When it says:
Give a constructional definition of the following sets

What does this mean?

Question

When it says:
Give a constructional definition of the following sets

What does this mean?

fibonaccichick666 · Answer

I need context, (what sets)

anonymous · Answer

sorry was afk @FibonacciChick666

fibonaccichick666 · Answer

I'm about to be afk myself I have class

anonymous · Answer

oh sorry im from the UK lmao

anonymous · Answer

I'm going to bed but ill be on when i wake so if you could answer it i would appreciate it @fibonaccichick666

fibonaccichick666 · Answer

So the issue I am having is that what you have linked to, is what I know as a constructional definition.  @ganeshie8 , any ideas on another interpretation?

ganeshie8 · Answer

I think it gives all powers of 2

fibonaccichick666 · Answer

no, it would include 9.  Right?

ganeshie8 · Answer

$3 | 9$ so 9 is out

fibonaccichick666 · Answer

ah division, I took it as n cannot be a prime

fibonaccichick666 · Answer

hmm, well, Does that mean the prime does not divide itself as well? If not, then I agree

ganeshie8 · Answer

does this look okay $$\mathbb{ \{ n \in Z  :  n=2^k \text{for some } k \ge 0\} } $$

fibonaccichick666 · Answer

Won't that end in a zero at some point?

ganeshie8 · Answer

2
4
8
6
2
...

fibonaccichick666 · Answer

ok, how about the ending in sixes, could they be divided by 3?

ganeshie8 · Answer

2^k  is never divisible by 3 cuz there is no 3 in its prime factorization ;p

fibonaccichick666 · Answer

I guess, can that be guaranteed for every k?

ganeshie8 · Answer

i think you're asking about proving fundametal theorem of arithmetic

fibonaccichick666 · Answer

oh gosh, no you're right

fibonaccichick666 · Answer

I was just thinking in terms of the last digit

ganeshie8 · Answer

that theorem guaranteees that the prime factorization is unique upto permutations.. 
so the prime power like : 2^k wont split into 3^m*5^n or some other thing ever...

fibonaccichick666 · Answer

but, for this we can say that by the (theorem that states all numbers have a prime factorization) we can express every number as $$n=p_1p_2p_3.....p_k$$
by setting the limitation of n not being allowed to be divisible by any prime greater than or equal to 3, we are limited in our prime choices leading us to any n must be able to be written as $$2^k$$

ganeshie8 · Answer

thats a nice argument

fibonaccichick666 · Answer

It was your catch that made it possible :)  I have an issue with reading division backwards when expressed with |, so I was like, well that's the definition of a prime set that includes 1 and 2,