discrete-math

[safe scroll]

that is what we are using right

n = 8 types

r = 6 donuts

if there are 0 jelly filled

then we only have 7 types, but all 6 donuts

if we have one jelly filled we still have 7 types but only 5 donuts

and if we have two jelly type, then we have 7 types but only 4 jelly donuts

so it would be

1596

[karmic nymph]

hm

i think i get it

[weary tiger]

anyone here

[lyric pumice]

Who knows?

[sour brook]

Hello, how would you prove AnA = A? (Ie which law would you use?)

[minor dagger]

reflexive property

[carmine grove]

Permutations
Order matters and repetition is Allowed is n^r

Order matters and repetition not allowed is

n!/(n-r)!

Combinations

Order doesn't matter and repetition not allowed

n!/(r!(n-r)!)

And combination with repetition

(r+n-1)!/(r!(n-1)!

That last one is he one I never know how to use. But is this what you're meaning?
@deep portal
Thanks!

[limpid citrus]

and 7 chairs, where seatings are considered to be the same if they can be
obtained from each other by rotating the table.
(a) How many ways are there? Show your derivation.
(b) How many ways are there if two people have to sit next
to each other? Show your derivation.
(c) How many ways are there if two people cannot sit in the
same table? Show your derivation.```

How do I do this?

[stray reef]

which part

[oblique bay]

help pls

[cursive lily]

f : N → Z.

someone help me

[pale epoch]

what have you tried

[oblique bay]

show that its one to one and onto

[cursive lily]

i havent tried anything @pale epoch

i will ask

how does one construct a bijection

its been so long i have forgotten

[pale epoch]

do you know what a bijection is

[limpid citrus]

which part
@stray reef all the parts....I figured out some things but I am not sure if that is correct.

at least 8 but no more than 12 characters, where each character in the
password is a lowercase English letter, an uppercase English letter, a
digit, or one of the six special characters ∗, >, <, !, +, =.
How many of these passwords contain at least one uppercase letter, one lowercase letter, one digit, and one special character?

For this, will we do (total no. of passwords available) - (passwords not containing uppercase + passwords not containing lower case + passwords not containing digit + passwords not containing special character)?

I think I'm missing out something in here.

[cursive lily]

yes

[sour brook]

reflexive property
@minor dagger thanks. checked it out and one of the textbooks calls it the idempotent law

[untold furnace]

hey guys

can someone please help me with a question?

I would really appreciate it

[unreal stump]

What have you tried?

[untold furnace]

To be quite honest

i honestly dont know how to start this

[unreal stump]

Expand both binomial coefficients you get a relation between t and s

[untold furnace]

so i did that but i dont know wat to do from here

how do i get a relation between t and s?

[unreal stump]

Just manipulate algebraically?

[untold furnace]

um i dont know how i would do that

sorry

[unreal stump]

Do you know t! means?

[untold furnace]

yeah the factorial

of t

[unreal stump]

Just expand the terms and cancel them

[untold furnace]

can you show me an example?

[pale epoch]

maybe it's easier if you multiply both sides by the reciprocal of either side

then you get an equation of the form "something = 1"

and you can start cancelling terms

i.e. you can cancel the t! on both sides

but also more if you think about how the factorial is defined

[untold furnace]

oh ok

thanks lemme try and ill let u guys know

[untold furnace]

hey so i got t as

[limpid citrus]

and 7 chairs, where seatings are considered to be the same if they can be
obtained from each other by rotating the table.
How many ways are there if two people cannot sit in the
same table?``` How do i approach this using permutation and combnation?

[pale epoch]

can you show your whole work @untold furnace

[limpid citrus]

Should I divide it into 2 cases saying, case 1: Let's say two people can not sit on table with 8 chairs case 2: Let's say two people can not sit on the table with 7 chairs

and what after that?

[untold furnace]

wait one sec

it got messed up lmao

[pale epoch]

eh

[untold furnace]

also i meant to put the ! outside of the bracket in the 3rd line

[pale epoch]

you turned (s-1)! into s!(s-1)

which is not correct

[untold furnace]

oh

wait

[pale epoch]

instead you should turn s! into s(s-1)! and cancel the (s-1)!

[weary tiger]

Yes

[untold furnace]

lmao my bad

[pale epoch]

you should arrive at something simpler

no squares

[pale epoch]

yeah but scrap the last line

and instead add 2s to both sides

actually there is still some mistake somewhere

you didnt distribute the 2 somewhere

[untold furnace]

omg

my mistake lmaooo\

[pale epoch]

anyways, you will arrive at 3s = 2t+2

[untold furnace]

yeah

[pale epoch]

which proves your result if you think about it

[untold furnace]

hmmm

so how would i word this?

[pale epoch]

word what?

[untold furnace]

like a therefore statment

[pale epoch]

therefore 3 divides 2t+2

[untold furnace]

oh wait

i just took that in

idk why i was thinking about 3/2t+2

its the way aroundd

ok makes alot of sense

thanks alot man really appreciate it

[pale epoch]

you're welcome

[vivid bison]

So I’m trying to prove that if (n choose m-1) = (n choose m) then n is odd. So what I tried was multiply the reciprocal of one to the other and got an answer but it doesn’t really make sense... can some give me some suggestions on how to do this?

if I set m to 1 then n becomes even which fails so idk what to do instead...

[naive saffron]

Use the formula $\binom{n}{m}=\frac{n!}{m!(n-m)!}$ to rewrite $\binom{n}{m}=\binom{n}{m-1}$

[vivid bison]

Hm ok I’ll give it a try

[naive saffron]

https://cdn.discordapp.com/attachments/589309951361679390/768871392178339850/454846421548531722.png

[vivid bison]

Hm sorta stuck... do I apply the symmetry rule as well? The (n choose m) = (n choose n-m)??

[naive saffron]

No not really

You're over thinking it

So $\binom{n}{m}=\binom{n}{m-1}$ can be rewritten as $\frac{n!}{m!(n-m)!}=\frac{n!}{(m-1)!(n-m+1)!}$

why is it taking so long to render

[vital dew]

Whoever:

[naive saffron]

@vivid bison now cancel

cancel the n!, (m-1)! and (n-m)!

[untold furnace]

hey can someone help me with this quesiton?

I tried using induction

but i got stuck at the inductive step trying to prove P(n+1)

could anyone help?

i would really appreciate it

[weary tiger]

can anyone help provide formal proof for the following?

∀xD(x) → ∃x[F(x) → D(x)].

∃x[J(x) → T(x)] → ∃xT(x).

two separate proofs

[untold furnace]

can someone please help me?

[green roost]

use contradiction

[untold furnace]

how would i approach it using contradiction

[untold furnace]

@green roost ?

