discrete-math

[cerulean ore]

5%6 = 5
10%6 = 4
15%6 = 3
20%6 = 2
25%6 = 1
30%6 = 0

That's why?

[faint narwhal]

Yeah, that's one way to look at it

[cerulean ore]

But, we cannot write {1,5,5^2,..}?

[pale epoch]

why not?

notice that your group operation is addition, not multiplication

[cerulean ore]

So, in this case the definition would be b = a*n where n is an integer?

[faint narwhal]

Oh yeah

When people write a^n, they mean the group operation applied to a , n times

They don't mean a to the nth power

[cerulean ore]

argh

[pale epoch]

it takes some getting used to

but in general multiplicative notation is much prefered

[cerulean ore]

My teacher told* us that definition

[faint narwhal]

Usually additive notation is used if the group is abelian

and multiplicative notation is used if the group isn't necessarily abelian

[cerulean ore]

Any specific reason why?

I am just 4 hours old in group theory ;-;

[faint narwhal]

convention mostly

And the fact that addition of things is usually a commutative operation

Whereas multiplication, like multiplication of matrices for example

Is not always commutative

[cerulean ore]

That looks good

[pale epoch]

tbh i'm just lazy and write $gh$ instead of $g \cdot h$, so multiplicative notation is natural

[vital dew]

Lochverstärker:

[cerulean ore]

I think I should stop for today and let the brain absorb the material now

[ripe ferry]

they arent axioms, theyre formula

P(A and B) = P(A) * P(B) when A and B are independent events

it really does answer your question
https://en.wikipedia.org/wiki/Independence_(probability_theory)

[hybrid crown]

They're the same formula
just that " P(A and B) = P(A) * P(B)"
is a special case of the first one, where the cases that satisfy both A and B are non-existent.

[distant hamlet]

What’s the formula to do this?

From 1000-9999, how many numbers are divisible by 5 and not by 7

[reef thistle]

@weary tiger hmm

@distant hamlet Try counting 5, then 5 and 7

@weary tiger okay so each triangle must use 3 different colours

considering picking each triangle in turn

[reef thistle]

@weary tiger pick any first triangle

then consider all possible second triangles

Since colours can be exchanged around, it must be the same number of possibilities for any first triangle

[obtuse lance]

might be able to do it by considering the triangles, I haven't thought it through completely

each triangle has to have each color on every corner, so there's 3! ways to place the leftmost triangle.

then there are 3 ways to place a triangle next to it so that it doesn't clash with colors

again 3 ways for the one next to that

so 3! * 3 * 3

[obtuse lance]

seems like a fun problem to try to generalize

for k triangles of course 3!*3^{k-1} but suppose instead it's a complete graph with n vertices connected in a similar string of k of them at n-1 places

I'd guess the solution is n!*n^{k-1} for that

oh it won't be, where the two complete graphs forms a complete bipartitegraph with n-1 vertices in each partition minus the equal pairs so there are (n-1)^2-(n-1) = (n-1)(n-2) not n

so that generalization would have solution n! * [(n-1)(n-2)]^{k-1}

[cerulean ore]

What is homomorphism?

I have read the definition but, unable to figure out what does it really means

[wraith tiger]

Are you referring to a group homomorphism?

[tranquil cargo]

a homomorphism is a function taht perserves

the structure

preserves*

[cerulean ore]

Yep, group one

[tranquil cargo]

if that function is bijective then its called an isomorphism

thats what ik tbh

[cerulean ore]

how the structure is preserved?

[tranquil cargo]

for groups depends on like the operation

[stray reef]

a homomorphism is a function between two groups that preserves the binary operation

like

[tranquil cargo]

^ listen to ann

[wraith tiger]

Like, say we have two groups $(G, \circ)$ and $(H, \star)$. Then a map $\varphi: G \to H$ is said to be a homomorphism if $\varphi(u \circ v) = \varphi(u) \star \varphi(v)$.

[stray reef]

a function f: G -> H is a homomorphism if f(xy) = f(x)f(y) for all x, y in the domain

[tranquil cargo]

cool af

[wraith tiger]

The idea is to generalize stuff like det(AB) = det A * det B and log(xy) = log x + log y

Composition in the inputs results in the composition of the outputs, that's the structure being preserved essentially.

[tranquil cargo]

i have a question about this ye

[cerulean ore]

Beautiful

[tranquil cargo]

why does this happena lot

why does this happen alot*(

like that condition

of morphisms

why is it like everywere in lower math

lol

[stray reef]

a good exercise is to check that all homomorphisms preserve the identity and inverses

ie the identity of the domain gets mapped to the identity and inverses get mapped to inverses

[cerulean ore]

What book did you guys use to solve questions on group theory?

[tranquil cargo]

ann's problem set

[wraith tiger]

Abstract Algebra - Dummit & Foote has quite nice exercises

[tranquil cargo]

i think just use youtube if you cant get textbook imo

like lectures or whatever

or just look for tests on google

[cerulean ore]

@stray reef Could you please send your problem set? (If possible, solutions as well)

[stray reef]

i remember linking it somewhere here

[cerulean ore]

I have heard good about Dummit & Foote.

Thank you very much for helping out guys! This server is one of the reason why I am progressing in Mathematics.

@stray reef If you find it, please send. Thanks!

[stray reef]

ok gimme some time

https://discordapp.com/channels/268882317391429632/486844829209198613/611552599551901733

difficulty ranking

@cerulean ore

the more stars the harder the problem is obv

[cerulean ore]

Thank you!

[wraith tiger]

Is 5 stars meant to represent an open problem à la Knuth?

[stray reef]

not really no

5 stars is something a good student of intro group theory would spend at least an hour thinking on

[wraith tiger]

I see

[tidal minnow]

Hello there, I failed discrete math twice in university, so this time I have to get it down.
how can I become a god at discrete math in the next 6 months?

is there a good platform with tons of excercises?

[pale epoch]

what does your discrete math class even cover

it's not like it's standardized

[cerulean ore]

I am on the step of checking if there is existence of identity

a*e = a = e*a

For a belongs to Q
a*e = a+e-ae = a e(1-a)=0

e!=0 since e also belongs to the set Q and Q is the set of non-zero rational numbers

So, the system is not a group, right?

