discrete-math

[errant bear]

so i can stay away from them

[faint narwhal]

A function is probably the most basic topic in all of math

There definitely isn't a single part of math that doesn't use functions

[errant bear]

i mean like this

[faint narwhal]

Same response

[errant bear]

bruh

[faint narwhal]

Most use them in way, way, way more complicated ways

[errant bear]

stats or accounting it is then (?)

[faint narwhal]

Nope

[errant bear]

F

[faint narwhal]

Stats is full of super complicated functions

And using these ideas to create new functions

[errant bear]

well yeah but do i really have to care if its bijective/whatever

[faint narwhal]

Oh yeah

[errant bear]

...

time to major in physics then

[faint narwhal]

lmao

That's even worse than stats

[errant bear]

welp

like what does proving these problems do, like whats the application

[faint narwhal]

Because you can use these ideas

Real world applications?

[errant bear]

i mean i guess? like why does a statistician care if a function is bijective, per say

and is not like they would create the functions by hand, it would be done by a computer

so you dont need to worry about invertability

[faint narwhal]

Because most of the time a statistician doesn't work with concrete numbers

They work with these same abstract ideas

Consider situations in which they have some data and they perform tests on them

[errant bear]

ok

[faint narwhal]

And they need to figure out situations in which this test will tell them certain information

You really shouldn't give up so easily

[errant bear]

it just seems like im doing all these proofs and what not of problems with no application, and everything is arbitrary so you dont have an example to base it off/ visualize

like all of this fancy notation and using sets, and ordered pairs to just say f(x)=x

ok rant over

[faint narwhal]

I mean you know that math and physics is all about abstraction

We need rigor to be able to do these things

To be able to rigorously talk about what a lorentzian manifold even is

Or how quantum states really works

We need to rigorously define things starting from the beginning

[errant bear]

i understand that this class is like building the foundations for upper maths

but like

idk

half the stuff just seems so extra

[faint narwhal]

Well, inverses of functions are important

You can see why that's true

And injectivity is basically the condition that you need for inverses to exist

[errant bear]

yeah

but like what function doesnt have an inverse

in the sense that it cant be broken up

like just use a piecewise or something for x^2

[faint narwhal]

No functions

But how do you break it up

For your example

You're basically going to say that 2 and -2 are the same thing, since they both map to 4

[errant bear]

hmm

i guess, but why does having 2 possibiliteis matter

[faint narwhal]

Oh I'm not saying it matters

we could have however many possibilities

The point is that

This is exactly an equivalence relation

You define an equivalence relation on your function such that two elements of the domain are equivalent if they map to the same thing here

so 2 and -2 are equivalent

Then, you can define a function from this set of equivalence classes to your codomain

And it will be injective

Sorry, I don't know if you know what an equivalence relation is

[errant bear]

yeah sym reflex trans

[faint narwhal]

Right

And this captures your intuitive idea of being able to break up a function so that it is injective

[errant bear]

so we need bijectivity (injectivity) as a property so we dont claim that 2 = -2

[faint narwhal]

Not sure what you're saying

[errant bear]

and these propertys are applied to other, higher level maths?

uh

oh nvm i see what u are saying

i didnt see the eq class

[faint narwhal]

Yeah no this idea is super super super super important

Like this simple idea might honestly be one of the most important ideas in math

And I'm really not exaggerating, so many theorems are just more complex versions of this idea

[errant bear]

ugh

im sad now

and i have to write a 5 page hist essay by noon tomorrow

thanks for talking and helping

[faint narwhal]

Good luck

If it helps

I was kind of in your situation

Well

In the sense that I got pretty stuck on some simple ideas that I was trying to learn

It also got me pretty disappointed and got me thinking that I couldn't do math

But I perservered and things started to click

Ideas started coming together and math is so interconnected that areas of math started coming together

[errant bear]

hopefully thats whatl happen

anyways im off to the gym gotta take my mind off school

[cursive lily]

so i kinda get it

but not sure how to display it

[mint salmon]

I'm trying to figure out how to go about this problem. Let A, B, C be sets. Prove or disprove this statement: A ∪ (B \ C) = (A ∪ B) \ C

[faint narwhal]

Well do you think it's true or not?

[static ivy]

try 'proving' it and see if any problems occur

according to the problems that might or might not occur you will maybe get an idea of a counter example that touches the problematic aspect of this statement

@mint salmon

[mint salmon]

I was able to sit down with a colleague who is better at it than I am

you can actually easily show that the two are equivalent if you write each piece of the statement in the form of its formal definition

[static ivy]

oh

interesting

i actually thought it was false at first glance

but now that u say that im gonna have to write it down

[faint narwhal]

wait

Uh

This is not true

[static ivy]

i think if u take {1} , {2} , {1}

as A B C, it wont work

[faint narwhal]

@mint salmon what was your proof?

[mint salmon]

1) A ∪ (B \ C) ⇔ a ∈ A ∨ a ∈ (B \ C) ⇔a ∈ A ∨ a ∈ B ∧ a ∉ C ⇔(a ∈ A ∨ a ∈ B) ∧ a ∉ C 2) (A ∪ B) \ C ⇔ (a ∈ A ∨ a ∈ B) ∧ a ∉ C

[faint narwhal]

Uh

Be careful with parentheses

[static ivy]

that aint it, chief

[faint narwhal]

$a \in A\lor (a \in B \land a \nin C)$

[vital dew]

Zopherus:

[faint narwhal]

is different from $(a \in A \lor a \in B) \land a \nin C$

[vital dew]

Zopherus:

[mint salmon]

well I guess that's what I get for jumping to conclusions

😂

how about taking the very first statement and using de Morgan's law on it

[faint narwhal]

Sure, what do you get

[mint salmon]

well I'm not yet entirely sure how to do it, but this is what I tried: ⇔ a ∈ A ∨ (a ∈ B ∧ a ∉ C) ⇔ a ∈ A ∧ (a ∉ B ∨ a ∈ C)

[faint narwhal]

Keep track of those parentheses

[mint salmon]

ok one question: when I use de Morgan's law, do I also need to flip the and/or that is outside the parentheses?

and yeah, I'll put the parentheses back in

[faint narwhal]

That's really not what demorgan's law says

[mint salmon]

I'll have to come back to this later, and I'll make sure to read up more on de Morgan's law first

[hollow osprey]

Quick question, what are the basic steps to finding how many numbers 2000 is divisible by the number 3? Is it like 2000/3 = 666 (I removed the decimals)

Idk what to do after that

[analog sonnet]

Yep, you're doing it right

Even the rounding-down part

[hollow osprey]

That's it?