[green roost]

let me try solve

[untold furnace]

alright

thanks

[green roost]

oh hang on

thats easier than i thoughht

@untold furnace

[untold furnace]

yeah

[green roost]

assume k exists

then let y = k

[untold furnace]

so hold up real quick

[green roost]

aight let me know if you get stuck

[untold furnace]

so the contradiction statement would just be that "there exists a postive integer k such that for all postiive integers y (equation) is prime?

and we prove this wrong?

[green roost]

yeah

[untold furnace]

therefore proving the original tru?

[green roost]

yes

[untold furnace]

i see

[green roost]

if you assume it exists and then prove it contradicts itself then it cant exist

[untold furnace]

so how does y=k contradict it?

is it because then we get k^2 +12k?

[green roost]

assume all y^2 + 11y + k = P

yep

[untold furnace]

ok i get it

thanks alot man

[green roost]

which is k(k+12)

[untold furnace]

yeah

[green roost]

and therefore p is a multiple of k

no worries

[untold furnace]

appreciate the help dude!

[green roost]

no worries

do you think you can send similar ones i have a test on monday and that kinda stuff will be in it hahha @untold furnace

[errant bear]

no worries

[untold furnace]

well this is kinda the only type of question i had of this lmao

[errant bear]

for every integer n, there exists an integer m such that k is

or is not

k

is

[green roost]

ah true all good lol

[sonic path]

can someone explain why 18 is a convenient choice to pick for k for part a

[wise dew]

Can someone help me with 4.6.8 a

it references 4.6.7 thats why i included it

[radiant shore]

Isn’t this 4^15?

[last timber]

@radiant shore no, it is definitely not 4^15

since 4^15 means that after you give one crayon, there is still 15

which is meaningless

[weary tiger]

are infinities the last points on the number line?
i mean in an extended real number system

[drifting sail]

In the extended reals yes.

can someone explain why 18 is a convenient choice to pick for k for part a
@sonic path You want to write 17x+11 < Cx^2 for large x (definition of Big O). So you come up with intermediate inequalities that get you there, while keepipng in mind you can assume x as large as you want. For instance you could write 17x+11≤17x+x=18x, assuming x≥11. But that's still not a bound in terms of x^2. So you assume that x>18=k to conclude that 18x<x*x=x^2, finally obtaining 17x+11<x^2 for x larger than k.

[fathom quiver]

could anyone help me with linear recurrence relations?

[potent mirage]

Does anyone know how the equation is going to look like? I started the graph but do not know how to go from there

[drifting sail]

,rotate ccw

[drifting sail]

Does anyone know how the equation is going to look like? I started the graph but do not know how to go from there
you don't need to write down equations for the lines, just assume "given n vertical and n horizontal lines, there are n^2 intersection points" as your induction hypothesis and then argue what happens when you add an additional vertical and horizontal line.

(hint: use the base case, which is given in the exercise.)

[radiant shore]

Why is it not 4^15? We have 15 unique crayons to distribute to 4 kids. We have 4 options for each crayon. That’s 4^15. @last timber

[potent mirage]

you don't need to write down equations for the lines, just assume "given n vertical and n horizontal lines, there are n^2 intersection points" as your induction hypothesis and then argue what happens when you add an additional vertical and horizontal line.
@derivada.schwarziana#0881 i see but I am little bit confused. Does it mean I am proving that after adding additional lines either horizontal or vertical, the total # of intersections should be less than or equal to n^2? Should adding another line be (n +1)^2 or n^(2+1)?

[blissful zenith]

can someone explain the discrete metric to me?

i just dont get

how all points are exactly 1 away from each other

doesnt that mean there has to be a limited no. of points?

im thinking 5

its obviously wrong but i cant visualize it

[blissful zenith]

<@&286206848099549185>

[weary tiger]

no there are unlimited numbers between numbers

the numbers of number is defined as infinity

is f(infinity)=limit to infinity of f(x) in the extended reals?

<@&286206848099549185>

[last timber]

how all points are exactly 1 away from each other
@blissful zenith we just say that distance between points is 0 if points are in the same set and 1 otherwise (if i remember corrctly)

it is kinda of indicator if you want

[weary tiger]

is f(infinity)=limit to infinity of f(x) in the extended reals?
@weary tiger someone please answer my question.

[weary tiger]

oh so i hadnt gone mental

thanks

for the help

[blissful zenith]

can someone help me prove

prove that d1 metric on R^n is a metric

<@&286206848099549185>

[serene basalt]

just prove each of the axioms

[sleek island]

Is there any reason why the terms "factor" is used for this?

[misty merlin]

Would someone be able to explain why he got -15? I understand how he got 15. This is my lecturer answering a students question*

[primal dome]

Could someone please explain this?

[deep portal]

Let's see. A is saying essentially that one of them is lying. If B is telling the truth then A is lying. But if A was lying that would mean both of them are lying. So I think A is the Knight.

[primal dome]

Thanks @deep portal

[stray reef]

But if A was lying that would mean both of them are lying.
no

A lying would mean A and B are either both telling the truth or both lying

[potent oracle]

Hey all! Not sure how this fully applies to discrete maths. I'm tryna help out a cousin with some problems he doesn't understand for school. (We have a test in a week or so)

[last sigil]

[potent oracle]

so... how does one show (~q and (p → q)) → ~p is a tautology.

without a truth table 😖

I read up on laws in the text book

still don't get it and was looking at some videos as well but to no avail

[main solstice]

that's propositional logic

[potent oracle]

ye I'm watching another video rn on it to try to grasp it
https://www.youtube.com/watch?v=itrXYg41-V0

just so far i've been gettin nothin from the books

[main solstice]

that's called modus tollens

if I'm not wrong

[potent oracle]

🙌 eh..ok? I'll look that up now and see if I get any progress

any sources? @main solstice

I just watched this video and I aint get anything pertinent to the questionhttps://www.youtube.com/watch?v=67x8NWKDctg

for reference everyone i'm still trying to solve this:
(~q and (p → q)) → ~p is a tautology. (without a truth table)
I've been told so far it's prepositional logic and I've read about it

so I get that I have to show both are true with some laws..I just don't know how. Anyone can help?

[main solstice]

here's the proof

the first one is the simplest

[potent oracle]

oop I trust you lol

erm alright I get the first one yea

ooo so now I think I see if it wasn't with the extraneous -q and the and

^

so what now? @main solstice
How is that applicable?

[main solstice]

p->q is logically equivalent to not-p V q

but given that q is false

the only conclusion is not-p

therefore (~q and (p → q)) → ~p is a tautology because, as I've demonstrated, it's always true

[potent oracle]

hmm lemme try to right that down

I lost myself
And what about the other laws?

@main solstice I don't think I get it

[main solstice]

Those are ways of proving the tautology, what do you mean by laws?

[potent oracle]

like in class we did this something like this:

@main solstice I'm talking about demorgans law etc..