[sour arrow]

Nope, not a group

[cerulean ore]

Aah, thank you!

[wintry rock]

(1) for i=1 to 4n
(2) ---- for j=1 to i
(3) -------- print (i,j); How many times is (3) executed? I'm having a little trouble with this one. I got...

It also says "Express your answer as an efficient formula in terms of n"

just in case that was important

whoops

that first total number was from something else

ignore that

[pale epoch]

tbh i have no idea what you calculated there

is "total number of combinations" supposed to be your answer?

[wintry rock]

yeah it was totally messed up, I did this problem over another problem I did earlier. Hopefully this is a little more clear?

I used the handshake principle

[pale epoch]

your final result probably shouldn't depend on i

[wintry rock]

that's what i was wondering, the online HW won't even let me type i

[pale epoch]

because that makes no sense, the answer only depends on n

as i is only used in the first for loop

so i'm not sure what you're calculating

[wintry rock]

oh

[pale epoch]

for questions like this it's always best to just do each loop individually starting with the inner most

e.g. in the innermost for loop the print statement is executed $\sum_{j=1}^{i}1$ times, can you see why?

[vital dew]

Lochverstärker:

[wintry rock]

yerr, j starts at 1 and goes to i; i = 1 the first time so it's executed 1 time. Then when i = 2 j goes to 2, so 1 + 2 +... etc. right?

[pale epoch]

well, yes

[wintry rock]

but

[pale epoch]

but at that point i wouldn't even think what happens for different i

actually wait

when i=2 this isn't executed 3 times

just the inner for loop

[wintry rock]

well I'm just adding up the total number of iterations. When i = 1 line 3 is executed 1 time. When i = 2 it's executed 2 times. Adding it up it's 3 total iterations so far

[stray reef]

@cerulean ore i'm in the process of writing a solution key to that group theory problem set i posted, if you're interested

[pale epoch]

well, yes but that's for the whole algorithm

i was talking about just the inner for loop

[wintry rock]

so line 1 has no relevance?

[pale epoch]

not right now

first step we just want to know how often the print statement is executed in the inner for loop

at this point the answer can also depend on i

[wintry rock]

if it's j = 1 to i then what's i

[pale epoch]

consider it fixed

just as you would consider n for the outer loop

[wintry rock]

yeah

so... line 3 is executed i times?

[pale epoch]

yeah

exactly

so every time the inner for loop runs, line 3 is executed i times

[wintry rock]

right. and line 3 is executed i times 4n times?

[pale epoch]

nah

not so fast

[wintry rock]

oh okay

[pale epoch]

so the outer for loop basically just does one thing

it executes the inner loop

and the outer loop runs from i=1 to 4n

and every time it runs, it executes the inner for loop and line 3 is run i times each time

[wintry rock]

right

[pale epoch]

so in the first iteration of the outer loop, you get line 3 run 1 time, then 2 times, then 3 and so on

[wintry rock]

mhm

[pale epoch]

until the last iteration, then it's executed 4n times

because that's how long the outer loop runs

[wintry rock]

ohhh

so when adding up the total iterations it's 1 + 2 + 3 + ... + 4n?

[pale epoch]

yes

and this you should be able to solve

[wintry rock]

should I use the handshake principle in this case too?

ah. (4n)^2 / 2 ?

[pale epoch]

i am not sure what you mean by handshake principle

[wintry rock]

err it's how my professor and my book came to the formula n(n - 1) / 2 sorta

hold on

n - 1 at the top right would be 4n in this case i think

[pale epoch]

yeah, that's what you should use

(4n)^2 / 2 is not correct though

[wintry rock]

was it close?

[pale epoch]

eh, yes

the general formula is $1 + 2 + \dots + n = \frac{n(n+1)}{2}$

[vital dew]

Lochverstärker:

[pale epoch]

and you correctly observed, that you have to replace n by 4n in this case

[wintry rock]

err I can't seem to find an error...

i feel really close though

[pale epoch]

you are missing one term

[wintry rock]

• 1?

[pale epoch]

the last 4n

[wintry rock]

i was thinking that lol

so that + 1

[pale epoch]

depends where you want to add the +1

[wintry rock]

((4n)^2 / 2) + 1?

[pale epoch]

no

tbh just use the general formula above

[wintry rock]

oh

oh i get it

thanks

[pale epoch]

or if you look at your image

you missed to add a 4n in the second line

because the idea is kinda to add every term twice

that's why you divide by 2 after

[wintry rock]

ye

[pale epoch]

so you have 1+2+3+...+4n

and add that again backwarda

4n+(4n-1)+(4n-2)+...+1

[wintry rock]

ah that's where I made the error

I did 4n + num but not for the last term, just num

[pale epoch]

then you get 4n+1 = (4n-1)+2 = .. = 4n+1

[wintry rock]

• i mean

[pale epoch]

and there are 4n of those terms

so you get 4n(4n+1)

and because you added every term twice, you have to divide by 2

[wintry rock]

right

it's correct thanks

[wintry rock]

how do you write a factorial of a number up to a certain point? Like 76543?

[pale epoch]

$\frac{7!}{2!}$

[vital dew]

Lochverstärker:

[wintry rock]

oh k thanks

[stray reef]

.>

[pale epoch]

reddit is a terrible platform to get help on math homework

[sour arrow]

Gotta love the "how far have you gotten" it weeds out the many

[pale epoch]

yeah

usually the first thing you do when trying to help someone is ask more questions

can't really do that on reddit and if i get no reply on discord it probably wasn't important

also a picture of homework is no question, so i guess the just wanted to share something 🤷

no

it's on my application list for phd

i still got some time until that though, maybe i will change my mind

but it's probably best german university for math

[weary tiger]

Hello

I wanna see if I'm understanding reflexive relations and symmetric relations correctly.

so, let's say you have the set

x = {a,b}

to have this symmetric, you do:

(a,b), (b,a) 

right?

and to have reflexive and symmetric you do:

(a,a), (b,b) (a,b) (b,a)

?

[faint narwhal]

For symmetry, you don't necessarily need those

A better example might be on the set {a,b,c}

The relation {(b,c), (c,b)} on {a,b,c} is symmetric

[weary tiger]

oh gotcha right right, but to make that reflexive, I would have to do (a,a) (b,b) and (c,c) right?

