anonymous
  • anonymous
Need help with a discrete math problem.
Discrete Math
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
For integers m, \let mZ = {n \in Z : m | n}. Show that 3Z \eta 5Z = 15Z.
amistre64
  • amistre64
``` \[ latex codes \] ```
anonymous
  • anonymous
lol I can't get the it written correctly

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
For integers m, \let mZ = {n \in Z : m | n}. Show that 3Z (upside down U) 5Z = 15Z.
anonymous
  • anonymous
The Z is also suppose to be all the Real numbers
amistre64
  • amistre64
\[let~ mZ = \{n \in Z : m | n\}. \text{Show that 3Z } \eta \text{ 5Z = 15Z} \]
amistre64
  • amistre64
so, in R not Z ?
anonymous
  • anonymous
wait no the set of integers, sorry, so that is right
amistre64
  • amistre64
what is the eta ?
amistre64
  • amistre64
did you mean to and them?
anonymous
  • anonymous
I was trying to do the upside down U
amistre64
  • amistre64
and = \cap or = \cup
amistre64
  • amistre64
\[and:~\cap\] \[or:~\cup\]
anonymous
  • anonymous
\cap
anonymous
  • anonymous
\[cap\]
amistre64
  • amistre64
so, show that mod3, anded with mod5 is equivalent to mod15 ....
amistre64
  • 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
  • amistre64
you might have to redress the setup ....
amistre64
  • 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
  • anonymous
Hmm, that is all the problem gives me. =\
amistre64
  • 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
  • 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
  • anonymous
I have no clue. lolz
amistre64
  • amistre64
youmight want to consider presenting the information as a picture or screenshot to make sure that nothing is getting lost in translation
anonymous
  • anonymous
ok I'll take a quick picture
anonymous
  • anonymous
Here it is.
1 Attachment
amistre64
  • amistre64
now it makes sense :)
anonymous
  • anonymous
:O
amistre64
  • 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
  • amistre64
they were simply defining the mod(m) sets is all:\[\Large mZ\equiv Z_m\]
anonymous
  • anonymous
ah ok
amistre64
  • amistre64
any idea how you would go about showing the "and" of the sets?
anonymous
  • anonymous
Maybe like this: {0, +- 15, +-30, +-45.....} ?
amistre64
  • amistre64
thats would be the final results, but you need to prolly include some thrms related to the least common multiple
amistre64
  • 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
  • anonymous
hmm, kinda makes sense.
amistre64
  • 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
  • amistre64
does that give you any ideas?
anonymous
  • anonymous
not really, I'm extremely new to all this. =\
amistre64
  • amistre64
im extremely rusty at this :)
amistre64
  • amistre64
3(2*15r) + 5(-1*15r) = 15r 6(15r) - 5(15r) = 15r 6r - 5r = r r = r
amistre64
  • 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
  • anonymous
Starting to make a little more sense now. :O
anonymous
  • anonymous
I will have to go over it a little more to understand all the beginning steps.
anonymous
  • anonymous
But looking a lot better :D Thanks for the help with this!
amistre64
  • amistre64
good luck with it :) im sure theres some easier way to go about it, but i never know the easy routes
anonymous
  • 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
  • anonymous
and practice some easier problems to build up to it
anonymous
  • anonymous
But my teacher likes to assign difficult problems right when class starts -_-
anonymous
  • anonymous
anyways, Thanks again. :)
amistre64
  • 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
  • anonymous
yeah definitely
amistre64
  • 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
  • anonymous
Wow, thank you so much. :)
amistre64
  • amistre64
good luck :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.