[analog sonnet]

Yep, that's it

Unless you count in the number 0 too

Because technically 0 is divisible by any integer (except 0) lol

In that case you should round up instead of down to account for this one extra case

[hollow osprey]

Ok now to counting methods, this is an example that my professor did, but I don't quite understand it

I know that the alphabet has 26 letters, that's about it

oh, OR means to add

[alpine goblet]

@analog sonnet I’m so sorry Mann

[analog sonnet]

How many 2-symbol combinations are possible, where the 1st symbol is a letter and the 2nd symbol can be either a letter or a digit?

[alpine goblet]

But can you also help me

Whenever you’re free 😭

[analog sonnet]

Is it discrete math related?

[hollow osprey]

Does that basically mean like a0,a1,a2..a9...z9?

[analog sonnet]

yes

but also aa, ab, ac, etc.

[hollow osprey]

Hmm I would assume it would be a lot

[analog sonnet]

Yep

But how many exactly?

[hollow osprey]

So we have 26 letters * 10 digits = (I haven't calculated this part yet)

Right?

[analog sonnet]

the 2nd symbol can be one of the 10 digits, but it can also be one of the 26 letters, so there are actually 36 possibilities for the 2nd symbol

[hollow osprey]

Oh wait so letter is 26, because it starts with a letter

The second symbol is a letter AND digit?

[analog sonnet]

letter or digit

can't be both at the same time, can it? Lol

[hollow osprey]

Oh noo

So thats how the 36 got there

[analog sonnet]

yeah

36 = 26 + 10

[hollow osprey]

26 = letter
10 = digits

when you add up the symbols you get 36

[analog sonnet]

[static ivy]

26 * 36

26 options for the first slot

36 for the second

[analog sonnet]

why are you in this channel @static ivy aren't you scared of combinatorics

[static ivy]

im trying to face my fears

😦

[analog sonnet]

Very brave

Godspeed

[hollow osprey]

So for 3 symbols, the 3rd symbol could be a letter or digit?

[analog sonnet]

yep

[alpine goblet]

Yeah

[hollow osprey]

So basically from 2 symbols and onwards, 26 will be always fixed, then * 36 by how many symbols there are?

[alpine goblet]

It is

I posted In the other chat

[analog sonnet]

Yes

[hollow osprey]

Ok thanks I understand it now

[analog sonnet]

np 🦈

[alpine goblet]

It’s just part 2

[analog sonnet]

My intuitive guess is that it's around about n/2

[alpine goblet]

Why that

If there’s 10 nodes

Then the average hop distance between two nodes is 5?

[analog sonnet]

Hmm you're right, I might be wrong about my guess

[alpine goblet]

What lol

Yeah

[analog sonnet]

Maybe even less than n/2, it could be closer to 3n/8

[alpine goblet]

But why though

[analog sonnet]

But I don't really have much time to calculate it now

Well, you could consider the extreme cases

If you're an endpoint node, your hop average is around n/2

If you're a midpoint node, your hop average is around n/4

it's gotta be somewhere between n/4 and n/2

But you could probably calculate this quite easily by exploiting some symmetries

[alpine goblet]

I don’t understand what it means by average hop distance

If you can explain that

I’ll prob get it

[quaint lichen]

would cardinality of union of mutually exlusive sets X and Y, | X U Y |
be |{null}|=1 or |null|=0?

[pale epoch]

why should it be either

[runic mauve]

Can anyone help me with this problem? My professor got it from a Bulgarian number theory textbook and he gave it to us for homework and the tutors have been working on it for days and can’t figure it out

I’m going into an exam in the next few minutes so I’ll have to read any responses in a bit, also we do not know induction yet just btw

[faint narwhal]

Think about the gcd instead

[runic mauve]

The gcd is always 1

[faint narwhal]

Are you sure about that?

Try it out for some n

[runic mauve]

They’re 4 consecutive numbers

Consecutive numbers are always coprime

[faint narwhal]

Then you can do it one at a time

The least common multiple of the first two is always n(n+1)

Then calculate the least common multiple of this and n + 2

[clever heath]

I'm really confused about this practice question and how to anwser it?

[runic mauve]

@faint narwhal that’s not workin out for us. We’ve got tutors here who have been working on this problem for hours to no avail so idk man lmao

[faint narwhal]

What's the gcd of n(n+1) and n+2?

[wintry rock]

I'm having a bit of trouble with this one question...

a1 = 7a0 + 2^n
a2 = 7a1 + 2^n = 7 (7a0 + 2^n) = 7^2 a0 + 7 * 2^n
a3 = 7a2 + 2^n = 7 (7^2 a0 + 7 * 2^n) + 2^n = 7^3 a0 + 7^2 * 7*2^n

an = 7 (an-1) + 2^n = 7 ( 7^n a0 + 7^(n - 1) 7^(n - 2) (2^n) + . . . + 2^n ) + 2^n

Now put a0 = 3

7^(n + 1) * 3 + 7^n * 7^(n - 1) * 7 * 2^n + . . . 7 * 2^n + 2^n (I distributed the extra 7 here)

that's what I have so far but not sure if I'm on the right track

[runic mauve]

@faint narwhal we already tried it that way, got stuck in a cf of unsolvable algebra, if you can find a solution that way that’s cool but I’m pretty sure the actual proof is mostly English. The professor has trouble solving problems of this sort and gave them to us expecting nothing. When I said he got it from a Bulgarian number theory book, that was not an exaggeration lol. The 3 problems that he gave us as homework can be thought of as extra credit worth 4 times a regular problem and many many times more difficult.

[last sigil]

What problem are you discussing?

[faint narwhal]

@last sigil it's just finding the lcm of n, n+1, n+2, n+3

@runic mauve you can do it this way, it's pretty straightforward

Just answer my question, what's the gcd of n(n+1) and n+2

[faint narwhal]

The final answer is either n(n+1)(n+2)(n+3) divided by 2 or 6 depending on the cases as described in the hint

[slender skiff]

If P(a) = 0,6 and P(b) = 0,4 and P( A anb B) = 0,2 would P(A or B) give 0.8 right?

[stray reef]

would it?

[slender skiff]

yea since 0,6 + 0,4 - 0,2 is 0,8

but the solutions to this exercise says its 1

[stray reef]

can you show the exercise exactly as stated

and its solution

[stray reef]

it's $P(B^c) = 0.4$ not $P(B) = 0.4$

[vital dew]

Ann:

[slender skiff]

This guy go the meaning of B^c wong

[stray reef]

$P(B) = 1 - P(B^c) = 0.6$

[vital dew]

Ann:

[slender skiff]

ohh

i see now

[stray reef]

$P(A \cup B) = P(A)+P(B)-P(AB) = 0.6 + 0.6 - 0.2 = 1$

[slender skiff]

its P(B)

[vital dew]

Ann:

[slender skiff]

okay i got it Thanks!