[faint narwhal]

Correct

[weary tiger]

Ok cool so I would only need to do (b,a) if I already set (a,b)?

otherwise, no need?

[faint narwhal]

Yeah

[weary tiger]

sorry for making it sound hard lol, just making sure I'm 100% with it.

thank you!

[stray reef]

the empty relation is symmetric

[weary tiger]

how is this set transitive and symmetric?

nvm symmetric makes sense

actually nvm it makes sense

[weary tiger]

Can anyone explain what determines if a set is anti symmetric?

[pale epoch]

@weary tiger this talks about relations, not sets

i guess a relation is a special kind of set

but it doesn't make sense to say a set is antisymmetric

also your book/notes should have a definition

[weary tiger]

@pale epoch yea relation, sorry

It does have the basic definition, just wondering if anyone could give an example and explain how it works the way it does

Book just says if a is related to b and b is related to a then it is anti symettric is a=b

[pale epoch]

an example would be less than or equal

on the natural numbers

if you pick any two natural numbers n, m

and you have $ n \leq m$ and $m \leq n$, then $n=m$

[vital dew]

Lochverstärker:

[pale epoch]

because a number can't be both smaller and larger at the same time

[weary tiger]

Ooo

I see

[pale epoch]

how did you even get 720 ways to arrange the lamps

this already seems wrong

because the lamps of the same color are indistinguishable

ok, yeah, my bad

[weary tiger]

Hey

[stray reef]

:?

what's your q

wait what

is the coin fair

,w C(7,4) (1/2)^7

[vital dew]

Display results online and refine query

Requested by Ann#0413

[stray reef]

4 heads on 7 flips does not happen with probability 1/2

yes you did

yes now that is actually 1/2

reason what

you can put the configurations of 7 coins with >=4 heads in one to one correspondence with the configurations with <4 heads

by turning all the coins over

so there are as many of the former as of the latter

and as such both events (heads >= 4 and heads < 4) have the same probability bc the individual configurations are equiprobable

you can't do the same thing with "exactly 3 heads on 6 flips"

i mean. (6C3) (1/2)^6 just isn't 1/2. that's it.

...

what

are you trying to get me to give you some sort of Universal Rule™ under which the probability of a certain event is 1/2 or what

there are as many outcomes that are part of the event as there are that aren't

a configuration with k heads can be reversibly transformed into one with 7-k heads by turning each coin over

it's not a "general rule" at all.

it's just a bit of symmetry specific to this problem.

[stray reef]

my group theory problem set, updated and cleaned up
see next message for answer key

@cerulean ore

[weary tiger]

in a set z = {1,2,3},

is

Z U {(1,2)}

Transitive because 1 --> 2 --> 2 ?

and 1 has a direct link to 2

[stray reef]

that doesn't make any sense

{1, 2, 3, (1,2)} isn't even a relation on Z

[weary tiger]

what do you mean

1 and 2 is in the set 1,2,3

Just looking at this and making sure I understand it

[stray reef]

Z is not {1,2,3}.

[weary tiger]

oh sorry

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

[cerulean ore]

@stray reef Thank you very much ann!

[hollow osprey]

could anyone help me start this proof? I tested it for x = 1 and n = 2 and yes, it is true because (1+1)^2 > 1 + (2)(1) or 4 > 3

This is what I am thinking so far
Base Case:
(1+x)^n > 1+nx for all x>0 and n>1

Inductive Hypothesis:
Assume P(n+1) true

Inductive Step
Algebra?

Am i going in the correct path or am I totally wrong?

[faint narwhal]

Look over that base case again

Not sure that's exactly what you want

[hollow osprey]

Would P(n+1) be the base case and P(n) be the inductive hypothesis?

[faint narwhal]

Not at all

[hollow osprey]

That means (1+x)^n > 1+nx for all x>0 and n>1 would be before the base case?

[faint narwhal]

That's the whole statement you're trying to prove

[hollow osprey]

Ok my starting point is (1+x)^n > 1+nx for all x>0 and n>1

Now I need to figure out what my base case is

Maybe the base case is to use x>0 and n>1 to show how its true?

[faint narwhal]

not sure what you're saying here at all

[hollow osprey]

What i'm basically saying is to plug in numbers for x and n

[faint narwhal]

I think you should think about what you want to induct on

[stray reef]

Ok my starting point is (1+x)^n > 1+nx for all x>0 and n>1

Ok my starting point is [the thing you're asked to prove in the first place]

[hollow osprey]

I'll come back later maybe by then I'll have an idea

[hollow osprey]

Dang I have no idea

All I know is to plug in a number then use n+1 somewhere to assume its true

[oak creek]

think of what a base case means again

it would be something like n = 2 or x = 1 or something of that form

hence the term base case

when you use induction, you're trying to show that, from one point, you can take a step past that, and that each step you take holds true.

so: you prove that it holds true at the base case

then, you assume that it holds true for some arbitrary integer

(that's the inductive hypothesis[I.H.])

then you use the I.H. to show that it holds true for one point past that arbitrary integer

[weary tiger]

Use the induction for n then for x

[obtuse lance]

here's a starting point, take (1+x)^n > 1+nx and then multiply both sides of (1+x). Since x is positive, it won't change the inequality sign. See what you can do from there.

[hollow osprey]

Thx guys I think I figured out how to do it

It should be similar to this now

P(2) 2^2 > 2+1

P(k) 2^k > k + 2

P(k+1) 2^k+1 > k+1+2

2k+2 > k + 3

Sorry just writing down some notes

2k+2 > k+3

[trail cobalt]

hi

i need help with a logic gate problem

ive been sitting here stuck on it for the longest time

i am a dead end

So circuits can branch. But each gate can only take in 2 inputs.

this takes more creativity than i can come up with

any guidance i can get for this?

[supple notch]

tf

[trail cobalt]

yea im struggling

i made a truth table

and like 30 different combinations

[pale epoch]

the requirement is basically (p AND q => NOT r) AND (p XOR q => r)

and you can rewrite this just using AND, OR and NOT

which will give you a solution

for example (a => b) is equal to (NOT a OR b)

@trail cobalt

[trail cobalt]

thank u, ill see what i can get from that

I did it! Not with that advice, but thank you anyways!!!