[main solstice]

ah yes

[potent oracle]

right so I'm totally lost

I take it I have it wrong

[main solstice]

p->q <-> not-p V q by Material Implication

[potent oracle]

I still have no clue xc

maybe you got sources that can help or can right it down and explain in detail?

[main solstice]

I have some in italian (my first language)

for english sources I use wikipedia

there's already a lot of stuff there

[potent oracle]

aight I pmed you because you said a lot of stuff there but never said where and i'm still in need of help

if anyone else can help dm me asap

i'll be here readin about it to try to figure it out still and looking for videos

Again the question is to solve:
(~q ^ (p → q)) → ~p is a tautology. (without a truth table)

[open narwhal]

Anybody know any information that can help with hamiltonian graph proofs? I'm so lost.

Like i know what a hamiltonian and euler circuit are, i just suck at proofs

[naive saffron]

What's the problem

[open narwhal]

its in #help-0 @naive saffron

[misty merlin]

<@&286206848099549185>

https://discordapp.com/channels/268882317391429632/496785905474994186/769195516767895573

[zinc swift]

may i know if part B is symmetric? i cant seem to find any counterexamples to disprove it

[stray reef]

do you think it is not symmetric?

what if i told you |a-b| = |b-a| for any real numbers a and b, from which it follows immediately that |a-b| ≤ 3 is equivalent to |b-a| ≤ 3?

@zinc swift

[zinc swift]

thanks so much, appreciate it 🙂 @stray reef

[weary tiger]

aSb = |a-b|<=3 <=> |b-a|<=3 = bSa.

therefore aSb = bSa, symmetry.

[fallen sable]

Hi to all,
I have an enumeration problem on which I make a mistake, but I can't figure out where.

Statement: How many ways can a group of 5 people be formed from a group of 4 adults and 7 children if the group must include at least 3 adults and if a particular adult and a particular student cannot be in the group together?

My demonstration :
Case 1: The particular adult is there. So two other adults are chosen, then the remaining adult and the 7 children. Except that a child cannot be present, we choose two persons among 8 - 1, so 7 persons.

Final formula: C^2_3 * C^2_7

Case 2: The particular adult is not there. So we choose three adults, then we can only choose among the 7 remaining children.

Final formula: C^2_7

Total formula: C^2_3 * C^2_7 + C^2_7

So my result is : 84

[violet kayak]

huh why did you take 8-1 instead of 7-1?

and I think you still need to look at the cases with 4 adults in the group

[fallen sable]

Because I have an adult + 7 childrens, so 7 + 1 - 1

Because I have only 3 adults and I can add a 4th

[violet kayak]

oooh

right sorry

[fallen sable]

it's just minimum 3

but I can get 4

[mystic moss]

@fallen sable you're overcounting in case 1.

[fallen sable]

Yes, I finally solve this problem, so, without overcounting in case 1

||final result is 74||

[burnt scroll]

Can anyone help me on 5 (b) and (c)

[stray reef]

try letting P(x) or Q(x) be something stupid

[burnt scroll]

i....is that really necesarry?

[stray reef]

try it

[burnt scroll]

i dun actualy understand what the question wants

[keen turtle]

Two sorted sequences lengths 9 and 7 are given: (1,2,3,...9) and (a,b,c,d,e,f,g). We want to
interleave them into a sequence of length 16 such that numbers 1-9 remain in relative order, and
also literals a-g remain in relative order. How many ways are there to do this? Example valid
sequences are 1a2bc34d56efg789, 12345abc678de9fg, and a1bcdef23456789g.

I'm thinking (16 C 9) * (7 C 7) but I'm not sure that's right

[glass condor]

it's equivalent to the number of ways to put 7 balls into 10 bins, I think

bins being placements among the digits (before the first one, between 1 and 2, ..., after 9), and balls being letters

[keen turtle]

hmm

how are you getting 10 bins?

and how would you make sure the balls are ordered in the same way?

[mystic moss]

I'm thinking (16 C 9) * (7 C 7) but I'm not sure that's right
i think this is good

[glass condor]

after 1, ..., after 9 is 9 bins, and also there's one before 1.

[mystic moss]

with balls and bins, you don't care about the order of the balls in a particular bin

[glass condor]

and how would you make sure the balls are ordered in the same way?
That's why I'm not counting their orderings - only how many letters are there between some numbers.

[keen turtle]

ahh gotcha

[mystic moss]

hm, i should say that differently

wait, maybe that works.

[glass condor]

basically, 1a2bc34d56efg789 would be (0,1,2,0,1,0,3,0,0,0)

and a1bcdef23456789g is (1,5,0,0,0,0,0,0,0,1)

[mystic moss]

ah, and that's counted exactly the same way.

because we have (10 + 7 - 1 choose 7)

[glass condor]

since the number of balls-in-bins is C(n+k-1,k-1), we have C(16,9)

[keen turtle]

ohhh okay

thanks guys

[weary tiger]

what is pumping length

no formal proof for pumping lemma describes it clearly

what is this blyat

[potent oracle]

chee I thought my internet went bad but dang

I dunno if I will ever understand this ngl

I bout to do not so well on this test..if anyone is free hit me up

[potent oracle]

is a valid conclusion. I have to "state all rules"

this was an example of another one in class

can any of you guide me in the right direction here?

[weary tiger]

There seems the be a discrepancy between the premises in your proof and the premises in the problem

[potent oracle]

hmm

not sure what you mean by that

expound please

[weary tiger]

Well anyways here is the proof

1. ~p ∨ (t ∧ q) (premise)
2. r → p (premise)
3. ~t (premise)
4. ~r → s (premise)
5. s → b (premise)
6. p → t ∧ q (rewriting 1. as implication)
7. ~t ∨ ~q → ~p (contrapositive on 6, distributing ~ into ∧)
8. ~t → ~p (∨ elimination in the antecedent in 7.)
9. ~p (modus ponens on 8. and 3.)
10. ~p → ~r (contrapositive on 2.)
11. ~r (modus ponens on 9. and 10.)
12. s (modus ponens on 4. and 11.)
13. b (modus ponens on 5. and 12.)

[potent oracle]

hmm mkay

let me write this down and try to onderstand

off the bat I get the implication

what do you mean for 7

it's contrapositive and the distribution law? @weary tiger

i see the contra

it's actually makin some sense which is good

[weary tiger]

Well we can use the contrapositive on 6. to get ~(t ∧ q) → ~p

~(t ∧ q) is equivalent to ~t ∨ ~q

[potent oracle]

oo so you are doing both steps at onnce

[weary tiger]

Yes

[potent oracle]

aight gotchu so far

(∨ elimination in the antecedent

never heard of this

because of the various laws are there many other ways of doing this?

[weary tiger]

These rules may be called differently in your course

[potent oracle]

also do you see the finish from the start or are you just arguing until you find a pattern?

[weary tiger]

It's just that when a ∨ b → c then also a → c

[potent oracle]

It's just that when a ∨ b → c then also a → c
@weary tiger right is there another name for that?