[floral wind]

Question,

Let A = {
k
k+1 |k ∈ N}. For each of the following, state whether it exists,
and if it does, determine its value:
inf A, max A, min A,sup A.

been a while since i did sets

k / (k + 1) btw

I don't have solutions but i kinda wanna verify im right

inf -> 0.5
max -> 1
min -> 0.5
sup -> 1

also question
p -> q
is only false if 0 -> 1 right?

and p <-> q only if they're not the same?

[stray reef]

uh

$A = \left{ \frac{k}{k+1} \mid k \in \bN \right}$

[vital dew]

Ann:

[stray reef]

this?

@floral wind

[floral wind]

yes

[stray reef]

you claim max(A) = 1

is 1 in A at all

[floral wind]

well if there's a sup

then isn't max the same?

or was it the other way around

if it's the other way around then max does't exist

cause it approaches 1

Also i have another question after those 2 ^ lol

p -> q
is only false if 0 -> 1 right?
and p <-> q only if they're not the same?

sorry for the spam

[stray reef]

no 0->1 is true

1->0 isn't

anyway no just because sup exists doesn't mean max does

but if max exists then sup = max

[floral wind]

yeah ok

how about <->

[stray reef]

yes

[floral wind]

only if they're not the same, cool

last question for right now haha

4. Consider the set A of numbers consisting of 4 digits, where each digit is from the
   set {1, 2, 3}. For example 1311 ∈ A

Odd numbers in A:
(2/3) * 3 ^ 4

right?

And then numbers that are either odd, or start with a 1 -> 1/3 * (3 ^ 4) + 54 (answer from prev question)

but what if thye're both odd and start with 1, then it shouldn't count towards the total right

cuz this would just mean that ALL the numbers are either odd or start with a 1

is it like 1/3 * 1/3 * 81 + 54? (2/3 of those numbers also end with a 3 or 1)

[floral wind]

?

[finite portal]

I want to prove the following subset of R is closed under addition: [5, inf) . Is this proof correct? Let x,y be arbitrary elements of [5,inf). This means 5<=x<inf and 5<=y<inf. Add the three expressions in 5<=x<inf by y to obtain 5+y <= x+y < inf. Since we know y<inf then we can conclude [5, inf) is closed under addition.

[reef thistle]

Let's not <inf

[finite portal]

?

[reef thistle]

sum of two real numbers is another real number

[finite portal]

agreed

then, does my proof work?

[stray reef]

rewrite it

[finite portal]

Want to show if x,y in [5, inf) => x+y in [5,inf). Let x,y be arbitrary elements of [5,inf). This means 5<=x<inf and 5<=y<inf. Then 5+y <= x+y < inf. Hence x+y in [5, inf)

since y < inf & x < inf and since 5 < 5+y

Does that re-written version work?

[stray reef]

Let's not <inf

and how do we know 5+y <= x+y?

or rather

how can we conclude x+y in [5,inf) from that?

[finite portal]

so we have 5 <= x and y <=5 where x,y in [5, inf). then 5<=x -> 5+y <= x+y since we have 5 < 10 <= y+5 <= x+y ?

so 5 < 10 <= x+y

[stray reef]

y <= 5?

[finite portal]

y in [5, inf) so y<=5

we have x,y in [5, inf), thus 5 <= x and y <=5. Then 5<=x implies 5<10<=5+y<=x+y, so 5< 10 <= x+y

[stray reef]

y in [5, inf) so y<=5

read that again

[finite portal]

oops

5<=y

GOOD catch!

re-writing...

y in [5, inf) so y<=5
we have x,y in [5, inf), thus 5 <= x and 5 <= y. Then 5<=x implies 5<10<=5+y<=x+y, so 5 < 10 <= x+y

[stray reef]

first line

but other than that this is ok

[finite portal]

sorry my brain is being weird mixing them up

ty

one question about that proof though, that convinces me that x+y is within the "lower bound" part of the set (don't know how to precisely say this) but doesn't convince me necessarily that it doesn't "go outside of the upper bound" part of the set (i.e. the inf part). how do I know not only is x+y in contained in the "lower parts" of the set but that also x+y isn't greater than infinity and thus goes outside the "upper bound" parts of the set? i know this is completely obvious but i want to be rigorous about it

[stray reef]

what upper bound

there is no upper bound

the right end of your interval isn't a number, it's infinity

[finite portal]

ahh i see

okay, that makes sense.

ty again

[real perch]

can someone tell me how this inverse mod works? i try to search some youtube but it make me more confused instead.

im fine with the whole top thingy

but the last

i get the answer s= -1839

but the inverse modulo whack my brain so hard

Helppp

[floral wind]

Anyone?

[analog sonnet]

3961 = 5800 - 1839

[real perch]

@analog sonnet but how? i just minus the Ø will do? In the rsa encryption case?

i mean add

[analog sonnet]

I'm not sure what that question means

[real perch]

Its this question

which is an encrypt/decrypt type

[floral wind]

Uhm I don't quite get this

can someone explain what's going on

[finite portal]

question about closure, say you have two sets: f(x)=3k, k in N for the set {3,6,9,12,...} which is closed under addition and a function f(x)=x^2, x in N for the set {1,4,9,..} that is not closed under addition -- what is going on a higher level here that makes the first set closed and the second set not closed? is there some sort of higher level property that describes this observation?

[robust mango]

@finite portal In the second set, lets say you take x=1. you get f(x)=1. If you take x=4, you get f(x)=16. if you add them both, that equals 17. But that's not in our set

[finite portal]

@robust mango -- indeed that is the case. is there a property we can abstract away about functions to say something about them wrt closure under addition in general?

or do you have to look at them individually to determine

[robust mango]

depends on the function I think.

[finite portal]

ok

[robust mango]

In the first set, the domain is multiples of 3. And the function itself multiplies again by 3.

And we know if we take any multiple of 3, add it to another multiple of 3, it's still a multiple of 3.

But in the second function, it's not necessary that we get a number which is a perfect square.

So in the first one, we don't have to take any numbers individually, but it's evident in the second one.

I think that is how it would be done, I'm not an expert.