[trim oak]

I've read and read through the book to understand how to answer this, and cannot figure it out. Intuitively it looks like its n^3, because it would be n^3 asymptotically right?

If that's a word

So k = 3...

It feels like I am wrong through, and even if I'm right I don't understand how to get 3 without just looking at it and knowing that n^3 grows the fastest

[faint narwhal]

Just prove from the definition that it is O(n^3)

It might help to note that log n < n

[dapper rose]

then prove anything below 3 won't work

[drifting fulcrum]

Any recommandations on books or resources for discrete math?

Kinda like Strang for linear algebra

[tranquil cargo]

rosen discrete math is the standard ig

but you can find many playlist lectures on discrete math

on youtube cuz like discrete math is popular for some reason

[drifting fulcrum]

Yeah, I found the MIT 6.042J fall 2010 lectures

Thanks 😄

[tranquil cargo]

gl hf

[obtuse minnow]

Insulting Ancient Greek Math ... Dis-Crete Math 🙂

[wintry rock]

I really need help with this one question... how many 7 letter permutations can be formed from 5 identical H's and 2 identical 2's?

[wintry rock]

I just figured out the formula... why do you put the factorials of the numbers of duplicates on the bottom?

[pale epoch]

well, what is the formula you found?

[wintry rock]

whoops

[pale epoch]

$\frac{7!}{5! 2!}$?

[vital dew]

Lochverstärker:

[wintry rock]

n! / r1! * r2! * r3! * etc when n is the number of elements to choose from and r is the number of certain duplicates

yeah that

why are 5! and 2! on the bottom

[pale epoch]

ok, let's go through this on your example

first you answer me why there is 7! on top

[wintry rock]

because when you have 7 elements in a list you have 7 different ones to choose from, then when you pick the first one, 6, 5, 4, etc.

[pale epoch]

ok, yeah

but this only applies when the 7 elements are all unique

[wintry rock]

right

i think

[pale epoch]

so let's assume our 2's in the example are not unique

i will call one of the 2's 2 and the other 2'

then you would consider HHHHH22' different from HHHHH2'2

because our different 2's have swapped place

thus those are different permutations

but in your example the 2's are identical

so we don't want to count those as separate permutations

now if you imagine any permutation of 5 H's and 2 2's

if the 2's are considered different, they can always switch places, to create a different permutation

so in the 7! you are actually counting twice as many permutations as you want

hence you divide by 2! = 2

the same applies to the H's

[wintry rock]

ah

[pale epoch]

because there are 5! ways to arrange 5 H's

and in each permutation you can arrange rearrange the H's and 2's independently to create a "new" permutation

[wintry rock]

got it, makes sense

[loud sage]

How do I simplify this, my book is not helping, @ me when you respond since I'm doing a million things at once

[pale epoch]

well, what have you tried @loud sage ?

[loud sage]

A lot of different stuff that makes it hard to explain through discord

[pale epoch]

that makes it difficult for me to help you through discord, sorry

[loud sage]

Alright, well let me try to put it into words

I've tried ((k+1)(5(k+1)-4))/2 + (2(k+1))/2 in a couple different ways
Working on trying (k(k+1))/2 + (2(5(k+1)-4))/2 right now

[pale epoch]

well

how did you even arrive at the last line

[loud sage]

(k(k+1))/2 + (2(5(k+1)-4))/2 ..?

Using this

[pale epoch]

what

how is this at all relevant to the task

[loud sage]

To try and simplify it?

[pale epoch]

simplify what

i am asking how you arrived at P(k+1) = ...

[loud sage]

Oh

Like what made me use k+1? I'm confused what you're asking

[pale epoch]

there is something written to the right of "P(k+1)="

who wrote that

and why

[loud sage]

Ah

[elfin lagoon]

Is it possible to prove that 2x-3y does not equal 0 if x and y are integers and both x and y do not equal 0?

[faint narwhal]

No because that's not true

[elfin lagoon]

why is it not true?

[pale epoch]

2*3-3*2 = 0

because multiplication is commutative

[loud sage]

I did, I believe I did that wrong, as I did it in my head and hadn't gotten scrap paper out yet, and don't know how I got to there

[pale epoch]

ok, then first try to correct that

because it makes no sense to try and simplify a wrong statement

[loud sage]

well that was the simplified result

but I also believe I did the wrong method there

[elfin lagoon]

if u have an existential statement like (E x,y,z in Naturals) , x , y, z can all be equivalent?

[pale epoch]

then let's start at the beginning, what is P(k+1)?

[faint narwhal]

Yes of course

[pale epoch]

without simplifying it

[loud sage]

What is it? It's the thing we're trying to prove is true

[pale epoch]

yes, but you know what P(n) is

[loud sage]

Isn't that the equation?

[pale epoch]

you know $P(n) = \frac{n(5n-3)}{2} = \frac{5n^2-3n}{2}$

[vital dew]

Lochverstärker:

[pale epoch]

now i want you to write down P(k+1)

because your general goal is to take P(k+1) and then write it down as P(k) + something

so you can use your inductive step

[loud sage]

Wouldn't p(k+1) be basically replacing n with k+1 ?

[pale epoch]

actually wait

i am way too tired for this, sorry

i told you wrong stuff

[elfin lagoon]

"For all x,y,z in integers, if x/y and x/z , then x/(3y-5c)" How would I prove this directly?

[pale epoch]

terribly sorry

[loud sage]

I have no idea what's right and wrong at this point

[pale epoch]

well, i can try again

but i would need you to tell me the first line of your equation

like what did you start with, when you arrived at your $\frac{5k^2+k+1}{2}$

[vital dew]

Lochverstärker:

[loud sage]

Well I did the wrong thing (I think) but I did $\frac{(k+1)(5(k+1)-4)}{2}$ - $\frac{2(k+1)}{2}$

[vital dew]

Rainy:

[loud sage]

When I should have done (again I think)

$\frac{k^2 + k}{2}$ + $\frac{2(5(k+1)-4)}{2}$

[vital dew]

Rainy:

[pale epoch]

well ok, that is both wrong

tbh this question is terribly written

i am not even 100% sure it is asking what it should be asking

[loud sage]

Here is the rest, but I figured this was a given

It also says this

"Prove the statement using mathematical induction. Do not derive them from Theorem 5.2.1 or Theorem 5.2.2."

