Need help with a discrete math problem.

- anonymous

Need help with a discrete math problem.

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- anonymous

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

- amistre64

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- anonymous

lol I can't get the it written correctly

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## More answers

- anonymous

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

- anonymous

The Z is also suppose to be all the Real numbers

- amistre64

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

- amistre64

so, in R not Z ?

- anonymous

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

- amistre64

what is the eta ?

- amistre64

did you mean to and them?

- anonymous

I was trying to do the upside down U

- amistre64

and = \cap
or = \cup

- amistre64

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

- anonymous

\cap

- anonymous

\[cap\]

- amistre64

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

- amistre64

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
^ ^ ^

- amistre64

you might have to redress the setup ....

- amistre64

let mZ = {n in Z : m|n}. Show that (3Z and 5Z) = 15Z
this feels a little light on pertinant information to me

- anonymous

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

- amistre64

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} ??

- amistre64

lol, on the right .. we have the specifics.
How does the anding of 3Z and 5Z relate the the set defined as mZ ?

- anonymous

I have no clue. lolz

- amistre64

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

- anonymous

ok I'll take a quick picture

- anonymous

Here it is.

##### 1 Attachment

- amistre64

now it makes sense :)

- anonymous

:O

- amistre64

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
\[\{0,\pm5,\pm10,\pm15,\pm20,...\}\]

- amistre64

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

- anonymous

ah ok

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

hmm, kinda makes sense.

- amistre64

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

- amistre64

does that give you any ideas?

- anonymous

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

- amistre64

im extremely rusty at this :)

- amistre64

3(2*15r) + 5(-1*15r) = 15r
6(15r) - 5(15r) = 15r
6r - 5r = r
r = r

- amistre64

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

- anonymous

Starting to make a little more sense now. :O

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

and practice some easier problems to build up to it

- anonymous

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

- anonymous

anyways, Thanks again. :)

- amistre64

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

- anonymous

yeah definitely

- amistre64

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

- anonymous

Wow, thank you so much. :)

- amistre64

good luck :)

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