[finite portal]

okay, what i was thinking ty

[robust mango]

I tried though! 🙂

[finite portal]

yup

ty

[robust mango]

Actually in the second one

you do get a perfect square, but if you add two perfect squares, then it's not necessary that the result is also a perfect square

[finite portal]

makes sense

eg 5 isnt a perfect square

[robust mango]

to have 5, we put square root of 5 in the second function. but we can't, it's not in the domain.

but you get the idea

[finite portal]

1^2 + 2^2 = 1 + 4 = 5, which isn't in the set, and yeah

[robust mango]

1^2 is okay, but you can't do 2^2

because x=2 wasn't in the set

you can try x=9.

[finite portal]

oh you're right

well

[robust mango]

so 1^2+9^2=82

now 82 isn't a perfect square

[finite portal]

the set is S={x^2| x in N}

sorry

brb

[robust mango]

function f(x)=x^2, x in N for the set {1,4,9,..}

[finite portal]

I meant to say the set is {1,4,9,16,25,...}

[robust mango]

yeah i get it

[finite portal]

f(x)=x^2, x in N

1^2 + 2^2 isn't in S, is my counterexample

oh i see what youre saying

[robust mango]

wait

[finite portal]

nvm

[robust mango]

{1,4,9,16..}, that's the domain right?

[finite portal]

thats the set

[robust mango]

oh

[finite portal]

generated by f(x)=x^2

x in N

[robust mango]

i see

[finite portal]

i was saying that set isnt closed under addition

[robust mango]

then yes, your example was right

[finite portal]

and counterexample yeah

k

there are things that would be in it, eg, 3^2 + 4^2

but i just need 1 counter example

[robust mango]

yes but we just made one

1^2+2^2 = 5

which is not in the set

[finite portal]

agreed

which is why it isn't closed under addition

[robust mango]

yes

[finite portal]

k, ty

[robust mango]

np

[atomic sapphire]

Guys quick question, are there 26+26^2+26^3 ways to arrange 3 lowercase letters?

[robust mango]

3 lower case letters?

[atomic sapphire]

Like making 3 letter words, but only using lowercase letters

I may have to ask this question later I don't have the math book with me right now and I don't think I'm asking the question right

[robust mango]

whats the difference between any letter and lowercase letter

i mean there's still 26 options right?

or am I missing something?

[atomic sapphire]

Including the lowercase letter and uppercase letter would be 52 options

[robust mango]

i see

so you need 3 letters

and repetition is allowed?

[atomic sapphire]

Yes

[robust mango]

so that's 26^3

26 x 26 x 26

since repetition is allowed, and we hve 26 options for lower case

[atomic sapphire]

Thanks

Oh I just realized why my other answer was wrong

I included the case for 1 letter and 2 letters also

But the question only asked for 3 letters

[faint narwhal]

What did you try?

why doesn't the line work?

Why doesn't the equation of the line work

Okay, well can you find the equation of that line?

That's the equation of a line yes

[south snow]

hey if I have a karnaugh map, which is the appropriate representation?

[faint narwhal]

You can't have both of those circles

blocks cannot overlap

[south snow]

this is from my TB

lol

[faint narwhal]

Hm idk the way I learned it is that they can't overlap but maybe I'm misremembering

For your circles at least

It would be the first one

[south snow]

We were told that the circles can't be contained within others, but overlap is fine

thanks!

[faint narwhal]

Yeah, I think that is right, I misremembered

In fact, you want them to overlap to mae the circles as big as possible

[dull tinsel]

Hey guys, I need help figuring out whether a directed graph with indegrees and outdegrees can exist.

I know that for simple graphs, you add all the degrees up. If the sum is odd, then the simple graph cannot be made, and if the sum is even, the simple graph can be made.

But what do I do for direct graphs (with indegrees and outdegrees) to see if they can be made or if such a graph can't exist?

[mint salmon]

is this a good place to ask about boolean algebra if I get stuck?

in my lectures yesterday there was quite a bit of overlap between the boolean algebra in computer systems and the formal logic in math

[stray reef]

sure

[mint salmon]

ok cool

[hollow osprey]

Would a) be 1800 different meals?

I just multiplied all of the numbers together

[stray reef]

yes

[hollow osprey]

For b) then it would be 90 dishes?

I multiplied everything together except 20

[stray reef]

,,, no

||you can view (b) as adding another option, "none", to the soups, appetizers and desserts||

[hollow osprey]

So this would be a lot of "or" options

20

• 20 * 5
• 20 * 6
• 20 * 3
• 20 * 5 * 6
• 20 * 5 * 3
• 20 * 6 * 3
• 20 * 5 * 3 * 6
  I am thinking it's this

[weary tiger]

Yeah precisely, you have 20 main dishes every time so you just permute the rest of the options, and the number of possibilities add

[sour arrow]

No, like now you have 6 soups, 7 apps, 20 mains, 4 desserts

because "none" is one of the options for soups, apps, and desserts

[weary tiger]

Why'd you say no lol both are correct

[sour arrow]

Fair, didn't follow the logic

[hollow osprey]

I just listed them like subsets then added them

[sour arrow]

I think Sir missed the simpler solution

[hollow osprey]

Kaynex could you explain how you did that?

[sour arrow]

It's the exact same as part a, except you have one more soup, app, and dessert

Because "none" is that extra option

[hollow osprey]

Oh so it adds an option to them

But how did you know there was an added "none" option?

"Don't have to have anything" means to add a "none" option?

[sour arrow]

You have an extra choice, and that choice is not to have one

You can still get a soup and dessert, but not an app, for example

I see your logic now, just split it into cases where someone might pick none

Still the right answer

[hollow osprey]

Ah I understand your way of doing it now

It's just adding an option

Ok 2 more questions, I know that (a) would be 26^3 ways but I have no idea what distinct means in (b)

[weary tiger]

Yeah that's just kinda poor wording. I would assume distinct within the word, so 'a' wouldn't be distinct in 'aba', but it would be in 'abc'

[gentle nova]

then its just 26c3

[weary tiger]

They didn't say pairwise or otherwise so it's vague

[hollow osprey]

So we basically can't use the same letters in a word?

[gentle nova]

or n!/k!(n-k)!

[sour arrow]

"distinct" means "never used multiple times*

[gentle nova]

which is 26!/3!(26-3)!

[hollow osprey]

The 3 is for the 3 letter cases?

[weary tiger]

why are you just spoiling the answer without letting 'em think 😂

[hollow osprey]

Ikr I was going to attempt to use the formula 😦

[gentle nova]

because 26 choose 3 should be the very first thing you learn in counting

or n choose k

[sour arrow]

It's fine to have the answer, much better to have the reasoning

[weary tiger]

Guide them to it then yeah

[sour arrow]

Do you know what the nCr function is?