But these are the only induction theorems in my book...

[pale epoch]

but your goals should be to show that $1+6+11+16+ \cdots + (5k-4) + (5k+1) = \frac{(k+1)(5(k+1)-3)}{2}$

[vital dew]

Lochverstärker:

[pale epoch]

and you are allowed to as a fact that $1+6+11+16+ \cdots + (5k-4) = \frac{k(5k-3)}{2}$

[vital dew]

Lochverstärker:

[pale epoch]

so, because i fucked up earlier i can try walk you through this

(and in the process give you the solution)

(at least mostly)

[loud sage]

If I can understand how to solve it through walking through that's all I need, screw the problem, I just want to understand how to solve this

[pale epoch]

to be honest my suggestion would be to check an easier induction proof first and see if you fully understand that

by easier i mean less notation

i hope your book proved $1+2+\cdots+n = \frac{n(n+1)}{2}$

[vital dew]

Lochverstärker:

[pale epoch]

i mean this is basically the same thing

but more notation

and tbh there are a probably a ton of videos on this, and video form is probably easier than discord

second best to irl

[loud sage]

I understand the basic version

the book did prove that $1+2+\cdots+n = \frac{n(n+1)}{2}$

[vital dew]

Rainy:

[pale epoch]

ok, then let me try to walk you through your current problem and pointing back to this

[loud sage]

kk

[pale epoch]

so for the "simpler" proof

[vital dew]

Lochverstärker:

[loud sage]

Right

[vital dew]

Lochverstärker:

[loud sage]

and and (n+1) to the end of it right?

[pale epoch]

and arrive at $1+2+\cdots+n+(n+1) = \frac{n(n+1)}{2} + (n+1)$

[vital dew]

Lochverstärker:

[pale epoch]

so yes

then we simplified that further

so back to your problem

here we start with $1+6+11+16+ \cdots + (5k-4) + (5k+1)$

[vital dew]

Lochverstärker:

[pale epoch]

and we know that $1+6+11+16+ \cdots + (5k-4) = \frac{k(5k-3)}{2}$

[vital dew]

Lochverstärker:

[pale epoch]

so we get $1+6+11+16+ \cdots + (5k-4) + (5k+1) = \frac{k(5k-3)}{2} +(5k+1)$

[vital dew]

Lochverstärker:

[pale epoch]

this is what in your question is the left side

so if you simplify that further, you get the answer

is this what you tried to do?

[loud sage]

I did this: $1+6+11+16+ \cdots + (k+1) + (5k-4) = \frac{(k+1)(5(k+1)-3)}{2}$

[vital dew]

Rainy:

[loud sage]

Which I see is wrong

[pale epoch]

ok

this is actually your right hand side

$\frac{(k+1)(5(k+1)-3)}{2}$ this

[vital dew]

Lochverstärker:

[loud sage]

Alright

So I did that right

[pale epoch]

and i gave you the left hand side above

i suggest you to look at how i derived it again at some later time

and compare it to the "simpler" proof

anyways

[loud sage]

I see that after 5k-4 on the left side you added 5k+1

Where did you get that from?

[pale epoch]

ah

[loud sage]

I thought it would just be k+1

[pale epoch]

the original statement

it's adding every fifth number

[loud sage]

ooooh.

[pale epoch]

it starts with 1+6+11+16 and so on

[loud sage]

Oh okay

[pale epoch]

so if you look at the statement for "one more number"

you add (5n-4+5) = (5n+1)

[loud sage]

I hate this book because it did all of its examples in cases of ones, so when I get to this, I don't know what carries where

So I'm trying to simplify $1+6+11+16+ \cdots + (5k-4) + (5k+1)$ now

[vital dew]

Rainy:

[pale epoch]

are you learning from a book exclusively?

[loud sage]

yeah, online course was the only available option

[pale epoch]

ok, that sucks

[loud sage]

I agree lol

[pale epoch]

well, i did the simplifcation above for you, but yeah

[loud sage]

Everything else has been easy enough so far

[pale epoch]

this is the "left side" the question asks you to simplify

[loud sage]

Oh the right side of that is the simplification

[pale epoch]

not completely done, but yes

[loud sage]

Yeah I need to convert it to this

[pale epoch]

i would suggest you to walk through it yourself either way

either now or at a later time

[loud sage]

$\frac{k(5k-3)}{2} + \frac{2(5k+1)}{2}$

[vital dew]

Rainy:

[pale epoch]

yes

good

[loud sage]

Then distribute each thing and go from there

kk

[pale epoch]

exactly

as mentioned your "right side" is $\frac{(k+1)(5(k+1)-3)}{2}$

[vital dew]

Lochverstärker:

[pale epoch]

and i think you can take it from there

[loud sage]

so the left equals $\frac{5k^2 - 3k +10k +2}{2}$

[vital dew]

Rainy:

[loud sage]

which simplifies to 7k

Alright, I think I got it now

Thanks!

[pale epoch]

yeah, seems correct

sorry for the confusion at the beginning

i hope i could at least somewhat help now

(if you want extra training you can "invent" similar statements btw, like the sum of the first n odd/even numbers)

[loud sage]

That did help a lot, I have a firm grasp on that concept, but next problem is throwing fractions in there, would it be the same rules?

Same questions, just with this equation

[pale epoch]

the overall idea will always the same

like, you take that sum plus "the next term"

then use the hypothesis to simplify

and arrive at the correct right side

[loud sage]

This one is going in increments of 1 so it should be simpler in that aspect, I was struggling at first but I realize that now

[pale epoch]

what would you say is "the next term"?

[loud sage]

1 over 3*4

[pale epoch]

i mean after $\frac{1}{n(n+1)}$

[vital dew]

Lochverstärker:

[loud sage]

Oh you mean what does it turn into, I'm still working on that

oh wait

it would be n+1 wouldn't it

[pale epoch]

i think i'll let you work on it

and i should probably go to bed, i hope someone else can help you further

[loud sage]

Thanks for the help

[loud sage]

I'm not getting this one, this one has me stumped

[loud sage]

How do I go about this one

I've tried setting the right to $\frac{k+1}{k+2} + (k+1)$

[vital dew]

Rainy:

[loud sage]

I've tried setting it to be $\frac{k+1}{k+2} + \frac{(k+1)(k+2)}{k+2}$

[vital dew]

Rainy:

[loud sage]

I have no idea what I'm doing wrong

[loud sage]

<@&286206848099549185>

[last sigil]

Are you confused on the right side of P(k+1)?