[weary tiger]

also do you see the finish from the start or are you just arguing until you find a pattern?
Usually these are not hard, you start by looking at what premises are given that actually say something, i.e. here it is ~t and then look at implications containing t, from that you chase implications

@weary tiger right is there another name for that?
I'm not sure

We can probably prove it

[potent oracle]

Usually these are not hard, you start by looking at what premises are given that actually say something, i.e. here it is ~t and then look at implications containing t, from that you chase implications
@weary tiger ah mkay

hm mkay let's try one more practice questions and then can you send me some to practice?

[weary tiger]

Sure

[potent oracle]

i'd say to send me 3 or so. I really don't see how to properly convert

aight lemme find one sir sent before

depending on how you do this one I will either understand the principles or cry myself to sleep

[weary tiger]

Well how would you start?

[potent oracle]

take pictures?

I can write it down now

first I put the 5 steps as hypotheses

or as premises as you call em

[weary tiger]

Yeah, but what do you do then?

[potent oracle]

then I'd scratch my head for a few hours and look at the laws and see if I can apply any of em

if I can I do but I don't know how i'll get there

[weary tiger]

Look at these logical ors

Do these look sus?

[potent oracle]

eh? wym?

[weary tiger]

For example ~r ∨ w, does that look suspicious?

[potent oracle]

not really

saying that implies it is

but I don't know in what way

[weary tiger]

You can rewrite this as r → w

[potent oracle]

ye but I mean what good would that do?

you can't find t in one line but how do you know you are getting progress

can you not rewrite some of the other premises?

[weary tiger]

Yes

[potent oracle]

right..so how do I know if I'm getting progress?

[weary tiger]

The given here is ~(p ∨ w)

What can you make out of this?

[potent oracle]

that we can rewrite it as an implication as you said

[weary tiger]

Well we can do that, but we don't want an implication out of a given

This is something that tells us actually about the truth if p and w

We can use this

Maybe some other way to rewrite this

[potent oracle]

hmm I don't think i'm getting this process with 'truths'

how do I know if something is producing a truth of an expression?

[weary tiger]

Well the way I see it is this: the implications aren't saying things, they are just "if this was true possibly, then..." but we can't use that to prove a statement like ~t, so we need to start with things that actually say how things are, for this we need to look for statements that are not implications

[potent oracle]

ah alright I get you

mkay so then what are we looking for after 4?

we imply r → w? if so then what?

[weary tiger]

Well most of those are clearly implications

What is the one thing that is not an implication?

we imply r → w? if so then what?
Yep that was to muster out one of the implications

[potent oracle]

alright so now we mustered out that for 5 what do we do for 6?

[weary tiger]

Wdym by 5 and 6?

[potent oracle]

oo i'm numbering them as you do for the questions on the sheet

i thought that was one of the steps

[weary tiger]

I'd have suggested that we take a closer look ~(p ∨ w)

[potent oracle]

hmm mkay

[weary tiger]

The one thing that is not obviously an implication

Can you rewrite this into something useful?

[potent oracle]

well distribution

no?

[weary tiger]

Yes

So we get ~p ∧ ~w

Which gives us two statements that are very useful with the implications given

[potent oracle]

aright now what?

[weary tiger]

Off the bat what can you do with ~p?

[potent oracle]

erm.. modus ponens?

[weary tiger]

On?

[potent oracle]

i'd say from the distribution?

I feel like that's wrong tho

[weary tiger]

Yes but what two statements do you use the modus ponens on?

First one is ~p

Now we need some implication

[potent oracle]

i'm a little lost. I should have wrote this down to see where we are 😢

I guess we can do an implication on 4

[weary tiger]

4 was ~p → s?

[potent oracle]

not p -> s

ye

[weary tiger]

Yes

So we have concluded s now

Sooo what useful things might we be able to do with s?

[potent oracle]

man yo makin this seem like a magic trick ngl 🤣

[weary tiger]

Take a look at the first premise

[potent oracle]

mkay

[weary tiger]

There is an s in there

[potent oracle]

eh isn't that another implication?

then modus ponens?

[weary tiger]

So if we proved that ~r is true, we get s → ~t and with that ~t

[potent oracle]

to get the t?

[weary tiger]

So now we want ~r

Where do you see an r?

[potent oracle]

in the second line

[weary tiger]

Right

Have you rewritten it as an implication?

[potent oracle]

we did didn't we?

r->w

[weary tiger]

Right so r → w

[potent oracle]

ye

[weary tiger]

In this form it is not incredibly useful

What can we do?

[potent oracle]

hmm

contrapositive

[weary tiger]

Yes

[potent oracle]

how am I guessing these? 🤣 🤣

[weary tiger]

Because these are actually very obvious

So we get ~w → ~r

Remind me, do we have a ~w?

[potent oracle]

we do no? from the distribution on 3

[weary tiger]

Correct

So what rule do we use?

[potent oracle]

uhh

[weary tiger]

To conclude something from an implication with the first part given

[potent oracle]

I could only think of the contra

[weary tiger]

There is a name for it

[potent oracle]

contra positive?

you said with the first part given

hmm

[weary tiger]

Contrapositive says

1. r → w
2. ~w → ~r

[potent oracle]

oop ye that's wrong

it's already negative so modus ponens?

[weary tiger]

Yes

So we use modus ponens to conclude ~r

Now going back to our original goal, take a look at
~r → (s → ~t)
what can we do here?

[potent oracle]

wait that's the first line

[weary tiger]

Yes

[potent oracle]

i'm confused now? wym do there?

[weary tiger]

Think of it like the usual implication

We just found that ~r is true

So what can we do with that?

[potent oracle]

uh

i lost myself xc

I dunno

;oop

[weary tiger]

Okay. Let's try this: Given are a → b and a, what rules apply?

[potent oracle]

the implication as you said

and that's the only one I could think of

maybe can you send me the laws and I can see?

I'n not sure tbh

so not

wait

[weary tiger]

Modus ponens says that

1. a → b
2. a
3. b (modus ponens on 1. and 2.)

[potent oracle]

bruh that was it?

[weary tiger]

Yes

Now going back to our original goal, take a look at
~r → (s → ~t)
what can we do here?

[potent oracle]

aight can I watch a video on this pleae?

[weary tiger]

Okay

[potent oracle]

you know a specific video?

[weary tiger]

No

[potent oracle]

general theme is what then?