@hollow osprey

[hollow osprey]

Kinda, It's like out of n numbers you choose r numbers right?

[stray reef]

So this would be a lot of "or" options
20
+ 20 * 5
+ 20 * 6
+ 20 * 3
+ 20 * 5 * 6
+ 20 * 5 * 3
+ 20 * 6 * 3
+ 20 * 5 * 3 * 6
I am thinking it's this

you're overthinking it

[sour arrow]

Yeah. Let's say you have a pile of 26 things, and you pull 3 of them out to make a new pile

There's 26C3 ways your new pile can look

[stray reef]

also if we're doing 6b

then it's not 26C3

26C3 would be the number of ways to pick 3 letters without regard for their order

[gentle nova]

super helpful principle and one that you need to master if you continue on in discrete math

its used in statistics as well

[sour arrow]

Anyway, yes nCr counts the number of ways your new pile will look without considering the order you made the pile

I think you'd agree that abc is not the same word as cba

[hollow osprey]

Yes

[stray reef]

$C_n^r = \left|{X \subseteq {1, 2, ..., n} : |X| = r}\right|$

[vital dew]

Ann:

[sour arrow]

So the order is important. In that case, nCr is the wrong function.

[hollow osprey]

Now we have to account for the distinct letters right?

[sour arrow]

26P3 is the number of ways you can take from 26 letters, and make an ordered list of 3 of them

This is considering distinct letters. Once we take a letter from the pile, it's no longer in the pile; we can't take it again

[stray reef]

i'd rather write it as 26 * 25 * 24

nPr is a lie

[sour arrow]

Oh yeah, it might be easier to explain that way

[stray reef]

26 options for the first letter, 25 for the second, 24 for the third

[sour arrow]

Where if you can keep using the same letters over and over again, it's 26*26*26

[hollow osprey]

Oh so you like remove an option for the ones after the first letter

[stray reef]

well,

yes,

[sour arrow]

If you use c as the first letter, c is no longer an option for the second letter

[hollow osprey]

Ohh I get it now, for the third letter it's 24 letters because the first and second letters are used up

Which means for the third one, you have 2 less letters

I really need to go to sleep now but from what I understand so far, 26/3 is what you keep because you are choosing 3 letters out of 26 letters to make a word, then the next step is to account for the distinct letters which is 26/3(26-3) where the (26-3) is so the letters are distinct

[stray reef]

26/3?

[hollow osprey]

Sorry I mean 26C3

I need to get used to this notation because I keep thinking it of a fraction because it's a number on top of each other

Thx for the help everyone and good night

[static ivy]

dam it, im late

[mint salmon]

I'm having trouble figuring out what I can and can't put together in boolean algebra

since AND and OR are commutative and distributive, does it mean that parentheses don't matter regardless of the operator? no right?

if I have ('x2 * x0 + x2 * 'x2 + x1 + 'x0) can I just add parentheses like this: ('x2 * x0 + (x2 * 'x2) + x1 + 'x0)?

and then get rid of that term because NOT x2 AND x2 is false?

[pale epoch]

usually its agreed that AND binds stronger than OR

just like with multiplication/addition in the integers

[mint salmon]

so it would be possible to reduce in that way

but now I see that there's actually a term of the form A + 'AB in that term

and that reduces further to A + B

the hardest part is knowing what to write

this course doesn't require strict proofs, but on the other hand, I don't know if I can even write a simple proof for this problem

[pale epoch]

if you reduce a logical expression to some other expression with only logical equivalence signs between them, it is a proof

(that the expressions are logically equivalent)

[mint salmon]

good point

[mint salmon]

my intuition tells me that ('x2 * x1 + x2 * 'x1) should reduce to zero

but intuition doesn't really help me

I need to use axioms and properties, but I'm totally unsure what I should try

ok I guess not

they're both prime terms

[mint salmon]

I just spent an hour minimizing the terms and that's not even what i was supposed to do

how in the hell am I supposed to get the canonical disjunctive normal form??

[weary tiger]

Hi,
discrete math questions

if you have
r -> ~q
q -> ~r

what does that become?

It was the rules of detachment, but I should use
q
q-> ~r

sometimes I am confused if I can use the rules when I have not before the symbols, but it seems like I can

[jagged grove]

So this is a proof of the bernoulli inequality my question would be, can u do that leap from the step in the middle to the last one and if so why?

[opal crescent]

if $ka² \geq 0$, you agree that $$1+(k+1)a+ka²\geq 1+(k+1)a$$

[vital dew]

emeric75:

[opal crescent]

right? @jagged grove

[jagged grove]

yes

what i dont understand is why is that he can take that out of the inequality

[opal crescent]

well if i have thing1 >= thing2, and thing2 >= thing3

[jagged grove]

wait im reading this more carefully nvm

[opal crescent]

i should have thing1>=thing3

ach

[jagged grove]

so hes just using the fact thing2=>thing3 to put thing3 in the right side of the inequality?

and thing3 just so happens to be the n+1 term of the induction?

@opal crescent

[opal crescent]

yeah

[jagged grove]

ok thx

[stray reef]

<@&268886789983436800>

[tranquil cargo]

XD

[rare barn]

those are indeed the rules @weary tiger

[weary tiger]

Yikes

Don’t ban me pls

I’m sorry

[pale epoch]

just try both and see what works

(or what you like better)

imo there is no need to split it into individual cases, but you might have a different opinion

both would work, yes

it's just a matter of how many steps you are able to think ahead

or what makes for a "nicer" argument

[runic mauve]

https://gyazo.com/8663782bd98e3233081cc757c3283231
any ideas on how to establish that inequality?

i was able to get somewhere with the inequality but i can't really get to that point

[faint narwhal]

Draw a picture

[runic mauve]

oh lmao ok i get it

that was pretty obvious with just 3 words oof

[olive dragon]

Hello I have a question, could someone please help me

Is this going to be x^3+3x^3+x^3 = 5x^3 ? where k = 1 and C = 5

This is a Big O problem

[sour arrow]

@olive dragon
You have to show that x² + 3x + 1 < cx³ for some c and values of x

As per the definition of big-O

[olive dragon]

Yes, I understand that. I usually solved this type of problem: 4x^3+x^2-x+10 < O(x^3)

I would choose x>1 and where k = 1 and do |4x^3+x^2-x+10| < 4x^3+x^2+x+10 < 4x^3+x^3+x^3+10x^3 = 16x^3

k = 1 and c = 16

@sour arrow isn't this one way of doing this?

[sour arrow]

Actually, that's neat

You need to specify what values of x this is true over

[olive dragon]

Yes, the examples in the book did it using lot of things

x > 1 ?

