Question

Need help with a discrete math problem.

Answer 1

For integers m, \let mZ = {n \in Z : m | n}. Show that 3Z \eta 5Z = 15Z.

Answer 2

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Answer 3

lol I can't get the it written correctly

Answer 4

For integers m, \let mZ = {n \in Z : m | n}. Show that 3Z (upside down U) 5Z = 15Z.

Answer 5

The Z is also suppose to be all the Real numbers

Answer 6

\[let~ mZ = \{n \in Z : m | n\}. \text{Show that 3Z } \eta \text{ 5Z = 15Z} \]

Answer 7

so, in R not Z ?

Answer 8

wait no the set of integers, sorry, so that is right

Answer 9

what is the eta ?

Answer 10

did you mean to and them?

Answer 11

I was trying to do the upside down U

Answer 12

and = \cap
or = \cup

Answer 13

\[and:~\cap\]
\[or:~\cup\]

Answer 14

\cap

Answer 15

\[cap\]

Answer 16

so, show that mod3, anded with mod5 is equivalent to mod15 ....

Answer 17

well since 3 and 5 have a LCM of 15 ....

0 3 6 9 12 15 18 21 24 27 30
0 5 10 15 20 25 30
^ ^ ^

Answer 18

you might have to redress the setup ....

Answer 19

let mZ = {n in Z : m|n}. Show that (3Z and 5Z) = 15Z

this feels a little light on pertinant information to me

Answer 20

Hmm, that is all the problem gives me. =\

Answer 21

on the left, 3 and 5 both divide 15; so m relates to those; and 15 would relate to n

but what does it mean that: 3Z = {15, such that 3|15} ??

Answer 22

lol, on the right .. we have the specifics.

How does the anding of 3Z and 5Z relate the the set defined as mZ ?

Answer 23

I have no clue. lolz

Answer 24

youmight want to consider presenting the information as a picture or screenshot to make sure that nothing is getting lost in translation

Answer 25

ok I'll take a quick picture

Answer 26

Here it is.

Answer 27

now it makes sense :)

Answer 28

:O

Answer 29

the set 3Z contains all the integers that are divided by 3:
\[\{0,\pm3,\pm6,\pm9,\pm12,...\}\]

the set 5Z contains all the integers that are divided by 5
\[\{0,\pm5,\pm10,\pm15,\pm20,...\}\]

Answer 30

they were simply defining the mod(m) sets is all:\[\Large mZ\equiv Z_m\]

Answer 31

ah ok

Answer 32

any idea how you would go about showing the "and" of the sets?

Answer 33

Maybe like this: {0, +- 15, +-30, +-45.....} ?

Answer 34

thats would be the final results, but you need to prolly include some thrms related to the least common multiple

Answer 35

for some n,k,r in Z. 3n is in 3Z, and 5k is in 5Z

3n + 5k = 15r

you should have some thrm that relate this

Answer 36

hmm, kinda makes sense.

Answer 37

some things that im trying to recall ... there are integers n,k such that:

3n + 5k = gcd(3,5)

5 = 3(1) + 2
3 = 2(1) + 1 <-- gcd is 1
2 = 1(1) + 2


0
0 1
0+1(1) = 1
1+1(1) = 2, let n=2

3(2) + 5k = 1; therefore k = -1

3(2) + 5(-1) = 1 , multiply thru by 15r

3(2*15r) + 5(-1*15r) = 15r

Answer 38

does that give you any ideas?

Answer 39

not really, I'm extremely new to all this. =\

Answer 40

im extremely rusty at this :)

Answer 41

3(2*15r) + 5(-1*15r) = 15r

6(15r) - 5(15r) = 15r

6r - 5r = r

r = r

Answer 42

this should amount to it ... maybe in a longer way than someone more proficient would take :)

we defined some abitrary elements of 3Z = 3n, and 5Z = 5k

we found values of n and k which satisfies the euclidean stuff:

3n + 5k = 1
3(2) + 5(-1) = 1, but we need to show that this is equal to some arbitrary
element in 15Z = 15r

6(15r) + 5(-15r) = 15r; solving for r would show us if we can obtain
ALL elements in 15Z

6(r) - 5(r) = r

(6 - 5) r = r

1r = r

r = r ; therefore we have no restrictions and can obtain all the elements of 15Z

Answer 43

Starting to make a little more sense now. :O

Answer 44

I will have to go over it a little more to understand all the beginning steps.

Answer 45

But looking a lot better :D Thanks for the help with this!

Answer 46

good luck with it :) im sure theres some easier way to go about it, but i never know the easy routes

Answer 47

haha, well I think I am starting to understand it a little now. I think I need to do some more reading on the topic

Answer 48

and practice some easier problems to build up to it

Answer 49

But my teacher likes to assign difficult problems right when class starts -_-

Answer 50

anyways, Thanks again. :)

Answer 51

something simpler tends to relate to "and" being represented as multiplication:

taking our arbitrary elements: 3n(5k) = 15(nk)

addition relates more to the "or" of 2 sets. But its something to think about im sure

Answer 52

yeah definitely

Answer 53

as i lay awake staring at the ceiling last night I thought this:

We need to determine for what elements do 3Z and 5Z have in common? or to put it another way, when does 3n = 5k , for some integers n and k.

If 3n = 5k, then 5 divides 3n. Since 5 does not divide 3, then 5 has to divide n. Let n = 5s for some integer s.

If 3n = 5k, then 3 divides 5k. Since 3 does not divide 5, then 3 has to divide k. Let k = 3r for some integer r.

3(5s) = 5(3r)

15s = 15r , therefore the only elements they have in common are the elements that compose 15Z

Answer 54

Wow, thank you so much. :)

Answer 55

good luck :)