[loud sage]

I'm confused with both, I just have been attempting the right because it seemed easier

If I got an answer right, I could then try to get towards an answer on the other side

[last sigil]

@loud sage

[loud sage]

?

[last sigil]

Add $\frac{1}{(k+1)(k+2)}$ to both sides

[vital dew]

pocofrosty12:

[loud sage]

after converting n to k+1 right?

[last sigil]

No because if you add the fraction above, there is no need to convert n to k+1

[loud sage]

Hm

[last sigil]

You are going from P(n) to P(k+1) not by substituting both sides, but by adding what the next term would be

On the left hand side, it will be

[vital dew]

pocofrosty12:

[last sigil]

Then the right will be:

[loud sage]

So why $\frac{1}{1*2}$

[vital dew]

Rainy:

[loud sage]

That's the first term

[last sigil]

I don't know how to do the ... notation xd

Pretty bad at LaTeX

• should be a multiplication sign

[loud sage]

You say that to $\frac{1}{(k+1)(k+2)}$

[vital dew]

Rainy:

[loud sage]

Meaning the left side? I'm not sure what part the 1 / 1*2 plays in this

[last sigil]

Take the sequence on the left hand side and add 1/(k+1)(k+2)

Property of equality means we can do the same on the right side, so the 2 sides stay equal

[vital dew]

pocofrosty12:

[last sigil]

Which you can simplify algebraically to something

Try to do it

[loud sage]

Wouldn't that be $\frac{k(k+2)+1}{(k+1)(k+2)}$

[vital dew]

Rainy:

[last sigil]

Yep and you can simplify further by distributing the top and factoring that

which is?

[loud sage]

$k^2 +2k+1$

[vital dew]

Rainy:

[loud sage]

And the bottom you can distribute too

[last sigil]

no

[loud sage]

Why not?

[last sigil]

Don't distribute bottom, just factor k^2 + 2k + 1

Pro tip: In fractions with polynomials, it is generally better to leave all the "factors" undistributed so you can cancel them more easily

what do you get at the end

[loud sage]

k+1 over k+2

after factoring

[vital dew]

pocofrosty12:

[last sigil]

What does that look like

[loud sage]

The right side

[last sigil]

Yep and the proof is complete

[loud sage]

Thank you, there's one more that my book doesn't even go over (I will never take an online math class again)

[last sigil]

Which one?

[loud sage]

This

[last sigil]

What did you try for the formula

@loud sage

[loud sage]

For the top one? random numbers with no reasoning as I had no idea what it was asking

[last sigil]

Try doing partial sums

So like what if n=1 (so it only goes to the 1st term), or n = 2(only the first times the second), etc.

Try to look for a pattern

[elfin lagoon]

"For all x,y,z in integers, if x/y and x/z , then x/(2y-3z)" How would I prove this directly?

[last sigil]

is the c supposed to be a z

[elfin lagoon]

yes, my bad

First I assume that x,y,z are integers and that x/y and x/z.

Then... idk. I was thinking about somehow proving that 2y-3z never equals 0 but apparently that's impossible bc its not true

[last sigil]

does x/y mean x is divisible by y

[loud sage]

Would the pattern be $\frac{n+1}{n}$

[vital dew]

Rainy:

[last sigil]

Why?

n = 1 is (1 + 1) -> 2

n = 2 is (1+ 1)(1 + 1/2) -> 2(3/2) -> 3

[loud sage]

Oh right....it multiplies

it would just be n+1 then wouldn't it?

[last sigil]

yep simple formula

But you should check up to n = 5 just to make sure

[loud sage]

See that's super easy, but the way the question asks that leaves me puzzled as to what it is asking

[last sigil]

"Compute the values of the product for small values of n" is basically saying try the smaller cases (like what I did before)

[loud sage]

Hm

[last sigil]

Conjecture is basically to hypothesize a formula

Mathematical language needs to be unique in order to be very precise

Over time you will get used to it

[loud sage]

it also asks for a left and right

but there is no = sign so what determines left or right?

the three ... s?

[last sigil]

Formula I am guessing should be on the right side

What grade/class are you in?

[loud sage]

Discrete math online

Junior in college

This specific question wasn't gone over in the online textbook, leaving me to try and figure it out online without knowing what to search

[last sigil]

... notation just means that it is excluding something, usually just to show that it is infinite sequence

[loud sage]

Like there wasnt anything remotely close to this

[last sigil]

Like if I write 1 + 2 + ... + n, then what I mean is the sum of numbers from 1 to n

[loud sage]

I know what they mean, I meant if they determined which side was left or right

Like if they were the divider

[last sigil]

No, I would never try to think of them as a divider

[loud sage]

Oh I gotcha now, that n+1 is the right side

That just clicked

sorry, I'm exhausted haha

[last sigil]

No problem, I gtg do my hw now so cya

[loud sage]

cya

ty

[elfin lagoon]

@last sigil yes

[elfin lagoon]

is there any way to negate the divides by symbol | ?

or should i just use a tilde/not in front of the expression

[vital dew]

euler/fermat:

[elfin lagoon]

would it be incorrect to use a not operator though?

[torpid pawn]

F(1) = 1
F(n)=nF(n-1) = n(nF(n-2) = n(n(nF(n-3)
n^3F(n-3) = n^kF(n-k)
k = n-1 so n^(n-1)F(n-(n-1))
n^(n-1)F(1) = n^n-1

Assuming F(k) = k^(k-1)
Proving = F(k+1) = (k+1)^k

F(k+1) = (k+1)F(k+1-1) = (k+1)F(k) = (k+1)^(k-1)
Where am I going wrong on this proof?```

hrm

ahh i think I see what you're talking about, lemme work it and hopefully it works, thanks

dang that didn't work either, I get a final step of = 2 for my hypothesis

[torpid pawn]

correct, apparently I was way off base

some kind of factorial, I think

at least according to the book

I think I got it

[elfin lagoon]

How do i prove that one of two consecutive numbers cannot be a perfect square?

