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is there a difference?

It turns out the compositions are associative, but what exactly is the difference between them ?

Hint: Function composition is associative

there is no difference only be careful about range and domain

order might affect on that only.

Hmm okay, domain and codomain are valid in both cases
```
h : X -> Y
g : Y -> Z
f : Z -> W
```

take this example
h : {b,1,2,3}->{1 }
g : {1,2,3}->{4,5,a,c}
f : {.4,1,2,3}->{a,b,c}

how does that work ?
say
\(f(x) = x\)
\(g(x) = 2x\)
\(h(x)=3x\)

\((f\circ g)\circ h(2) = ?\)
what does it mean to compose \(f\) and \(g\) first, here ?

I think I get it! thanks a lot for breaking it down !

I kind of like to leave the x off sometimes

ur doing abstract now ?

what?!? you really are doing abstract algebra?

\(f:\mathbb{R}\rightarrow \mathbb{R}, f(x) = 3x+2\) is a function

I thought you already graduated and stuff. You know everything. :p

\(f:\underset{x\mapsto 3x+2}{\mathbb{R}\rightarrow \mathbb{R}}\)

I really like that last one, but it takes so long in latex

hmmm yeah using this notation is better

nah some times u are perfect in math gane.

so we could say \(f= \{(1,2),(2,4),(4,5)\}\) but \(f(x)= \{(1,2),(2,4),(4,5)\}\) makes no sense.

and in this situation we could say \(f(\{1,2,4\})=\{2,4,5\}\).

some people would want \(f[\{1,2,4\}]=\{2,4,5\}\)

i thought this is also correcto
\(f (\{(1,2),(2,4),(4,5)\} )\)

this is the 'difference' in order, @ganeshie8