[sour arrow]

Note that x² < x³ doesn't hold for all x. But yes, if x > 1

Yeah, I don't see a reason why that wouldn't work here

[olive dragon]

Okay, thank you

[olive dragon]

Hey, another question I need help with, someone please help, thanks

I think the big-O estimate should be O(n) since we only have 1 for loop and it runs from 1 to n

[sour arrow]

It's literally n additions

But yes O(n)

[olive dragon]

Okay, thank you

[sweet island]

can anyone help with pigeon hole principle

[sour arrow]

Go for it

[stray reef]

[mint salmon]

I'm trying to prove that this R := {(x,y) ∈ M x M | x < -y} either is or is not a function

[stray reef]

what's M

[mint salmon]

right, M := ℤ \ {0}

[stray reef]

ok

[mint salmon]

so my thinking is that if I just try to prove that it's possible for x to be larger than -y, then I will have shown that R cannot be a function

since it isn't true for every (x,y) in M

[stray reef]

overthinking

(1, -7) ∈ R, (1, -8) ∈ R, -7 ≠ -8

therefore R violates the definition of a function

[mint salmon]

so giving a single example would be enough?

I mean, I immediately thought of that and said to myself " this is clearly not a function. now I need to prove it."

when can I be sure that "common sense" is enough to be a proof or to disprove something?

[stray reef]

disproof by counterexample is a 100% valid proof method.

[mint salmon]

I'm glad

[little nacelle]

hey

can someone help me

[solar hearth]

Isn't it ab ac ad bc ba cd

[little nacelle]

it is

thats the answr

but i w ant someone to explain why thats the answer @solar hearth

why arent ba and b elements

or cc or ca or cb or cc elements

[sweet island]

Max has 5 weeks to prepare for a final exam. His tutor will tutor him for either 15min or 30min every day until the test but not for more than 15 hours total. Show that during some period of consecutive days, max and his tutor will study for exactly 8.75 hours. How can I use the pigeon hole principle to solve this?

[last sigil]

Find the "pigeons" and the "holes"

What do you think they are?

@sweet island

[sweet island]

i think there are 95 pigeons and 70 holes

the pigeons would be quater hours and the holes would be sum of quarter hours (35) + sum of quarter hours shifted by 8.75 hours(35)

[last sigil]

Look at it in terms of days and how many minutes per day he can study

[runic mauve]

Hey guys I’ve got a definition of a Carmichael number in this textbook but I’m having trouble understanding it, any help?

I can post a pic of the definition in a sec

[analog sonnet]

Did you read and understand the example below? It should be illuminating

[finite ginkgo]

heyo

so my math TA said I should take a course in discrete math

I am wondering if I could just learn it myself instead? or is a course very recommended?

also I am sorry if this is inappropriate place to ask and this is more for questions about the math itself

thank u! ❤

[tranquil cargo]

it depends imo

if ur doing pure math

you probably need like

just the first few chapters as they likee

are the first time u may see proofs and logic

so it can serve as a transition

to more advanced math

if ur doing cs

then i think its p importnatt xd

thats just imo.

@finite ginkgo

[viral merlin]

hi i need a bit of help on homework with combinations...

A major software company is arranging a job fair with the intention of hiring 6 recent graduates. The 6 jobs are different, and numbered 1 through 6. No candidate can receive more than one offer. In response to the company’s invitation, 136 candidates have appeared at the fair.

How many ways are there to extend the 6 offers to 6 of the 136 candidates?

im guessing its (6!)(136 C 6) but that doesn't seem right to me

[gentle nova]

its right

[viral merlin]

How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer, but we do not know which?

[orchid jasper]

Fuck

If I have

$\displaymode \prod_{i=1}^n {p_i ^{min(a_i, b_i)}$

[vital dew]

ratsella:
Compile Error! Click the reaction for details. (You may edit your message)

[orchid jasper]

and

$\displaymode \prod_{i=1}^n {p_i ^{max(a_i, b_i)}$

[vital dew]

ratsella:
Compile Error! Click the reaction for details. (You may edit your message)

[orchid jasper]

can I say

$\displaystyle \prod_{i=1}^n {p_i ^{min(a_i, b_i)} \cdot \prod_{i=1}^n {p_i ^{max(a_i, b_i)} = \prod_{i=1}^n {p_i ^{max(a_i, b_i) + min(a_i, b_i)}$

[vital dew]

ratsella:

$\displaystyle \prod_{i=1}^n {p_i ^{min(a_i, b_i)} \cdot  \prod_{i=1}^n {p_i ^{max(a_i, b_i)} =  \prod_{i=1}^n {p_i ^{max(a_i, b_i) + min(a_i, b_i)}$
```Compile error! Output:

! Missing } inserted.
<inserted text>
}
l.11 ...}^n {p_i ^{max(a_i, b_i) + min(a_i, b_i)}$

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]

yo anyone here good at recurrence relations?

[sour arrow]

Go for it

[weary tiger]

@sour arrow https://gyazo.com/4c0248914ae569e26735791671ae5c46

https://gyazo.com/dc62a8ad6ea2db00c883e63ff49a2a11

can you explain to me how he got from

an = an-1 + 0.09an-1

to (1.09)An-1

[sour arrow]

What's x + 0.09x?

[weary tiger]

I believe the 9 percent interest annually?

[sour arrow]

It's collecting like terms

2m + 3m = 5m
A + 0.09A = 1.09A

[weary tiger]

oh dam

lmao

@sour arrow thanks

[sour arrow]

Np lol. Feel free to ask if you need anything else

[static ivy]

how many 4 digit codes can you make under the following condition:
-the digit 5 has to appear, and no other digit in the code can be greater than it```

ok so i tried it in 2 different methods

and i was hoping someone would be able to tell me why i got different results

method 1: i have 6^4 codes built using the numbers 0,1,2,3,4,5 , i substract all the numbers that are only formed from the digits 0,1,2,3,4 (cuz 5 has to appear) which is 5^4

so method 1 gives me 6^4 - 5^4

method 2:

i make sure 5 appears by first choosing one of four spots for it, and then choose 3 more numbers that each have the options 0,1,2,3,4,5

so 4 * 6^3

ty in advance

[stray reef]

the second thing overcounts codes with more than 1 five

[static ivy]

I suspected but can't really narrow it down to why

[stray reef]

5352 can be constructed in two different ways using your second method

either pick the first slot to be a 5, and set the other three to 352

or pick the second to be a 5, and set the other three to 532

[static ivy]

Oh dam

Well said

Is there anything I can do to get rid of those extras using this method?

Nvm it's too complicated and isn't worth

Thx btw

[static ivy]

another one that i cant wrap my head around...

i have 13 boxes and 7 identical boxes, im asked to distribute them so that one box contains exaclty 2 balls and the rest contain 1 ball at most

so i did 13 * 12choose5 , which is first choosing one box out of the 13 to contain 2 balls, and then 5 places out of the 12 boxes to place the 5 other balls

however my teacher did 13 * 16choose5

which is distribute 5 balls to 12 boxes without restrictions (?)

which of us is correct

[unreal berry]

can someone help me check a proof?

@name plz

[tranquil cargo]

@unreal berry just show

i hope i can help

[unreal berry]

right

It says here let n be any number bigger than 1

n is a prime number if and only if the statement below is true

let's say n is 4 which is not a prime

if we follow the logic below basically the only thing r can be is 1

since it's 1 v s = 1

which means s must be 4

4 = 1 times 4

fair enough

except it's not a prime

[stray reef]

4 = 2*2

4 = 2 * 2, but (2 = 1) OR (2 = 1) is false

@unreal berry

[unreal berry]

=> r = 1 OR 2 = 1

[stray reef]

no

[unreal berry]

r does not have to equal s

[stray reef]

of course it doesn't

but it doesn't mean it can't

like

"4 is prime" is the same as "for any positive integers r and s, if rs = 4, then either r is 1 or s is 1"

this is disproved by presenting the counterexample r=2, s=2

[unreal berry]

Huh?

[stray reef]

what

[unreal berry]

r always has to equal 1

[stray reef]

no

$r = 1 \lor s = 1$ isn't $r = (1 \lor s) = 1$

[vital dew]

Ann:

[unreal berry]

r = 1 V s = 1 => r = 1

[stray reef]

$1 \lor s$ doesn't even make any sense

[vital dew]

Ann:

[stray reef]

no

this is not an equality chain

this is $(r = 1) \lor (s = 1)$

[vital dew]

Ann:

[stray reef]

$1 \lor s$ does not make any sense

[vital dew]

Ann:

[stray reef]

$\lor$ is not an arithmetic operation

[vital dew]

Ann:

[unreal berry]

it's a logical operator right?

[stray reef]

it's a logical operator and it connects two STATEMENTS, in this case the equalities "r=1" and "s=1"

[unreal berry]

I mean not really.

s can be different from r

[stray reef]

these equalities are connected by an OR

not an AND

have you not heard of the use/mention distinction

[unreal berry]

I never said that

[stray reef]

or are you deliberately ignoring it just to be thick

gdi

[unreal berry]

This statement was never presented as a counter example

[stray reef]

hhhhhhhhh

[unreal berry]

So I was working under the assumption of Modus Ponens

[stray reef]

...

you're being incredibly dense or i'm being incredibly unclear despite all my efforts

and i can't tell which

[unreal berry]

Speaking of or. Would you consider or to be mutually exclusive or as V?

like OR or XOR

since the math book says x V y but both are correct

but semantically or is always mutually exclusive as far as I know.

[stray reef]

mathematical OR is always inclusive.

[unreal berry]

all I am saying is that if we have 1 V s = 1 then we know that s can basically be whatever

but r always has to be 1 because r = 1 V s = 1

is the same as r = 1

well I guess we aren't talking about equality here

which makes this entire statement interesting if we aren't looking at it from a logical perspective

[stray reef]

NO!!!!!!!!

WE DON'T HAVE "1 V s = 1"!!!!!!!!!!!!

YOU ARE COMPLETELY MISUNDERSTANDING ABSOLUTELY EVERYTHING!!!!!!!!!!!!!!!!!!!!

[unreal berry]

fair enough.

can you translate this into english then?

[stray reef]

i already did!

"n is prime" is defined as "for any positive integers r and s, if rs = n, then either r is 1 or s is 1"

[unreal berry]

then what exactly is 1 v S called? if it's not a premiss then what is it?

[stray reef]

THERE IS NO "1 V S"!!!!!!!!!!!!!!!!!!!!

THERE IS NO "1 V S"!!!!!!!!!!!!!!!!!!!!

THERE IS NO "1 V S"!!!!!!!!!!!!!!!!!!!!

THERE IS NO "1 V S"!!!!!!!!!!!!!!!!!!!!

THERE IS NO "1 V S"!!!!!!!!!!!!!!!!!!!!

THERE IS NO "1 V S"!!!!!!!!!!!!!!!!!!!!

THERE IS NO "1 V S"!!!!!!!!!!!!!!!!!!!!

HOW MANY MORE TIMES DO I NEED TO REPEAT THAT?????

[unreal berry]

ok what if I say n = 5?

[stray reef]

5 is prime.

[unreal berry]

how can I deduce that from the statement?

[stray reef]

5 is not divisible by anything, except itself and 1

it's not divisible by 2, not divisible by 3, not divisible by 4, and not divisible by anything greater than 5

[unreal berry]

from the statement

[stray reef]

what statement

[unreal berry]

the one I asked you about.

[stray reef]

that's a DEFINITION, not a STATEMENT

[unreal berry]

but it says n is a prime if and only if

[stray reef]

it says

DEFINITION

right fucking there

in plain text

sigh

you're impossible

i'm out

[faint narwhal]

How can you write 5 as the product of two things

[unreal berry]

only if one is 1 and the other is 5

if we know r = 1

then s must equal 5 since 5 = 5 x 1

still doesn't really help me to know if n is a prime or not.

maybe if it had said something like if s = r then n is not a prime.

but then we have 6

and since s always has to be n we always get n = n x 1

[faint narwhal]

Why does s always have to be n

[unreal berry]

cus it's the only way for n = rs

since r has to be 1 us 1 = blablabla = r which is the same as r = 1

s can be whatever and r will always be 1 anyways

but I guess this is not logic so idk what this actually says

shrugs

[faint narwhal]