[weary tiger]

Yeah it's just chasing the implications

[potent oracle]

also then

Yes
@weary tiger right so the modus ponens would just be t

[weary tiger]

No it wouldn't be

Applying modus ponens to ~r → (s → ~t) and ~r gives us s → ~t

[potent oracle]

you mean not t

i*

[weary tiger]

a → b and a with a=~r and b=s → ~t

[potent oracle]

ah alright

[weary tiger]

Ok but we showed earlier that s is true

[potent oracle]

alright

[weary tiger]

So what do we do now

[potent oracle]

modus ponens again

[weary tiger]

Yeah

So we get the end goal ~t

[potent oracle]

oooo

that's why I did it twice in my head thinking it wasn't done

I thought we were lookin for t 🤦‍♂️

hence why I said the modus ponens earlier was just not t

[weary tiger]

See if you can piece together a proof starting with ~(p ∨ w)
The rules you have to use are:
Modus ponens
Contrapositive
Distributing ~
Rewriting ∨ as implication

[potent oracle]

heh I found this awesome tutor for it

https://www.youtube.com/watch?v=rAxXcX_w5fE

ye I need to really go over this

so what would it look like together when all of them are applied?

because I see you cross referenced different lines to find the truth for a preposition

@weary tiger ye I askin what it would look like all together as I'm puttin it in juxtaposition with the one we did earlier

(can't believe I found this lady on youtube called kimberly brehn
It looks liike she can help me in the future with this course as well :D)

erm hopefully you can read this it's the let question which seems really really easy so I did em earlier today and wanted to know if they were right

here they are:

are they? :3

[weary tiger]

,rotate 180

[weary tiger]

What is that?

[potent oracle]

jaji? how did you?

anyway

nani*

it's expressing the propositions as logical connectives

is it right?

also what about the question we did?

@weary tiger ye I askin what it would look like all together as I'm puttin it in juxtaposition with the one we did earlier
@potent oracle ^^

[weary tiger]

How you would prove something with these premises?

[potent oracle]

wym? I not understandin

@weary tiger expound and ye I gotta learn faster lol
quiz at in 34 mins

[weary tiger]

I mean what exactly are you asking

[potent oracle]

was askin if my answers were correct

was also askin for you to send the second one we did with the steps so I can compare

again I wanna make sure I understand because if I don't i'm dead

I'm really confident about my answers so far

[potent oracle]

,rotate 180

[potent oracle]

what does the last question mean?

@weary tiger you headin to bed?

cause you the only one that here ._.

[weary tiger]

Yes

Maybe ask these logic questions in #proofs-and-logic

[weary tiger]

Hi everyone! Is this theorem equivalent to the english sentence above?

I just started studying Discrete Math : )

[naive saffron]

Yes

If you define N=positive integers

[nimble holly]

I am really bad at induction, any guidance in the right direction ?

[weary tiger]

Oh thank you! So I'm better not to use N since according to wikipedia it also includes 0 which is not a positive integer

Is induction even possible with Real numbers?

[nimble holly]

I believe so

[weary tiger]

Real numbers are basically not "discrete", which is what this whole branch of mathematics is about. But I don't know, I'm a student just like you

[unreal stump]

Just induct on n,ig

[weary tiger]

ooooh right

[last timber]

Is induction even possible with Real numbers?
@weary tiger well, there is transfinite induction

i mean there is theorem that every set can be well-ordered and once you have well-ordered set you can do induction on it

[misty merlin]

Hey Commander could u please help me

[last timber]

[misty merlin]

I'm trying to figure out how my techer got -15

[pale epoch]

the real induction on real numbers that might be useful is

show it holds for every real number in [0, 1)

[misty merlin]

Or Lochverstarker

[pale epoch]

then show if it holds for a real number x, it holds for x+1

[last timber]

@misty merlin looks like he just did +a-a

which is basically the same as adding zero

[pale epoch]

then you get all positive real numbers

[misty merlin]

So 15 - 15 = 0?

[last timber]

just reread this sentence carefully to get the idea

[misty merlin]

Thats correct but he used that to do the induction

Like I don't get how he got the -15

[last timber]

well, you can use anything for the induction if it is valid and helps

[misty merlin]

It seems he just got it from no where

[last timber]

you can take bounds, estimates, etc. literally out of your ass as far as they are valid

but for induction he did it to get form depending on inductive assumption

[misty merlin]

So he actually got it from his ass?

To solve the problem

[last timber]

ye

he states it here

[misty merlin]

I get the 15 x 15^k+1 is equivalent to 15^k+2

[last timber]

i mean it is natural to make something like b=b+a-a which can simplify a lot

and in the induction it often helps to relate P(n+1) with P(n) which is assumed to be true

like as you can see then he gets nice form which then proves the claim

[misty merlin]

How would I explain how he got -15 just from "flash inspiration"? Because that picture is a hint from him that I tempted to use cause I managed to solve the actual question from the second last line in the picture

It's just I can't explain and understand the -15

[last timber]

How would I explain how he got -15 just from "flash inspiration"? Because that picture is a hint from him that I tempted to use cause I managed to solve the actual question from the second last line in the picture
@misty merlin no need

i mean no need to explain how you get estimate if it is correct

[misty merlin]

He provided a template for us to use, and in the induction step we have to explain how we got from this line to this line

Different question to the one im asking

Actually ill just write 'to relate to 241', thanks

[last timber]

wait lol

am i right that you are confused by this line

then it is just algebra

i mean he just collapses expression here

[misty merlin]

no no the this one

How he got -15 or 'subtract 15'

Sorry to bother if ur busy*

[last timber]

ok so i was just working with expressions rn

to make it clear

ok so now

in order to simpllify

Let P(n) stand for the expression which is to be proven to be divisible by 241

i mean this

(P(n) is diff from P(k) just by notation)

we are assuming for the induction that P(n) is true

and want to prove P(n+1)

@misty merlin is it clear for you that if a is divisible by m, and b-ca is divisible by m, then b is divisible by m?

(where c is arbitrary integer)

[misty merlin]

I have no idea what u just said in the last 2 lines

[last timber]

i mean

let a be just any integer

and let m be fixed

and also, we want a to be divisible by m

is it obvious that if c is integer

then ca will be also divisible by m?

[misty merlin]

yes....?

cause any integer i can use

certain ones will work

[last timber]

?

