Can someone help me with this question, please

Can someone help me with this question, please

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any help would be much appreciated

There are n even numbers and n odd numbers. A function could map every number x to any one of n numbers (based on whether x is even or odd). There are 2n numbers to map. The possible number of functions is therefore n^(2n)? (Been a while since I did this stuff but I think that's right.)

Whoo! thinks was him...!?

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newfags can't math
only coom

ahh so its n^2n for both

there is n even numbers and n matches for each even number, vice versa for odd numbers. Therefore it is 2(n)^n not n^(2n)

but is it not n possible for each of the 2n positions,

Map 1 to any odd number (n/2 numbers). Map 2 to any even number (n/2 numbers). Map 3 to any remaining odd number (n/2-1 numbers). Continue. You get (n/2)!*(n/2)!

i. e. ((n/2)!)^2

Even: (2)n^n
Odd: (2)n^n + 1

Never done this math before but maybe this helps???

the question states that there are 2n numbers, therefore there is n odd numbers and n even numbers. the question never said shit about remaining numbers.
it is n possible for every number, and there are 2n, so the answer is 2n^n

function ends at a even number

Also should mention this isn't by any means a rigorous solution, just intuitive. But if you're not a math major, you could probably get away with it.

>it is n possible for every number, and there are 2n
What you're saying is n^(2n), not 2n^n.

(n possible) * (n possible) * ... * (n possible)
2n times
= n^(2n)

It doesn't specify you cannot reuse numbers (i.e. map all odd numbers to 1).
Also, there are 2n numbers meaning there are n odd numbers and n even numbers (not n/2).

Whoops, instead of n/2 it's n. So the answer is (n!)^2

are you fucking retarded there are 2n numbers and n possibilities for all those 2n numbers

so its literally 2n^n

Oh you're right. The solution only works for injective functions. The solution should be (2*n)^n then, since there are n even and n odd numbers (map even numbers to any of n even numbers and odd numbers to any of n odd numbers).

Are you trolling?
If you have 8 bits (0 or 1) the number of possible values is 2^8, not 8^2.

Meant n^(2n)...

wait it is even numbers n - to the power of the number of items in X which is 2n.

Hence it is n^2n

lemme explain to you you slow motherfucker
i have 1 which matches to n numbers
that means i have n numbers that matches to n numbers, which means that there are n^n matches
since there is n^n matches for odd numbers, there is also n^n matches for even numbers
n^n + n^n = 2n^n

You are not creating functions X->X, because your n^n is only mapping the evens, and your other n^n is only mapping the odds.

To create a function X->X you need to map all 2n numbers, which is why the answer is n^(2n).

try for n=1...

Your answer can easily be shown to be incorrect for n=1 and n=2.

It is n^(2n).

do any of u guys know relations

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linear is based

been teaching myself for fun

As long as you know what an equivalence relation is, this is extremely simple.

it is a valid equivalence relation

An equivalence relation is a binary relation that is reflexive, symmetric and transitive.

Reflexivity:
aRa holds for all a in S, because a=a holds for all a in S, reflexivity holds.

Symmetry:
Assume aRb.
This implies a = b or a = -b.
If a = b then b = a, so bRa holds.
If a = -b then b = -a, so bRa holds.
So aRb implies bRa, symmetry holds.

Transitivity:
Assume aRb and bRc.
This implies a = b or a = -b, and b = c or b = -c.
If a = b, then a = b = c or a = b = -c, aRc holds.
If a = -b, then a = -b = -c or a = -b = -(-c) = c, aRc holds.
So aRb and bRc implies aRc, transitivity holds.

All 3 hold and so it is an equivalence relation.

what would be the set of equivalence classes for this

you technically dont even need that. equality is already an equivalence relation so it holds

Yes, but aRb === ((a = b) OR (a = -b)). You are no longer dealing with a regular equality.

{0} and all {a,-a} where a in S and a > 0.