/sci/ - Science & Math

[Anonymous No.9842020]

If I add 3 to each element of the set of all multiples of 3, do I get the same set?

[Anonymous No.9842032]

>>9842020
You’re a big boy, what do you think?

[Anonymous No.9842034]

>>9842020
Is that all multiples by naturals, integers, fractions or reals?

Cause if it's naturals, you lose 3.

[Anonymous No.9842048]

>>9842032
>>9842034
Integers multiples.
This is what I think:

if S is the set of all integer multiples of 3, S is a group under addition. Let S' be the set of elements s + 3. Now since 3 is an element of S, for s belonging to G, s + 3 remains in S, hence S' is a subset of S. We also know that order(S) = order(S'). Hence S' = S.

[Anonymous No.9842049]

>>9842048
meant S, not G lol.

Sorry I am taking an abstract algebra course for fun and never took analysis.

[Anonymous No.9842053]

>>9842048
S is a subset of S' and S' is a subset of S so S=S'

[Anonymous No.9842055]

>>9842049
Just because the order of sets are the same it doesnt mean it's the same set.
And the answer to OP's question is no because you would lose 0 (or 3 if 0 is not included in your set).

[Anonymous No.9842056]

>>9842055
Equal orders does mean the same set if one is a subset of the other.

[Anonymous No.9842061]

>>9842056
If set A is a subset of B and B is a subset of A then it's sufficient to prove that they're the same why bring orders into question in the first place then. Also you failed to prove S is a subset of S'.

[Anonymous No.9842064]

>>9842061
You only need to prove that one is a subset of another if the orders are the same. Proving the other direction is a waste of time.

[Anonymous No.9842065]

>>9842061
Because I felt like it? Is my proof valid or not? didn't need to prove S is a subset of S', I showed that S' is a subset of S since given an element s+3 of S', we find that s+3 belongs to S as well since it's closed under addition of 3.

[Anonymous No.9842067]

>>9842064
That would work with finite sets but the cardinality of yours is N0 and therefore not finite

[Anonymous No.9842071]

>>9842020

Yes is all integer multiples. No, if all natural multiples (the new set would not include 3).

[Anonymous No.9842076]

>>9842064
>>9842065
With your fallacy you could literally prove that your set is exactly the same as N. Or you could prove that the subset of reals (0, 1) is equal to the entire R.

[Anonymous No.9842080]

>>9842067
>>9842076
So the set inclusion + equal orders isn't a valid proof of set equality for non-finite sets? Sorry dude I haven't gotten to cardinality yet

[Anonymous No.9842082]

>>9842080
Exactly. Many laws for finite sets don't apply to infinite ones. Good thing you're trying to learn though.

[Anonymous No.9842088]

>>9842082
Thanks for the help, I appreciate it

[Anonymous No.9842109]

>>9842020
They're isomorphic. [math]3\mathbb{Z}+3 = 3(\mathbb{Z}+1) \cong3\mathbb{Z}[/math].
>>9842034
You don't "lose" anything, tardo.

[Anonymous No.9842284]

>>9842109
>You don't "lose" anything, tardo.
So you think {n | n in N, 3 divides n} is equal to {n+3 | n in N, 3 divides n} tardo?

[Anonymous No.9842333]

>>9842284
don't reply to undergrads; it's in bad taste.

[Anonymous No.9842340]

>>9842020
No, 6+3 is 9, not 6

[Anonymous No.9842355]

>>9842340
Bait