ITT: brainlet filters

you forgot about the case where >1 pair is obtained.

Assume it is possible to take 3 socks out of the drawer and not get a pair. However only 2 colours exist. Showing that a contradiction exists in the premise.

the negation of an exact amount is everything below and above 1 pair of socks in this case.

do you mean 2D > 3D

We have a signature [math]\Sigma = (\text{blue}^1,\text{black}^1)[/math] plus equality, with a FO-theory [math]\Phi \equiv \Rightarrow (\exists x,y,z.~x\neq y \wedge x \neq z \wedge y \neq z)\wedge (\forall x.~\text{blue}(x)\vee\text{black}(x) \wedge \neg(\text{blue}(x)\wedge\text{black}(x)))[/math],

Your statement can be modeled as [math]\psi \equiv \forall x,y,z.~(x\neq y \wedge x \neq z \wedge y \neq z) \Rightarrow (\text{blue}(x) \Rightarrow (\text{black}(y) \wedge \text{black}(z)) \vee \text{blue}(y) \vee \text{blue}(z))[/math]

For a proof that [math]\Phi \Rightarrow \psi[/math] is valid, refer to a FO-prover of your choice (or do it by hand if you enjoy living in the 19th century), Z3 works fine for me.

If you have 3 socks and no two of these socks are a colored pair then that would imply each sock is a different color.
That implies there are at least three color types in the drawer.
Since we assumed there are only two color types in the drawer we've reached a contradiction.

I'm not doing your discrete math homework, you little cs brainlet, but lookup the pigeonhole principle.

There is no way you can possibly get anything but a pair if you pick three socks since there are only two colors.

lol just try it and let me know if you didn't get a pair

suppose you have 3 socks but no pair.
then assume without loss of generality that you picked a blue sock at the first go.
the next sock will not provide a pair. so it is not blue. meaning it is black.
the next sock will not provide a pair so it is not blue not black. but we have no socks that satisfy this property. contradiction.

>help me with my proof 101 homework
OP is a piggot (that's a portmanteau of "pig" and "faggot" -- it means pig-faggot)

filter yourself out

Why would I pick three? I only have two feet.

You're wrong. While you will get at least two socks of the same color, you are not guaranteed that you will get both a left and a right sock.

He didn't say they were socks for your feet.
>He doesn't have a dick sock

A complicated proof is not necessary. Since we only have 3 socks and need 1 pair, we can literally go through each draw one by one.

You draw [math] sock_1 [/math].
[math] sock_1 [/math] is either blue or black.
Case 1: [math] sock_1 [/math] is blue.
You draw [math] sock_2 [/math]. It is either blue or black.
Case 1.a:
[math] sock_2 [/math] is blue. We are done.
Case 1.b:
[math] sock_2 [/math] is black.
You draw [math] sock_3 [/math].
Case [math]1.b. \alpha [/math]: [math] sock_3 [/math] is blue. We are done.
Case [math]1.b. \beta [/math]: [math] sock_3 [/math] is black. We are done.

Analogous for Case 2.
q.e.d.

is this really what philosophers do all day?

Pigeonhole principle. How is this difficult?

I think this thesis should have been thrown out on the grounds that even if the nugget of real research into some niche field of algebra/nnumber field is acceptable, the fact that she wasted this much time writing a vernacular derivation of counting arithmetic all the way up to her research means that she could have instead been improviing her actual research by investigating next developments further, characterising use cases, linking cohesively with other research or other useful things that would have added to the body of humanity's mathematical knowledge.

it's shameful that she got a PhD with this.
at teh very least the supervisors should have told her to rewrite it .

>pick a sock
>rip it up
>pick another sock
>burn it to ash
>pick a sock

>that feel when no pair of socks

the first sock can be either blue or black
the second sock can be eitheer blue or black.
the third sock can be either blue or black.
so there are 2^3 possible sequences of socks that can be picked because each picking can be either blue or black.

so the possible sequences of socks picked are
black, black, black
black, black, blue
black, blue, black
blue, black, black
blue, blue ,black
blue, black, blue
black, blue , black
blue, blue, blue

as we can see all possible sequences either at least 2 blue socks or at least two black socks.

A = blue
B = black

AAA (AA)
AAB (AA)
ABA (AA)
ABB (BB)
BAA (AA)
BAB (BB)
BBA (BB)
BBB (BB)

if there are less than 3 socks
then a pair is not guaranteed
A
B
AA (AA)
AB
BA
BB (BB)

somehow you still picked a white sock and two left black socks, faggot

probability of picking 3 the same>0 for blue sock no.>3 or black sock no.>3

1st sock will be black(A) or blue(B)
2nd sock will be black(A) or blue(B)

If AA or BB, there's no need to pick a third sock. You already have a pair.

If you don't have AA or BB, then you either have AB or BA.

Picking the last sock will therefore mean completing one of these combinations:
ABA - Black Pair
ABB - Blue Pair
BAA - Black Pair
BAB - Blue Pair

Since there are only 2 colors, picking 3 socks by definition means picking a pair.

There aren't right or left socks faggot.

yes there are