To be precise, I need to prove that of two given consecutive numbers, one of the two is not a perfect square.

[faint narwhal]

Or in other words, at most one of them is a perfect square?

[reef thistle]

0 and 1 are consecutive

[faint narwhal]

And I think you want consecutive positive integers

[elfin lagoon]

yes but the claim was presented as i stated

[reef thistle]

which one is not a perfect square?

oh, if positive integers, then sure

[elfin lagoon]

the two numbers given are positive integers, yes

[reef thistle]

Two cases

the higher one is a square

the lower one is a square

then you can figure what happens

[stray reef]

you can rephrase this as "no two perfect squares differ by 1"

which is actually false! as is your original problem. there actually exists a number n such that both n and n+1 are perfect squares.

[reef thistle]

• positive integers

we established

[stray reef]

oh, well then

my other point still stands

"of two consecutive numbers at most one is a perfect square" is equivalent to "no two perfect squares differ by 1"

does that make sense

@elfin lagoon

[elfin lagoon]

yes it does

[stray reef]

are you able to prove the latter statement

[elfin lagoon]

im trying to figure it out

how would i prove that taking the sqrt of x^2 + 1 does not result in an integer ? Can i just say that this is true bc of algebra?

[stray reef]

no

that'd amount to avoiding the question entirely

so no you can't say that

there is a proof of this that doesn't even once mention the square root function

[upbeat minnow]

I'm sorry it's in Finnish, but I don't understand what the 1R∘S2 and others mean here, how am I supposed to use the 1 and 2 there

If anyone can help I'd be grateful but if the language barrier is too much I understand.

[faint narwhal]

Yes. There's a nice symmetry argument here

You're asking why the symmetry in the first problem doesn't hold for the second?

One nice method of solving problems, or thinking about problems, is to see what happens if you change the numbers

For example, you can see that 100 doesn't matter too much here, like any even number would really work right

So what if you do this with the integers from 1 to 2

How big are the groups here

Yeah

It's like saying, either I roll a 1 on my die or I don't roll a die, so it should be 1/2 for me to roll a 1

What do you mean?

There are no other cases

In either your example or mine

Is the probability of rolling a 1 on a die 1/2?

Then what's the logical error

Just think about this example

the die example

yeah

Do you understand what the error in logic is there though

I mean, in the die example

Why not?

Kinda

I think the better explanation is that you do have two separate events

And their probability definitely adds up to 1

But their probabilities don't have to be equal

yep

[oak creek]

please don't double ping

[olive hamlet]

and dont delete one of said pings

dude

[oak creek]

reread #❓how-to-get-help

[olive hamlet]

for fucks sake

[oak creek]

it says once

after 15 minutes

not once every 15 minutes

[olive hamlet]

you have been told this 100 times on this server

you already know this rule

[oak creek]

oh repeat offender?

[olive hamlet]

you ping helpers like 10 times a day, like hell you havent

lmao i have better things to do than fight with you

[weary tiger]

What is this channel for

[wraith tiger]

Discrete mathematics

[cerulean ore]

Guys, if we talk about idempotent property

A is a non-empty set and * is the binary operation then,
for every a belongs to A, a*a = a

doesn't this means that a is equal to e?

[stray reef]

no

there is no e

• isn't required to satisfy any of the group axioms

so there need not be any identity to speak of

[faint narwhal]

Nor are there inverses, which you would need to cancel things out

[cerulean ore]

Aah, got it 😄

Thank you guys! I got Maths exam tomorrow

[cerulean ore]

Are these properties enough for an algebraic structure to be ring?

[faint narwhal]

Depends

The answer is yes usually

[cerulean ore]

on?

[faint narwhal]

But some people demand that a ring has both multiplicative identity

And that multiplicative is commutative

[cerulean ore]

So, which one to follow? The one said by the teacher?

[faint narwhal]

Yeah

Everyone means at least these things when they say ring

[cerulean ore]

Thanks for the short response!

[faint narwhal]

It's just that sometimes people require more

[pale epoch]

sometimes a ring without multiplicative identity is called a "rng"

[cerulean ore]

lmao

[gentle nebula]

pronounced rung

[cerulean ore]

Checking for identity
there is an element e belongs to Q such that
a*e = a i.e. a+e - ae = a e*(1-a) = 0 or e = 0

So, identity element also exists

now, let y be the inverse of a then,
a*y = 0
a+y -ay = 0
y(1-a) = -a
y = a/(a-1)

What if there is "a" that gives an irrational number?

then, y won't belong to Q

[stray reef]

can a/(1-a) ever be irrational if a is rational

[cerulean ore]

I have never thought about that

[stray reef]

and this isn't even the primary concern

what about a=1?

does 1 even have an inverse under this operation?

[cerulean ore]

oops

Nope

So, it is not a group

😄

BTW, if a is rational, will a/(a-1) be always rational?

[obtuse minnow]

except if a = 1

[stray reef]

well, i don't know, will it?

[obtuse minnow]

hushes

[cerulean ore]

not when a=1 but, I don't know about the rest ;-;

[stray reef]

well then why don't you attempt to prove it

let $a \in \bQ \setminus {1}$, prove that $\frac{a}{a-1} \in \bQ$

[vital dew]

Ann:

[stray reef]

if you know a thing or two about the rational numbers, this should be obvious

[cerulean ore]

our teacher said that 22/7 is not rational as the last digits are non-repititive ;-;

[stray reef]

uh

[cerulean ore]

wtf

[stray reef]

what

22/7 is rational by definition

in fact, its decimal expansion has a period of 6

$\pi$, on the other hand, is not equal to $\frac{22}{7}$ at all

[vital dew]

Ann:

[stray reef]

your teacher should be fired for incompetence

people who unironically think π EQUALS 22/7, and hence conclude that 22/7, a RATIO OF TWO INTEGERS as it were, is NOT RATIONAL

are not the kind of people who should be teaching math

[cerulean ore]

fml

[stray reef]

$\frac{22}{7} = 3.\overline{142857}$

[vital dew]

Ann:

[stray reef]

22/7 and pi differ in the third decimal place already

[cerulean ore]

fuck this university

[stray reef]

idk about your uni in general but that teacher in particular is not even nearly qualified

[wraith tiger]

Your teacher said 22/7 is not rational? wtf

[cerulean ore]

What they follow is a local author's book

[stray reef]

does the book also say that

[wraith tiger]

Does the book say 22/7 is not rational

[cerulean ore]

not yet but, it says all other bullshits which our teachers also say

[stray reef]

what does it say

[cerulean ore]

Remember antisymmetric, asymmetric and nonsymmetric?