[misty merlin]

ca can be any integer and some of them can be divisible by m

[vital dew]

Commander Vimes:

Commander Vimes:

Commander Vimes:

Commander Vimes:

[misty merlin]

ok

[vital dew]

Commander Vimes:

[last timber]

it is not hard to prove, but if you want you can just accept it

[misty merlin]

I'll just accept it cause this whole induction really screwing my head

I just want to figure why the -15

not why where*

[last timber]

in words it means that

If n divides a and difference of b and multiple of a, then n divides b

now i can explain

Let P(n) stand for the expression which is to be proven to be divisible by 241
@last timber recalling this

what your teacher does

he estimates difference

P(n+1)-15P(n)

if you will then expand the last expression he gave and move 16^(2k-1) out of brackets

you will get

[vital dew]

Commander Vimes:

[misty merlin]

I know that i'm trying to get it to be 241, i got down to 16^2-15

[last timber]

16^2=256

[misty merlin]

yeah and -15 is 241

[last timber]

yes

that was the whole purpose of this -15

[misty merlin]

But how did he get -15 in the first place

just by guessing/estimating

[last timber]

prolly he did write it backwards

i mean he knew that this would allow him to get convenient form

[misty merlin]

So he just got that number out of no where and shoved it to get teh solution

shoved it to make the working out conclude at the solution*

[last timber]

well, i guess he firstly worked with estimates and then formalized the proof

it is how proofs are often done

we first work with estimates and see whether this action helps

and then formalize

[misty merlin]

This induction stuff really pisses me off

apologies

[last timber]

(when you will do analysis you will often meet proofs where they take some estimates which are from the first look taken from nowhere)

[misty merlin]

But thanks tho 🙂

[last timber]

yw

[long bronze]

Hi can someone help me to check if this is correct 😄

(x_1,x_2 )⪯(y_1,y_2 )  if and only if x_1≥y_1  and x_1+x_2≤ y_1+y_2  
Prove that ⪯ is a partial order.

For a relation to be a partial order it must satisfy three things and they are
    It must be reflexive (a≤a)
    It must be antisymmetric (a≤b) and (b≤a) then (a=b)
    It must be transitive (a≤b) and (b≤c) then (a≤c)

    Let us prove if it is reflexive.

Suppose (x_1,x_2 )≤(x_1,x_2). 
This means that x_1≥x_1 and x_1+ x_2≤x_1+ x_2  
∴ this is reflexive

    Let us prove if it is antisymmetric.
Suppose (x_1,x_2 )≤(y_1,y_2) and (y_1,y_2 )≤(x_1,x_2). 
Then x_1≥y_1 and y_1≥x_1
Thus, x_1= y_1

Similarly, x_2≥y_2 and y_2≥x_2
Hence, x_2= y_2

Since, x_1= y_1 and, x_2= y_2
      ∴ x_1+ x_2≤x_1+ x_2  and hence this is antisymmetric

    Let us prove if it is transitive.

Suppose (x_1,x_2 )≤(y_1,y_2 ) and (y_1,y_2 )≤(z_1,z_2 ).
Then x_1≥y_1 and y_1≥x_1
Similarly, x_2≥y_2 and y_2≥x_2
∴ (x_1,x_2 )≤(y_1,y_2 )

Then y_1≥z_1 and z_1≥y_1
Similarly, y_2≥z_2 and z_2≥y_2
Thus,(y_1,y_2 )≤(z_1,z_2 )
∴(x_1,x_2 )≤(z_1,z_2 ) and hence this is transitive.

Since this relation is reflexive, antisymmetric and transitive
∴ it is a partial order.```

[last timber]

nvm

ah no

nvm (2)

@long bronze

Suppose (x_1,x_2 )≤(y_1,y_2) and (y_1,y_2 )≤(x_1,x_2). 
Then x_1≥y_1 and y_1≥x_1
Thus, x_1= y_1

Similarly, x_2≥y_2 and y_2≥x_2
Hence, x_2= y_2``` prolly to show where ineq comes from

Then x_1≥y_1 and y_1≥x_1``` why it is in transitivity

ok lol proof with transitivity is much unclear

[long bronze]

To be honest I am not sure how to write the proof for transitivity😥

[last timber]

well look

you want to prove that

[vital dew]

Commander Vimes:

Commander Vimes:

[last timber]

@long bronze look, i want you to try to prove the transitivity on the second coordinate by urself

hint: no need it splitting the sum

[long bronze]

So will the second coordinate be something like this?
x_2 ≥ y_2 and y_2 ≥ z_2 and hence x_2 ≥ z_2

[last timber]

have you read the hint?

i mean reread how your order is defined

(x_1,x_2 )⪯(y_1,y_2 ) if and only if x_1≥y_1 and x_1+x_2≤ y_1+y_2

i have shown that (x_1, x_2) <= (y_1, y_2) and (y_1, y_2) <= (z_1, z_2) implies that transitivity holds for first part

i.e x_1≥z_1

i want you to prove that second condition is also true

[gritty crescent]

What is the contribution of a loop to the degree of a vertex?

[last timber]

@gritty crescent are we in directed or undirected graph

[gritty crescent]

I suppose undirected.

[last timber]

then two

+2

[gritty crescent]

Makes sense, thanks!

[last timber]

in directed it adds 1 to outer degree and 1 to inner degree

[gritty crescent]

Was trying to prove the theorem that for a finite group X, the sum of degrees of all vertices of X is equal to the twice the number of edges, so loop should've been contributing two but I wasn't sure.

in directed it adds 1 to outer degree and 1 to inner degree
I see. Thanks again!

[last timber]

yw

[gritty crescent]

Does this proof look okay?

[last sigil]

Yeah, looks fine to me

[gritty crescent]

Nice.

Thanks for taking a look!

[weary tiger]

Prove that for every positive k, the following is true:
For every real number r>0, there are only infinitely many solutions in positive integers to 1/n1 +...+1/nk =r

Does anyone know how to go about this question?

Ig we will prove it by induction on k

But I'm not sure how to

[stray reef]

"only infinitely many"

[weary tiger]

Oh I'm sorry mb

only finitely many solutions*

[stray reef]

yeah that's more like it

hmm

ok so here's my idea: first, consider only solutions where the denominators form a nondecreasing sequence

every other solution is a permutation of one of those

now denote by $S_{k,d}(r)$ the set of all solutions to the following system: $$\begin{cases} \frac{1}{n_1} + \frac{1}{n_2} + \dots + \frac{1}{n_k} = r \ d \leq n_1 \leq n_2 \leq \dots \leq n_k \end{cases}$$

[vital dew]

Ann:

[weary tiger]

oh ok

[stray reef]

im not done here, there's a bunch more stuff i want to state as lemmas

[weary tiger]

alright

[stray reef]

denote by $S_k(r)$ the set of all solutions to the above system but without the constraint $n_1 \geq d$

[vital dew]

Ann:

[stray reef]

actually, hold on. $S_k(r)$ will then just be $S_{k,1}(r)$. whatever.

[vital dew]

Ann:

[stray reef]

no wait

ugh nvm i've confused myself

[weary tiger]

oh

[stray reef]

to be clear i AM going with the idea of an induction proof

[weary tiger]

ah ok

[stray reef]

okay so we have those notations established.

so the first thing i will say is: if $d > k/r$, then $S_{k,d}(r) = \rien$.

[vital dew]

Ann:

[stray reef]

this is gonna be the key to finitude so i wanna make sure it makes sense to someone other than me.

