Question

I don't understand the following relationship:

Let M be a relation on integers such
that (i, j) ∈ M if i ≡ j(mod 5)
(0,0) ∈ M, (0,5) ∈ M, (5,0) ∈ M, (0,10) ∈ M
(1,1) ∈ M, (1,6) ∈ M, (6,1) ∈ M, (1,11) ∈ M
(2,2) ∈ M, (2,7) ∈ M, (7,2) ∈ M, (2,12) ∈ M
(3,3) ∈ M, (3,8) ∈ M, (8,3) ∈ M, (3,13) ∈ M
(4,4) ∈ M, (4,9) ∈ M, (9,4) ∈ M, (4,14) ∈ M

It then says:
[0]M = {...,−10,−5,0,5,10,15,...}
[1]M = {...,−9,−4,1,6,11,16,...}
[2]M = {...,−8,−3,2,7,12,17,...}
[3]M = {...,−7,−2,3,8,13,18,...}
[4]M = {...,−6,−1,4,9,14,19,...}
It doesn't work though, for instance [1]_m =/= -9=-4(mod 5)

Answer 1

looks like your getting into cosets, or partitions

Answer 2

you need to get the mod number above 0, add another 5 to it

Answer 3

-9 = 1 mod(5)

-9 = [1]

Answer 4

[1]M = {..., -9, -4, 1, 6, 11, 16, ...}
^ ^

-4 and 1 are both partitioned into the same set:

-9 = [1]
-4 = [1]
1 = [1]

Answer 5

wait -9 = 1 mod(5)? I thought a relationship would be
a Rb meant (a,b)∈ R therefore each numbers in the pair would be something like:
-9= -4 (mod 5) should leave an answer of 1?

Answer 6

but -9 = 1 ( mod 5) should give you a 2, no ? -9 - 1 / 5 ( unless I got this wrong)

Answer 7

you have your set of integers partitioned into equivalence classes

by convention, you (mod n) a value so that the remainder is between 0 and n

-9 = a (mod 5) gives us a value of a that is between 0 and 5

-9 = 1 (mod 5)
-4 = 1 (mod 5)
1 = 1 (mod 5)

therefore -9 (mod 5) = -4 (mod 5) = 1 (mod 5)

Answer 8

another way to look at it is to divide the integers like this

...
-10 -9 -8 -7 -6
-5 -4 -3 -2 -1
0 1 2 3 4
5 6 7 8 9
...

each column is an equivalence class.

Answer 9

I thought the modulo was the congruence of two numbers to their remainder ( right ? haha my teacher doesn't explain well) so -9 - 1 / 5 is congruent to 2 but -4 -1 /5 is congruent to 1.

But if I look at the first set, the equivalence class is 0 as -5 = 0 ( mod 5) will give you a remainder of 0. Unless I'm seeing this all wrong.

Answer 10

-9 = 5(-2) + 1

Answer 11

you need to find an integer k, such that 5(k) is smaller than -9

Answer 12

then the remainder defines the equivalence class by convention

Answer 13

and in my equivalence class, will be defined by the remainder that any given pair will use right?

Answer 14

the remainder will be in the same equivalence class.

-9 = 5(-1) -4
^^ remainder

-9 = -4

they are in the same class, the same partition, but we have named the partitions as:

[0] , [1] , [2] , [3] , [4]

there is no [-4] partition so we need to establish what sort of remainders work for us, namely the remainders from 0 to 4

Answer 15

Oh I see, the remainders aren't the equivalence class, but we decided to use them and we have to
"fit" any other number included in our set in the equivalence class.

Answer 16

since -9 = -4, where does -4 reside?

-4 = 5(-1) + 1
^^
-9 and -4 are in the partition [1]

Answer 17

correct enough lol

its best to find some integer multiple of 5, that is just smaller then the number we are modding for.

negative number get smaller as the get 'bigger' :)

-9 is bigger than -10

-9 = -10 + 1

-9 = 5(-2) + 1

-9 = 1 (mod 5)

Answer 18

For that last example, you used k = -2 but that k resides in another equivalence class, are you allowed to do that?

Answer 19

-4 = -5 + 1

hence: [1] = 5(k) + 1

Answer 20

Im still trying to wrap my head around it but we're going for a = b ( mod 5 ) to a = 5(k) + 1 type thing, so I'm not seeing where the k comes from. Sorry for not understanding fast :/

Answer 21

its an integer multiplier, and yes there is some connection to it that you ahvent gotten around to discussing yet.

will any [2] equivalent work?