They all are same according to the book

[stray reef]

hkdkhjfkg

yikes

this one isn't as massive a thonker

but still

[cerulean ore]

I don't read that book else I could've told something spicier :3

such as the order of a finite group is defined as the no. of elements in G

[stray reef]

that one's true, though

the order of a group is just the number of elements in it

[cerulean ore]

distinct elements?

[stray reef]

...i fail to see your point?

why would you count an element twice

[cerulean ore]

so, my teacher is more incompetent than the author. Yay!?

[faint narwhal]

I mean it's pretty rare that you have a teacher that's better than the author of the textbook

[cerulean ore]

The way they deduct my marks, makes it impossible for me to pursue MS in Mathematics in future on scholarship.

[obtuse minnow]

That is a travesty.

You should try to keep the scholarship you have and also apply to other scholarships in case your original one get dropped. Sorry to hear that.

[slender skiff]

Hello, I'm pretty lost on this exercise, what am I supposed to do? and how? (is It determine the value of a and use it to calculate c? if so how do I do this? )

[faint narwhal]

You know the value of a (mod 13), that's all you need

[slender skiff]

Not sure what you mean by value of "a (mod 13)"

How is that a "value"

a (mod 13) is the same as saying a mod 13 isn't it?

how can a mod 13 have a remainder ?

[faint narwhal]

Yes it's the same

[slender skiff]

But isn't their multiple values of a (mod 13)

depending on a

[faint narwhal]

But they're telling you what it is

[slender skiff]

well by writing a ≡ 4 (mod 13) they are saying a mod 13 = 4 mod 13

i don't see how they give the value of a

[faint narwhal]

It doesn't

"You know the value of a (mod 13), that's all you need"

[slender skiff]

hmm

that was dumb

So I have to find the value c that makes the statement c ≡ 9a (mod 13), true?

[faint narwhal]

Yes

[slender skiff]

aka determine the value of c that makes c (mod 13) = 9a (mod 13) true

okay

but what I'm not getting is, how can c mod 13 be true?

by doing for example 2 mod 4

you get 2

so I have to match the remainders on both sides of equals?

[slender skiff]

So here is what I did:

To get the values a, a = k * n + b = k * 13 + 4

for when k is 0,1,2,3

a would be {4,17,30,43....}

9 * 4 = 36

because its mod 13, 13 * 2 = 26, 36 - 26 = 10

then the answer is 10

Can anyone confirm this? 😄

[pale epoch]

it's fine

[slender skiff]

@pale epoch What I don't get is, that by doing this we get the remainder don't we?

[pale epoch]

remainder of what

[slender skiff]

9a (mod 13)

aka

9a mod 13

[pale epoch]

you calculated the remainder of 9a when dividing by 13, yes

[slender skiff]

but if 10 is the remainder

[pale epoch]

(that's what the question is asking)

[slender skiff]

wait really

so by writing the notation a ≡ b (mod n)

its asking to calculate the reminder of b (mod n)

which is a

[pale epoch]

not really

in general there are infinitely many solutions

[slender skiff]

that's what I though

[pale epoch]

but the question sepcified c in a range

if that wasn't the case 23 would also be a solution

and -3

a ≡ b (mod n) means that a and b have the same remainder when dividing by n

[slender skiff]

yea I ge thtat

[pale epoch]

or more formally, that a-b is a multiple of n

[slender skiff]

but since a ≡ b (mod n) is a mod n = b mod n

aren't we signing a value to the variable c?

because c is a number

[pale epoch]

what do you mean

c is a number

[slender skiff]

so c is mod n, and a*k mod n

currently by getting 10

we found the reminder of ak mod n

does that mean that the remainder of c mod n

in a)'s case

is also 10

[pale epoch]

if I understand you correctly yes

but the question says c is between 0 and 12

and there is only a single number in that range with remainder 10 when divided by 13

[slender skiff]

I think the thing that is confusing me is ≡, in normal programming we would declare a variable with equals(=), but here it is using ≡ to say that the value of c is something

[pale epoch]

it's just notation

[slender skiff]

the symbol itself

isn't declaring c to anything

[pale epoch]

why should it

neither is the equals sign in equations

this isn't programming

there are no declarations

the notation a ≡ b (mod n) just means that a-b divides n

[slender skiff]

so by the definition of the notation, the reminder of ak (mod n) is c?

[pale epoch]

well, by the definition of the notation c and 9a have the same remainder when divided by 13

[slender skiff]

oooooohhhhhhhh

so by calculating right sight

we automaticly know

the other side

[pale epoch]

yeah

[slender skiff]

ahhhhhhhhhhh

because they are the same

[pale epoch]

well, at first step you only know the remainder of c divided by 13

so it could be 10, 23, 36 and so on

because they all have the same remainder when divided by 13

[slender skiff]

yea

but because of the 0 < c < 13 thingy

it can only be 10

[pale epoch]

yeah

[slender skiff]

omg thanks!

I think I get it now 😄

Just to confirm that I know this, for c ≡ 11b(mod13).

it would be 8 right?

because b = k * 13 + 9 = {9,22 .....}

11 * 9 = 99

13 | 99 = 8

[pale epoch]

your solution is fine

[slender skiff]

GGGGREAT

Omg I'm so happy! Thank you so much @pale epoch

[pale epoch]

13 | 99 = 8, this notation is ugh though imo

you're welcome

[slender skiff]

should it be 99 mod 13 = 8?

isn't | used to show the remainder aswell ?

[pale epoch]

not that I know of

[slender skiff]

oh

that's divisble

[pale epoch]

yes

[slender skiff]

if 13 is divisible to 99

ahh okay

thanks!

[pale epoch]

9 | 99 is true bcs 9 divides 99

13 | 99 is not true

[slender skiff]

and 9 | 99 is true because. 99 / 9 = 11 (integer)

99 / 13 = not an integer

therefor not divisible

[pale epoch]

yea