S_k,d(r) is the set of all (sorted) solutions to 1/n_1 + ... + 1/n_k = r with the additional constraint that n_1 (and by extension, all n_j) are greater than or equal to d.
because the n_j form a nondecreasing sequence, the terms 1/n_j are nonincreasing, and in particular 1/n_1 is the biggest.
if d > k/r then 1/n_1 ≤ 1/d < r/k, and so all the subsequent terms in the sum are also less than r/k, and thus the LHS is less than r and we have no solutions.

does this make sense?

i am not continuing until i get a confirmation that it makes sense

[weary tiger]

Yeah sorry I'm on the last part 1 sec pls

does this make sense?
yes

[stray reef]

aight good

so now

$S_1(r)$ is clearly either empty or a singleton depending on whether or not $r$ is of the form $1/n$.
and also $S_k(r)$ is empty for $r \leq 0$.

[vital dew]

Ann:

[weary tiger]

ok

[stray reef]

now assume we've already proved $|S_{k-1}(r)| < \infty$ for all $r$.

oh dear, the bot's dead.

this is basically my IH tho

[weary tiger]

oh ok

[stray reef]

i claim that $S_k(r)$ is equipotent with $\bigcup_{d=1}^{\infty} S_{k-1,d}(r - \tfrac{1}{d})$, and i will describe my bijection as such: each solution $(n_1, n_2, \dots, n_k) \in S_k(r)$ gets sent to $(n_2, n_3, \dots, n_k) \in S_{k-1,n_1}(r - \tfrac{1}{n_1})$.

[vital dew]

Ann:

[weary tiger]

Uh I'm not sure about that U shape

[stray reef]

infinite union

[weary tiger]

Oh ok

[stray reef]

have you seen the notation for infinite series?

it's like that but with unions, kinda.

the defn doesnt actually involve limits but intuitively it's like that.

[weary tiger]

Oh I'm just familiar with the sum notation

[stray reef]

yeah like take the sum notation and replace the terms with sets and the addition with union and you'll get a good intuitive idea of what im talking about

[weary tiger]

ah ok

[stray reef]

anyway, by my lemma, $S_{k-1, d}(r - \tfrac{1}{d})$ is empty whenever $d > \frac{k-1}{r - \frac{1}{d}}$

[vital dew]

Ann:

[weary tiger]

ah okay I guess I got it

[stray reef]

and this inequality actually turns out to be equivalent to d > k/r

via some algebra which i can reproduce but won't unless asked

[weary tiger]

Yeah

[stray reef]

point is, we can cut off the infinite sum and have the upper limit be ceil(k/r)
which will make S_k(r) equipotent with a finite union of finite sets

and hence finite

and hence we'll be done

[weary tiger]

thank you so much!!

[weary tiger]

as sets can be expressed in terms of logic, it is completely legal to convert statements involving sets (intersections, unions, etc.) to logical language and apply known tautologies in logic to come to conclusions about the sets?

e.g. trying to prove transitivity with set relations, converting all unions to logical ORs, all intersections to AND etc. Then showing transitivity holds by logical tautologies

I was convinced this was the case and still am, but we have gotten answers back from an assignment where a seemingly easier solution was available, but in which I did something similar to what I've said above

So want to make sure

[pale epoch]

you can solve questions like "is x in the set X" by using logic, yes

essentially truth tables

but thats often not the easiest way

[weary tiger]

Is Q8 just an induction exercise?

[stray reef]

what's Omega here?

[turbid cargo]

I have seen that Omega is used for the Universal set

but idk

[weary tiger]

what is this triangle

hey there

i have a question

is anyone online to help?

[minor dagger]

its delta @weary tiger

[weary tiger]

ok ty

ty

@stray reef Omega is a sample space or universal set

it looks like infinite intersection can be proved via induction, I don't see any other way (because I guess the statement about real numbers is also an induction proof)

[stray reef]

ok, is it known whether Omega is infinite or not

[weary tiger]

Not known

[stray reef]

the truth of your statement is equivalent to Omega being finite

[weary tiger]

It could finite or countably infinite, but it doesnt specify

[stray reef]

yeah great then assume $\Omega$ infinite for the sake of counterexample, and take a sequence of points $$\omega_1, \omega_2, \omega_3, \dots \in \Omega$$ and let $$A_n = {\omega_n, \omega_{n+1}, \omega_{n+2}, \dots}$$

the A_n form a decreasing chain so the intersection of any finite number of them is just the last set in the chain, but no point belongs to every single A_n at once

[vital dew]

Ann:

[weary tiger]

so infinite intersection will be empty if no point belongs to every single A_n?

like if it was finite A_n will have a point that is embedded in every single one form A_n to A_1

[stray reef]

$\bigcap_{n=1}^{\infty} A_n$ \textbf{by definition} consists of precisely those points which belong to every $A_n$ at once.

[vital dew]

Ann:

[stray reef]

but there is no such point therefore the intersection is empty.

[weary tiger]

cool, thanks!

@stray reef Oh, but there is a related questions for real numbers, surely there there always 'll exist b such that it is larger than infinite sum of a_i ?

[stray reef]

??

oh lol no

take a_i = 2^-i and b = 1

[weary tiger]

oh b doesn't change

i see

makes sense

thanks

[charred forge]

how would i set this problem up

https://s3.amazonaws.com/blackboard.learn.xythos.prod/57c93be5643d6/10075213?response-cache-control=private%2C max-age%3D21600&response-content-disposition=inline%3B filename*%3DUTF-8''ind1.PNG&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20201025T210000Z&X-Amz-SignedHeaders=host&X-Amz-Expires=21600&X-Amz-Credential=AKIAYDKQORRYTKBSBE4S%2F20201025%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=76619552ae5abedb0cbdbbb6b86082c96209673658d19e324f7c1d2fecdfb3e2

would I just plug in (k+1) for both i and n

[nimble cove]

@charred forge proof by induction question?

[charred forge]

yeah

reading teachers notes didnt really help me learn how to set it up lol

[weary tiger]

how can i use leibniz with demorgans law?

im really bad at understanding leibniz

[charred forge]

@nimble cove

[nimble cove]

sry

So the first case is n=0 {the trivial/ base case}

and you have to show that that is true

[lucid marlin]

How is it the case if pj | Q, then Q - p1p2 * ....* pn = 1. What steps were taken to get this? I know you start with Q = p1p2p2 * ... * pn + 1, but don't see how this is true because if pj | Q then you could write it as Q = pj * (m) not Q - p1p2..pn = 1, so could someone explain what is going on

[nimble cove]

then assume the statement is true for some n=k

and show that the statement being true for n=k implies that it is true for n=k+1

by using the assumption

[charred forge]

Yeah i did that first part where i plug in 0

so since n=k

we plug in

k+1 for both sides

or only the right side

[nimble cove]

you plug it in into the left hand side

and show that it equals the statement you get when you plug it into the right hand side

[lucid marlin]

any idea?

[charred forge]

so we start off with something like

4(k+1)^3 =((k+1)+1)^2 * (k+1)^2

and we work our way through the problem so that it equals to the right side?

