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Mathematics 19 Online
OpenStudy (anonymous):

Please help me... Determine which, if either, of the following sets form a group with respect to the given operation. I. {0, 1, 2, 3, 4}; addition modulo 5 II. the set of integers with respect to addition.

terenzreignz (terenzreignz):

Let's have a look at the first set :D {0,1,2,3,4} the set of nonnegative integers less than 5. When you add any of them and take the result at modulo 5, is it still going to be in that set?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so like, 3+2 = 5

terenzreignz (terenzreignz):

nu-uh... remember, after adding, you take the result modulo 5, and what's 5(mod 5) = ?

OpenStudy (anonymous):

5

terenzreignz (terenzreignz):

no...

OpenStudy (anonymous):

hum this is were I don't understand

terenzreignz (terenzreignz):

Recall modulo... a number modulo 5 is the remainder if you divide said number by 5. If you divide 5 by 5, what's the remainder?

OpenStudy (anonymous):

1

terenzreignz (terenzreignz):

Really? When last I looked, 5 was divisible by 5 :P

OpenStudy (anonymous):

oh duh. sorry

terenzreignz (terenzreignz):

Let's give that set a name, for future reference: \[\Large \mathbb{Z}_5=\{0,1,2,3,4\}\]

OpenStudy (anonymous):

thats right because we are using algebra

terenzreignz (terenzreignz):

now, when you add any two elements of \(\large \mathbb{Z}_5\) and then get the result, modulo 5, is that result still going to be in \(\large \mathbb{Z}_5\)?

OpenStudy (anonymous):

this is all new to me

terenzreignz (terenzreignz):

I just gave it a name, don't worry. Now, my question ^ Please give it a good think...

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

How do you know?

OpenStudy (anonymous):

Because we are dividing Z5 by Z5 and it divisible.

terenzreignz (terenzreignz):

not a very clear reason, is it? :)

OpenStudy (anonymous):

no

terenzreignz (terenzreignz):

Look, when you're dividing by 5 and getting remainders, it has to be one of 0,1,2,3,4 any more than that, it won't make sense as a remainder, yeah/

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

so that means that both form a group

terenzreignz (terenzreignz):

We haven't even gone to the second item yet, slow down LOL The rest is easy though... is addition associative?

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

are u good with symmetry because I have no idea how to do that and my next 6 questions is on that

terenzreignz (terenzreignz):

never heard of them :)

OpenStudy (anonymous):

yeah me either :)

OpenStudy (anonymous):

so was i right about that they both form a group

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