-9 = 5(5(3)+2) + a
^^^^^
its in [2]

-9 = 5(17) + a

-9 - 5(17) = a

-94 = a

is -94 in [1] ?

-94 = 5k + 1

-95 = 5k, it is :)

Answer 22

you can show later on that mod partitions form groups, and groups can be algebraically manipulated.

Answer 23

Oh I think I understand your point with the k. If I did -23 = 5k + 2, then 23 would be in the equivalence class of 2

Answer 24

only if -23 - 2 is an integer multiple of 5

-25 = 5k is valid so yes

Answer 25

-23 would be in the equivalance of [2]

Answer 26

23 - 2 = 21, which isnt an integer multiple of 5 :)

23 is not in the equivalence of [2]

Answer 27

So what does that have to do with the modulo. If I wrote -23 = 2(mod 5 ) , then the group (-23,2) would be in equivalence class 2?

Answer 28

yes

Answer 29

-23*2 = 2*2(mod 5)

-26 = 4 (mod 5)

(-26,4) is in [4]

Answer 30

-46 not 26 .... i forgot how to multiply lol

Answer 31

ahhh I see. But then, lastly, why are our equivalence class 0-4? Because those are not remainders, but more like integers to which if a number is added / subbtracted, it will be in the set of that number. So for instance Integer 5 will be in set 0 because 5 - 0 / 5 = 1, or am i seeing this wrong still

Answer 32

we have to have some convention whereby we can organize the data.

5k + 0
5k + 1
5k + 2
5k + 3
5k + 4

these are all the possible remainders since k can be determined quite readily such that all integers fall into one of these 5 partitions

Answer 33

are job is not to find some remainder that fits, but some k value that allows us to fit an integer into one of these 5 possible categories

Answer 34

-9 = 5k + (0,1,2,3,4)

-9 - (0,1,2,3,4) = 5k

-9-1 allows us to determine k

Answer 35

Oh I see... for that part haha. for me, remainders are ( like in programming) 5 % 4 = 1 because 4 * 1 + 1. so 5 - 4 = 1 or for instance 10 % 4 = 2, as 10 - 4(2) = 2 . In our case of b = a(mod 5) then a-b / 5 = 0,1,2,3 OR 4. If it equals to 0 ---> equivalence class of 0 for (a,b) and so on. This is what I'm understanding

Answer 36

(a,b) in [n] if

a = 5k + n
and
b = 5k + n

for some n=0,1,2,3,4

-9 = 5(-2) + 1
-4 = 5(-1) + 1

(-9,-4) = [1]

Answer 37

the value of k only needs to be some integer that makes it work

Answer 38

im sure youve got the idea :)

Answer 39

can you tell me if we partition the integers into groups of 8

93 = 8k + n, what is our equivalence class [n] going to be?

Answer 40

Oh I see where youre getting at, it just confuses me as my notes use modulus and you us k. K being the number b and the number on the left being the number a, hence
a = 5(b) + 1
a = 5(b) + 2
a = 5(b) + 3
a = 5(b) + 4 so is this right? and it will be... 5? as 93 % 8 = 5 , no?

Answer 41

yes :)

93 is between 88 and 96

take the smaller value so we can 'add' n to it

93 = 88 + n

n = 5, so 93 = 5 (mod 8)
---------------------------------

(a,b) is not to be thought of as:

a = 5b + 0
a = 5b + 1
a = 5b + 2
a = 5b + 3
a = 5b + 4

a and b are not related that way

Answer 42

(a,b) in [n] if and only if

a = 5k + n
b = 5j + n

for some integers k and j

Answer 43

the gcd of a and b follows something similar to your idea, but we arent finding the gcd

Answer 44

ah i see . thanks so much for your patience. id give you much more medals if i could !!!

Answer 45

your welcome

if youve covered gcd

gcd (-9,-4) follows an algorithm

-4 = 1(-9) + 5
1 = 0(5) + 1 <-- gcd = 1
0 = 1(0) + 0

Answer 46

we did learn it, id have to go over it again, discrete math is complicated at first :/ thanks a lot :) ( again)

Answer 47

not sure if that was wht you were confusing this with or not

Answer 48

good luck :)