[nimble cove]

yes

@lucid marlin the idea is that if pj divides Q

then it divides 1

[charred forge]

for sure

ill ask again if i made a mistake somewhere

[nimble cove]

yes

[lucid marlin]

if pj divides Q then Q = pj * m, i don't see how this divides 1?

Q - p1p2p3.. pn = 1 isn't what pj | Q means it means Q = pj * m

so this doesnt make sense

[nimble cove]

consider Q/pj

[lucid marlin]

Q/pj = m

there is nothing about p1p2 * .. * pn where

a | b means b = a k

oops

[nimble cove]

Q = p1p2...pn+1

so Q/pj = some integer + 1/pj

so pj must divide 1

[lucid marlin]

i dont see your steps from Q = p1p2...pn+1 to Q/pj = some integer + 1/pj to pj must divide 1

[nimble cove]

some integer = p1p2p3...pn/pj

it has to be an integer because 1 <= j <= n

[lucid marlin]

i still dont see what youre doing

where does pj come from and why do you have stuff divided by pj

the definition of a | b is b = ak

it doesnt talk about a/b

p1p2...pn+1 to Q/pj <--- dont know where pj comes from and what a/b means

[nimble cove]

if pj | Q then Q/pj must be an integer

like the division

[lucid marlin]

ok

where does pj come from?

are you dividing both sides by pj?

Left Hand Side / pj = Right Hand Side / pj <--- is this what you are doing

[nimble cove]

yes

pj | Q => Q = m*pj

so m = Q/pj

where m is some integer

[lucid marlin]

Q / pj = (p1p2 * ... * pn)/pj + 1/pj <---- i agree with this now, what is your next stemp

[nimble cove]

Q/pj = m (some integer)

because it divides

and pj has to be one of your primes

[lucid marlin]

Q/pj = m + 1/pj

the 1/pj is a remainder? so it doesnt divide?

[nimble cove]

yes

because none of the primes can divide 1

hence no prime that you found divides Q

[lucid marlin]

let m=p1p2..pn

[nimble cove]

Q has to be some primes multiplied together and add 1

9 is not the product of distinct primes

[lucid marlin]

pj divides Q if Q-m = 1, where m=p1p2p3..pn

is what the book is saying

[nimble cove]

no

wait

yes

nvm

[lucid marlin]

lets let 2 *3 *5 * 7=210 and let m=210

what is the book saying here

[nimble cove]

for any of the primes {2,3,5,7}

diving m+1 = Q = 211

by any of those primes is not possible

[lucid marlin]

so if x in {2,3,5,7} divides 211, you'd have x | 211 -> 211 - x = 1?

[nimble cove]

x | (211 - m = 1)

where m = 23 * 57

[lucid marlin]

so you have to check if 2 | (211-210 =1) which means check if 2 |1, 3|1, 5|1, 7|1

i feel like i am close to understanding but still slightly confused

what does it mean for m | a-b, a - b = mk yeah?

[nimble cove]

if m | x

and x = 2m + 3 for example

then m | 3

[lucid marlin]

im not getting what Q - x = 1 is suppose to meant if m | Q - x then that means Q -x = mk

but we dont care what Q-x is, we want to know if pj divides Q

Q = pj * k

so if pj | Q then it should be Q = pj * k

i dont see where the substraction of Q - x is coming from and the 1

from Q = pj* k definition of divides

where x = p1p2 * ... * pn

[nimble cove]

Q = pj*k = x+1

k = x/pj +1/pj

x/pj is an integer because x is the product of all pi

so k - x/pj = 1/pj

which also has to be an integer

1/pj is some integer means pj | 1

I'm not sure how to explain this any better, I hope you understand what's going on

[lucid marlin]

shouldn't Q = pj*k + 1 ?

[nimble cove]

from Q = pj* k definition of divides
@lucid marlin

[lucid marlin]

what does pj*k = x+1 mean? pj * k = x ?

and where is the 1 coming from

no one in pj * k

[nimble cove]

Q = x+1

[lucid marlin]

what is x

[nimble cove]

by the definition of Q

[lucid marlin]

and where is +1 coming from

Q = pj * k

[nimble cove]

where x = p1p2 * ... * pn
@lucid marlin ^

[lucid marlin]

no x and no 1

[nimble cove]

they are 2 separate equations for Q

we equate them because they both equal Q

[lucid marlin]

what is x?

ok i see

x = p1p2 * ... * pn

[nimble cove]

yes

[lucid marlin]

Q = pj*k <--- definition of pj | Q, Q = (p1p2p2 * .. * pn) + 1 = x + 1, where x = p1p2 * ... * pn

i understand the first line now

[nimble cove]

yes

[lucid marlin]

Q = pj*k = x+1
k = x/pj +1/pj --- are these the same ks?

[nimble cove]

yes

pj*k = x+1

divide by pj

[lucid marlin]

okay i see how we get k = x/pj +1/pj because you set pj*k = x+1 equal to each other and divide by pj

[nimble cove]

yes

[lucid marlin]

but what does k = x/pj +1/pj mean

[nimble cove]

what you you mean by that question?

[lucid marlin]

i see how we got the expression k = x/pj +1/pj but dont know what it means or how it helps us prove if pj | Q then Q-x = 1

where x = p1p2..pn

[nimble cove]

k is some integer, do you get this

[lucid marlin]

yes

[nimble cove]

x/pj is some integer

[lucid marlin]

yes'

[nimble cove]

so k - x/pj is an integer

[lucid marlin]

how does k being that integer mean if pj | Q then Q-x = 1

[nimble cove]

wait

do you agree that k - x/pj is an integer

[lucid marlin]

yes

[nimble cove]

so k -x/pj = 1/pj

from the expression above

agreed?

[lucid marlin]

i dont know what expression you mean

im very confusrd

i want to show if pj | Q then Q-x = 1

dont see how none of this does

[nimble cove]

okay i see how we get k = x/pj +1/pj because you set pj*k = x+1 equal to each other and divide by pj
@lucid marlin ^

just wait

that equation

[lucid marlin]

i see how k -x/pj = 1/p now

[nimble cove]

so k -x/pj is an integer

[lucid marlin]

yes

[nimble cove]

and hence 1/pj is an integer

[lucid marlin]

yes

[nimble cove]

let 1/pj = n

for n being an integer

1 = n*pj

which by definition of | means pj | 1

[lucid marlin]

how is 1 = n*pj

[nimble cove]

we let 1/pj =n

where n is some integer

[lucid marlin]

i see k-(x/pj ) = 1/p which means 1/p divides k-(x/pj )

[nimble cove]

yes

it is equivalent to it

[lucid marlin]

i dont understand what you are doing still

i need to see it written out completely

and dont see how to proves if pj | Q then Q-x = 1

[nimble cove]

its proving if pj | Q then pj | 1

I've got to go sleep, I hope someone else can come and explain this to you

[lucid marlin]

okay, thanks for your help