[/quote]
,
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
,
,
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
,
and non-negative reals 
,
Here 
and 
denote the median and arithmetic mean of 
,
![$f \in C^1([0,1]), g \in C^2(f([0;1]))$](http://latex.artofproblemsolving.com/f/5/2/f52369171ef8571734950120c43aacac10154fcd.png)
.
?
be an isosceles triangle with 
,
an arbitrary point on side 
.
meets the circumcircle of 
again at 
,
meets it again at 
.
,
and 
are concurrent.
be non-negative real numbers. Prove that
be a triangle such that 
.
lie on segments 
,
and 
are the angle bisectors of triangle 
.
.
satisfy 
prove in equality.
and circumcenter 
.
with center 
intersects the segment 
and 
,
,
are all different and lie on line 
and 
be the points of intersection of 
be the second point of intersection of the circumcircle of triangle 
and the segment 
.
be the second point of intersection of the circumcircle of triangle 
and the segment 
.
and 
are different and intersect at the point 
.
.
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
,
,
and 
could be chosen to satisfy 
and 
.
,
are positive integers and 
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
,
be the incenters of triangles 
,
,
,
,
meet at a point.
satisfying
Prove
and 
.
with multiplication of matrixes operation is making an isomorphic-group structure with 
.
.
Y by
Which one of these functions are invertible? or both are invertible?
Let
and 
be defined as
,
I think both of them are invertible, for
its trivial for 
is where it gets confusing.
The condition for invertibility of a function according to Wikipedia is
>The function is invertible if and only if it is bijective. This is because the condition
implies that is injective, and the condition
implies that is surjective.
My friends whom I discussed this problem with say that isn't invertible because isn't surjective since never reaches values 
since when I make
in 
the domain it won't produce values for for 
, but I think that argument is wrong but I can't point it out. The only argument I could provide was
Is it necessary for surjectivity here? and if so why is it contradicting the defintion? where am i going wrong?
The question asks, "Is it invertible?", which I interpret as, "Will an inverse exist?"
YES, if I restrict the codomain to the range.
NO, if I keep the codomain as it is.
It never said anything about whether I can restrict the codomain or not, so I should be allowed to restrict it.
Restricting the codomain doesn’t change the actual input-output behavior of the function — it just changes how we describe the function.
The mapping rule remains the same. For example, if
, then 
and 
, regardless of whether the codomain is or 
.
It is valid to restrict the codomain to the range when discussing invertibility.
This doesn't alter the nature of the function — it just makes the description precise and allows an inverse to exist by making the function surjective. Many textbooks do this without issue.
If restricting the domain or codomain changes the function, then technically those functions don't have inverses.
But we came up with functions like
or 
precisely to define inverses — so maybe it's fair to come up with an inverse here too.
Yes, strictly speaking, changing the codomain defines a different function in the formal sense.
But since the input-output rule doesn't change, and the goal is to determine if an inverse exists, it's mathematically acceptable — and often necessary by many textbooks — to restrict the codomain to the range.
Since the question was to analyze invertibility, it makes sense to say the function can be made invertible in that context.
Let
and 
be defined as
,
I think both of them are invertible, for
its trivial for 
is where it gets confusing.The condition for invertibility of a function according to Wikipedia is
>The function
is invertible if and only if it is bijective. This is because the condition
implies that is injective, and the condition
implies that is surjective.My friends whom I discussed this problem with say that
isn't invertible because isn't surjective since never reaches values 
since when I make
in 
the domain it won't produce values for for 
, but I think that argument is wrong but I can't point it out. The only argument I could provide was
Is it necessary for surjectivity here? and if so why is it contradicting the defintion? where am i going wrong?
The question asks, "Is it invertible?", which I interpret as, "Will an inverse exist?"
YES, if I restrict the codomain to the range.
NO, if I keep the codomain as it is.
It never said anything about whether I can restrict the codomain or not, so I should be allowed to restrict it.
Restricting the codomain doesn’t change the actual input-output behavior of the function — it just changes how we describe the function.
The mapping rule remains the same. For example, if
, then 
and 
, regardless of whether the codomain is or 
.It is valid to restrict the codomain to the range when discussing invertibility.
This doesn't alter the nature of the function — it just makes the description precise and allows an inverse to exist by making the function surjective. Many textbooks do this without issue.
If restricting the domain or codomain changes the function, then technically those functions don't have inverses.
But we came up with functions like
or 
precisely to define inverses — so maybe it's fair to come up with an inverse here too.Yes, strictly speaking, changing the codomain defines a different function in the formal sense.
But since the input-output rule doesn't change, and the goal is to determine if an inverse exists, it's mathematically acceptable — and often necessary by many textbooks — to restrict the codomain to the range.
Since the question was to analyze invertibility, it makes sense to say the function can be made invertible in that context.
it is surjective only if 
but this do not exist for the 
so the function it is NOT surjective hence it is not bijective, so have not inverse function.
result 
, hence the function will be surjective. So we see the injectivity: we will verify 
hence the function it is injective. So the function it is bijective, and have inverse function 
.
and
are two different functions. That happens because the domain and the codomain are part of the definition of the function.
is bijective, because we make it surjective by restricting its codomain to 
and we make it injective by restricting its domain to 
