleek swallow]

Oooh kinky

You’re welcome.

[weary tiger]

i wish u were my professor

im also learning boolean algebra in another class, when its using 0s and 1s its much easier

[sleek swallow]

Ew disgusting

There’s no reason to ever use $\varnothing$ and ${\varnothing}$

🙂

[vital dew]

Abhijeet Vats:

[sleek swallow]

In all seriousness, all this just comes with practise lol. I’ve not necessarily mastered it tbh

[weary tiger]

idk its just not clicking with me yet

maybe its the lack of numbers and regular math operators lmfao

[sleek swallow]

uh try fundamentals of mathematics by bernd schroder if you wanna understand it better

It worked very well for me, so hopefully it works well for you

The logic and discrete math book i recommended by conradie and goranko is good but the introduction and focus on logic is way more massive than in the book by schroder

[weary tiger]

thanks ill check them out

[sleek swallow]

Others have recommended how to prove it by velleman. I don't like his style so i've never read it. But it's still there to try as a book if you think it might work for you

[weary tiger]

I've read How to prove it, Book of proof, Kenneth Rosen's discrete math book and lots of other material from university notes etc

and i strongly reccomend

how to prove it

very intuitive and beginner friendly

and i also strongly recommend grinding through the exercises, that's where i learned the most

[weary tiger]

Theory without practice is useless 👍

[sleek swallow]

No u

Prank

Relax

Memes

Chill

[dapper rose]

Calm down

[sleek swallow]

No u

[dapper rose]

jkjk

[sleek swallow]

Yer mum was jkjk

Prank

[dapper rose]

No u

[sleek swallow]

What, are you 9?

[dapper rose]

I know you are but what am I

[shut fjord]

Hello everyone, I’m having an issue trying to prove the first one false:

[reef thistle]

@shut fjord As in, you want to show that there exists a P such that it is not true?

[shut fjord]

yes

sorry I got caught up with work

But we are suppose to write our reasoning in English using the General terms provided

So while a predicate example would be nice, it wouldn’t be enough to be considered a valid answer

I wrote this:

[sleek swallow]

‘But this cannot be true because there are cases where our x may have a different y’

Uh explain this a little bit more

[tired axle]

would anyone know any discrete math areas/problems (such as the Chinese Postman problem) that would be interesting to explore for a first year uni project

[weary tiger]

you looking for open problems or ones that have been already solved?

[wild flame]

Hng.

https://i.imgur.com/NHEt6WG.png

I thought propositions only became "if and only if"

if the conditions were both necessary AND sufficient

[stray reef]

what's p and what's r and what's with this insistence on not giving things intuitive names

"to get an A in this class, it is necessary for you to get an A on the final" doesn't translate to "A in class <-> A in final"

[wild flame]

Sec.

https://i.imgur.com/w2uTLZI.png

@stray reef
That wasn't my answer. That's the answer I got from chegg (since they don't give the answers of evens in the book)

[stray reef]

what's the proposition you were asked to write again

"to get an A in this class, it is necessary for you to get an A on the final"?

[wild flame]

To get an A in this class, it is necessary for you to get
an A on the final.

Yes

would the proper proposition be.

P --- > R?

[stray reef]

no

[wild flame]

Wait no.

P <--- R

[stray reef]

er

wait

okay first off

math is case-sensitive

A in class -> A in final

r -> p

[wild flame]

Yeah. But its definitely not a bi-conditional no?

Guess this is a case of chegg being wrong

[stray reef]

i can assure you that it's not a bi-conditional.

[wild flame]

Fucking chegg.

[stray reef]

don't trust everything you see online lol

[tacit marlin]

Can somebody explain this to me?

I don't understand why we now have (n+1)(n+1)! on the right side

When I look at the question, for my induction step I want to do
= (n+1+1)! - 1 but thats not what they do

[sleek swallow]

Your last step involves you manipulating the left-hand side in order to obtain something on the right-hand side that would make the entire statement equivalent to P(n+1)

So, you want to get (n+1+1)!-1 on the right-hand side but you gotta do it by manipulation of the left-hand side.