That's the only way?

No other way to write 6 as the product of two numbers other than 1 times 6?

[unreal berry]

you can write 2 x 3 obviously

but r has to be 1

so only 1 x 6 is viable.

[stray reef]

no, r does not have to be 1

[unreal berry]

hmmmm well aactually

=> means that everything past it could be false

so it's actually worthless for everything to the left of it

[faint narwhal]

Uh

If the thing on the left side are true, then the thing on the right must be true too

Is what => means

[unreal berry]

however if we want everything to the right of => to be true we need to set 1 x 6

yeah

but

since we can get the left statement to be true through 2x3 we can also get everything in the right row to be false

so this definition is trash.

since it doesn't actually accomplish anything

[faint narwhal]

No

It just says that 6 isn't prime

Since 6 doesn't satisfy the definition

[unreal berry]

but the definition is r in Z+: s in Z+:

[stray reef]

"in", not "c"

[unreal berry]

right

[stray reef]

the criterion for $n$ to be prime is $\forall r, s \in \bZ^+: n = rs \implies ( (r=1) \lor (s=1) )$

[faint narwhal]

What did you even try to say

[vital dew]

Ann:

[unreal berry]

so what it says if that if we can get n through rs then n is a prime IF => is true?

[stray reef]

IF this criterion is met, THEN n is called prime. IF this criterion is NOT met, THEN n is NOT called prime.

so what it says if that if we can get n through rs then n is a prime IF => is true?
no

[unreal berry]

but => literally means that if the first row is true then so is the second.

and that is a lie since we can have 2x3 which is true

but still doesn't make the right row true

[faint narwhal]

And so that's why 6 isn't prime

Because it doesn't work

[unreal berry]

so if the definiton is true then n is a prime

[stray reef]

definitions can't be TRUE

definitions can be SATISFIED

[unreal berry]

:I

[stray reef]

IF n satisfies this definition THEN we say n is a prime number.

and no, i'm not being pedantic.

[unreal berry]

well I understand how to use this now. Thanks.

[stray reef]

i did nothing worth thanking for, but you're welcome i guess

[unreal berry]

well you tried

which is what matters

🙂

[unreal berry]

I have a question regarding amounts

need to check which of these assumptions are ccorrect

I have a few questions regarding the structure itself

jämt mean even number

x is divisible by three in s and 6 in T

does R = {} mean all of the things according to the definition?

like . . . -2, 0 2, 4 . . .?

I keep seeing these = but there is not much of an explanation of how it's treated here.

anyways I feel R in T is wrong since 6,12,18 . . . are divisible with 3

well actually all of those numbers are in R

T = 3,6,9,12

S = 6,12,18

yeah got this nvm

I do have a question regarding an examination

which is kind of similar to this one

[analog sonnet]

What is it?

[weary tiger]

Question: https://gyazo.com/dc62a8ad6ea2db00c883e63ff49a2a11

https://gyazo.com/4c0248914ae569e26735791671ae5c46

Does anyone know why this person multiplied 0.09 by An-1?

[pale epoch]

because 0.09 is 9%

[weary tiger]

@pale epoch yea I get that part but why multiply it by the recursive call?

[sour arrow]

Balance + Interest
@weary tiger

You could also say that (1.09)A is a 9% increase

[errant bear]

can someone help me with this problem

using induction, i either get

these two expressions, which are wrong

like i need the numerator for exp 1 but i need the denom from exp 2 to get n+1, am i just messing up the algebra or am i doing something wrong

[sour arrow]

Is the inductive hypotheses used anywhere?

Oh yeah I see it

[errant bear]

im using it to sub in for sigma n, then manually putting in 1/2^(n+1) for sigma n+1

yea

[sour arrow]

That's an i in the numerator, not a 1

[errant bear]

you right

im dumb and blind

thx

[sour arrow]

Lol sry. Hope it goes well

[errant bear]

im putting n+1 for i, correct

[sour arrow]

If you pull a term off, that term would be (n+1)/2^(n+1)

This should follow by adding the fractions together, I can't imagine it would be any harder than that

[errant bear]

the 1 in the numerator should be n+1 here, is what i meant

yeah ik

[sour arrow]

Yus

[errant bear]

im getting 2 - (3n+5)/(2^(n+1))

idk

if i had to subtract n-1 then i would get the answer in the right form

wait its negative nvm

[maiden forge]

I apologize if this is the wrong channel, not sure where to put it. Are restrictions on uncountable infinite sets bijections? I wouldn't think so but math be kinda wacky like that sometimes so I wanted someone at a higher level to either verify my intuition or correct me, thanks

[hearty crane]

define restrictions

[maiden forge]

a function which removes or "restricts" part of a domain

ie the floor function on R would be a restriction from R->N because it is removing all of the elements in R that aren't in N

[hearty crane]

Then no, they aren't bijections.

Not necessarily, at the very least

Like you said, the floor function restricts to a countable set. That's a surjection

[sour arrow]

They can be. Take R → (0,1) which can be a bijection

There's nothing special about the output being a subset of the input

[errant bear]

if im understanding this correctly, 15) is asking for a bijection that switches the x and y components of ordered pairs between two sets

so just like f(a,b) = {(b,a) | a in X and b in Y?

[faint narwhal]

Is that how functions work?

[errant bear]

uh

[sour arrow]

You have the right idea. f(a,b) = (b,a)

[errant bear]

functions take an something from the domain into a unique member of the codomain

[sour arrow]

Then f undoes itself

[errant bear]

wait so is the set thing not necessary

[sour arrow]

A function outputs a unique member of the codomain, not a set

[errant bear]

right

i was thinking of the image/preimage

my b

and yeah f composed w its inverse of (b,a) will just lead back to a,b

[unreal berry]

what exactly is C?

¨:I

C = ?

[stray reef]

C is a set

it's defined right there

it's a set of points in R^2

aka the plane

[unreal berry]

shouldn't it be (x,y) in C?

[stray reef]

no

you can't use C when defining C

[unreal berry]

Why not?

[stray reef]

bc it'll be circular

you can't use the thing you are defining when you are defining that thing

C the set of all points (x,y) in R^2 for which x^2 + y^2 = 1

C is a subset of R^2

a point (x,y) from R^2 is in C if and only if x^2+y^2=1

[unreal berry]

I thought R X R was a subset of C

[stray reef]

no

it is not

(0,0) is in R^2 but is not in C

[unreal berry]

I see

so (x,y) need to be in RXR so that blablabla becomes 1 which is C

so C = 1

[stray reef]

...

no

no

no

no

C is not 1

you're missing the point again

C is a subset of R^2

it is a set of points

for example, (0, 1) is in C

(-0.6, 0.8) is also in C

(5, 2) is not in C

but C is a set of points. C is not a number.

[unreal berry]

can't we simply drop C = and just do the {}

[stray reef]

"droo"?

[unreal berry]

Yeah C is just in the way

[stray reef]

i mean, you CAN just write ${ (x,y) \in \bR^2 \mid x^2+y^2=1}$

[vital dew]

Ann:

[unreal berry]

Ha I knew it

[stray reef]

$C$ is just the name you're giving to that set

[vital dew]

Ann:

[stray reef]

it's convenient to have names for sets

[unreal berry]

So it's just a pleb. Have it shot by a firing squad.

[stray reef]

what

what

what

what

what

what