/sci/ - Science & Math

[Anonymous No.3606976]

Help /sci/!

I'm preparing for an exam and I just don't understand this type of exercise:
>Determine the number of functions f:{0,1,2,3}->{0,1,2,3}, which have the property that f(0) is uneven.
The answer is either 4^3 or 4^4. I don't get it. How in the world do you do this type of thing? I really don't remember doing it at school in the past years. I'm a senior in highschool btw.

[Anonymous No.3607062]

>2011
>not understanding algebra I

[Anonymous No.3607076]

try mod 4

[Anonymous No.3607090]

128

[Anonymous No.3607104]

But how? Just how do you get to this?? I'm going crazy here. I can't find any exercise of this type or answer on the internet.

[Anonymous No.3607110]

>>3607104
By learning these mysterious things called facts. You will master many of them in coming years.

[Anonymous No.3607112]

f(1), f(2) and f(3) are irrelevant to the property you want (f(0) is uneven)
and they each have 4 possibilities as result (0,1,2,3)
So that's 4^3 possibilities there

Then you have f(0) which has to be uneven, so either 1 or 3
So it's 2 * 4^3

But don't worry about that it's a bullshit problem.

[Anonymous No.3607130]

I understand that f(1), f(2) and f(3) are irrelevant. I just don't understand what's with that 3 from 4^3. Where did that come from? I thought it was logical that f(0)=1,3 which are uneven, so there are 2 posibilities. What's with the 4^3?

[Anonymous No.3607152]

Brain, y u no work?

[Anonymous No.3607165]

options for f(0): 2
options for f(1): 4
options for f(2): 4
options for f(3): 4

total options: 2*4*4*4 = 2*4^3

you get where the 3 is from now?

[Anonymous No.3607178]

>options for f(1): 4
>options for f(2): 4
>options for f(3): 4

Why 4 options for each???
f(1)=/=f(0) so f(1) may or may not be uneven
The same for all the others...

[Anonymous No.3607205]

The exercise doesn't say anything about the other functions, just about f(0). Why are the other functions added to the result?

[Anonymous No.3607208]

[Anonymous No.3607219]

what do you mean why are there four options?

a function must map each number in the first set to a number in the second set.

so to fully describe the function you must describe what 0,1,2,3 are mapped to.

for 1,2,3 they can map to any of 0,1,2,3

[Anonymous No.3607221]

>>3607178
>f(1)=/=f(0)
Nope.

f is, by your definition:
> f:{0,1,2,3}->{0,1,2,3}
So for each element x from the left stuff
>{0,1,2,3}
f associates a value y to the right stuff
>{0,1,2,3}
So f(1) can either be 0, 1, 2, 3 depending on f. 4 possibilities for 4 different f.
Same thing for f(2) and f(3).
You can have f1(1) = 0, f1(2) = 0,... or f2(1) = 1, f2(2) = 0, ... If you count them all, it's 4^3, and *2 if you add f(0)

[Anonymous No.3607233]

> select all the functions
4 x 4 x 4 x 4
> from these, select only the ones where f(0) is odd
2 x 4 x 4 x 4
> done

[Anonymous No.3607255]

>>3607221
I thought I only need to find how many f(0) are there that are uneven. That's what I don't get:
>So f(1) can either be 0, 1, 2, 3 depending on f. 4 possibilities for 4 different f.
f(1) is not what I must count, only the f(0)'s that are uneven

[Anonymous No.3607259]

How is f(0) odd in this: f(1)=something->where's f(0)???

[Anonymous No.3607267]

no, you must count all different possibilities of f(0), f(1), f(2), and f(3) together, because these are what define the FUNCTION. f(0) does not define the FUNCTION. f at EACH INDIVIDUAL VALUE IN THE RANGE defines the FUNCTION. the question is asking you to count the different possibilities for the FUNCTIONS, not f(0).

[Anonymous No.3607276]

>>3607267
Oh, so f(3)=2 has the property that f(0) is odd?

[Anonymous No.3607278]

Huh. I never did anything like this in high school.

Then again I got an inflated B grade for advanced functions.

[Anonymous No.3607282]

>>3607267
*DOMAIN

[Anonymous No.3607290]

math....

[Anonymous No.3607295]

>>3607276
IT DEPENDS WHAT f(0) IS FOR WHATEVER FUNCTION YOU ARE TALKING ABOUT.

you haven't defined a function. you've given the function at a single value. you are extremely stupid.

[Anonymous No.3607316]

>>3607295
I won't get this even if you'll kill me.
Bye, gonna go an hero now....

[Anonymous No.3607330]

this thread....

[Anonymous No.3607342]

[Anonymous No.3607411]

pasteb*n.com/2EY6JRsX