[tacit marlin]

is the reason they added (n+1)(n+1)! to the right side was because it was added to the left side when we were doing the n to n+1 step?

[sleek swallow]

Well, think of what P(n+1) is:

$P(n+1) \iff \sum_{k = 1}^{n+1} k.k! = [(n+1)+1]!-1$

In other to get that left-hand side, you need to add (n+1).(n+1)!

[vital dew]

Abhijeet Vats:

[obtuse phoenix]

Could somebody explain this to me? I originally thought it was a Russell’s paradox question but now I’m unsure

[ivory badge]

Well, what makes you think it isn’t a set @obtuse phoenix

Or, if you think it is a set, why?

[obtuse phoenix]

I was thinking it is a set because x is not an element of itself but it is stated that x is an element of s making it part of set S im pretty sure

Which would make T = s?

[ivory badge]

@obtuse phoenix Well, do you have the axiom (schema) of specification?

[errant bear]

proving a | b iff ma | mb is like 2 lines max, right

like its super basic, right

[faint narwhal]

yes

[errant bear]

ok, wanted to make sure i wasnt missing anything obv

[weary tiger]

Everyone who likes to hike has hiked with someone else

$\forall x \exists y (H(x) \implies (T(x,y) \land (x \neq y)) $

[vital dew]

Skwiid:

[weary tiger]

is this correct?

@sleek swallow Could u check this if you are free? ❤️

[barren apex]

Quick question with Big O notation. Do higher big O's hold true if a smaller big O holds true? In other words, if f(x) = O(x^2) is it also true that f(x) = O(x^3)?

Speaking purely in a mathematical definition and not computer science intuition (time complexity)

[faint narwhal]

Yes

If your big O's are as your function goes to infinity

[barren apex]

Cool, thanks

[weary tiger]

can someone check that problem i sent

just need some confirmation 😂

[sleek swallow]

Fucking hell i just woke up shit

Uh yea i think it looks okay

[weary tiger]

sorry 😂

[sleek swallow]

You don’t need to state x neq y, cos it’s already a part of the definition of T(x,y)

[weary tiger]

ahh thats the part i was confused about

thank you

oops

$ \forall x \exists y(\lnot{T(x,Tom) \land (x \neq Tom)) $

[vital dew]

Skwiid:

$ \forall x \exists y(\lnot{T(x,Tom) \land (x \neq Tom)) $
```Compile error! Output:

! Missing } inserted.
<inserted text>
}
l.54 ...sts y(\lnot{T(x,Tom) \land (x \neq Tom)) $

I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.

[weary tiger]

this correct too? or do i not need (x != Tom)

Nobody has hiked with Tom

[sleek swallow]

Uh looks okay but why is there an ‘there exists y’

[weary tiger]

oh true

[flat depot]

hey I am learning about binary heaps at https://en.wikipedia.org/wiki/Binary_heap#Building_a_heap and ran into a statement that doesn't seem to be true.

The only way this could be true is if it means the number of leaves at height h, rather than number of nodes.

[weary tiger]

u learn about the math behind heaps in discrete math?

[flat depot]

No but a google search indicated that binary trees are in discrete math

[weary tiger]

ah

[flat depot]

so I think this is the right channel

[weary tiger]

i remember learning about em in data structures but not the math stuff about em 😂

[sleek swallow]

No u

[flat depot]

I have a technical interview tomorrow so I'm refreshing myself about the math and I ran into this issue

[weary tiger]

ahh good luck man, ive applied to 46 internships and got no reply to any of them yet 😦

[flat depot]

well this is the last thing that I don't understand so if I figure it out I should be good to go

the only thing that makes sense is if it means leaves when it says nodes

[sleek swallow]

Oh fuck 46 internships? Wtf you’re in uni now right? Where are ya studying?

Jesus that’s a lot

[weary tiger]

a university in seattle lmao

im a sophomore tho

[flat depot]

should I repost in a more advanced channel?

[weary tiger]

maybe try a comp sci discord?

[flat depot]

fair enough

[faint narwhal]

Uh I think you might be misunderstanding what height means

@flat depot

[flat depot]

@faint narwhal yeah I actually figured it out. I thought it meant number of nodes in a tree of height h

but it meant number of nodes at height h. definitely my mistake

[blissful zenith]

how do i write the real numbers in terms of their decimal expansions if we Let f: R->N be defined by: f(x) = the third digit of x after the decimal point

[soft thorn]

Logarithms probably

Logarithms and ceil/floor

[blissful zenith]

ceil/floor?

any1

[soft thorn]

I mean

What even is the question

Do you actually even need to find a formula for the function?

[blissful zenith]

yes

[analog dagger]

for what

[soft thorn]

Why?

[blissful zenith]

write the real numbers in terms of their decimal expansins. Let f: R->N be defined nbyu f(x) equals the third digit of x after the decimal point

@analog dagger

[soft thorn]

Just use logs

[blissful zenith]

how

[soft thorn]

Actually

Let me think of something else

I may of been thinking of something else

But really

What is the question as it is written in your book?

[blissful zenith]

@soft thorn that is exactly the question

write the real numbers in terms of their decimal expansions. As usual, we do not allow a real number to end in all 9's repeating. if we Let f: R->N be defined by: f(x) = the third digit of x after the decimal point (this is called the michelle function)

i tried to find the michelle function online i could not find it

[soft thorn]

That doesn't really sound like a full question though

[blissful zenith]

that is exactly the question i copy pasted it

@soft thorn

write the real numbers in terms of their decimal expansions

[soft thorn]

You can just multiply the number be 1000

Then mod 10

[blissful zenith]

@soft thorn what do you mean

[soft thorn]

Multiplying the number by 1000 pushes the third digit after the decimal point to the ones place right?

[dapper rose]

btw f isn't well defined

nvm

[blissful zenith]

yes but how is it written out

the formula

[soft thorn]

Is there even a formula for this

[dapper rose]

what, you suggested f(x)=(1000x)%10

[soft thorn]

Yeah

That's the best I can do

Only problem being calculating mod 10 is not necessarily easy

But

That function does work

[haughty herald]

Hi could someone help me with this? I just cant figure out if its partial injection or partial bijection

[dapper rose]

probably f(x)=floor(1000x)%10+(sgn(x)-1)/2 is better

[blissful zenith]

1000x mod10

what is floor?

[soft thorn]

Ahh signs

Didn't think of negatives

Yeah, you're right

[dapper rose]

@blissful zenith round down

[blissful zenith]

what is sgn?

[soft thorn]

Signum

Basically just returns 1 if positive

-1 if negative

[dapper rose]

0 if 0

[soft thorn]

And I think 0 for 0?

Yeah ok

[dapper rose]

so my shit is bad for x=0

.-.-.-.

[soft thorn]

Meh

[blissful zenith]

wouldnt 1000xmod10 work

[soft thorn]

Piecewise define the function lmao

And throw the 0 case into a different branch

[dapper rose]

@blissful zenith well for say x=0.0111 we get 1000x%10 = 11.1%10 which I think is 1.1

[soft thorn]

Ah yeah, I forgot to floor

Rip

[dapper rose]

and uh i don't think this answers the original question

[wild flame]

I swear to god, Logic is the most infuriating thing

[sleek swallow]

It's fun lol

[loud basalt]

i think hes talking about the rapper ./s

[sleek swallow]

I like Logic and Discrete Mathematics by Conradie and Goranko

So you can use that if you want

It has a stronger focus on logic, though, since it has been written by Logicians

[weary tiger]

After 44 applications I got a phone interview 🙏

[still nexus]

why is this d and not b?

[gleaming zephyr]

if A then B

A-->B