# Condition for function

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If there is a function f, for which:

f(a*a)=a

What possible function could complement this condition , beside the f(x)=sqrt(x) function?

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What is f(4) ? Well, f(2*2) = 2, so f(4) must be 2. Similarly for any non-negative value instead of 4. So f(x) = sqrt(x) if x>=0, and it can be anything you want if x<0.

EDIT: I assume you forgot to say "for any a>=0".

Edited by Álvaro

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Well actualy, though sqrt has definition intervat x>=0, the a can be anything in that case, even negative. So any other function, no matter what its defintion interval is, wheather would complement that condition, for any a.

I forgot to mention a of Real number, but I am not even sure wheather I need all a, it woulbe be interesting to provide solution even for subset interval of real numbers of a.

What has suited me already is f(x)=(x*x)1/4    , but I am seraching for simplifaction over sqrt() function.

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If a can be negative then there is no solution.

f(4) = f(2*2) = 2
f(4) = f((-2)*(-2)) = -2

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If a can be negative then there is no solution.

f(4) = f(2*2) = 2
f(4) = f((-2)*(-2)) = -2

So would there be a solution for a being postive, other than sqrt()?

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If a can be negative then there is no solution.

f(4) = f(2*2) = 2
f(4) = f((-2)*(-2)) = -2

So would there be a solution for a being postive, other than sqrt()?

if you want it to be positive, also if the a is negative, use an abs before or after the sqrt.

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So would there be a solution for a being postive, other than sqrt()?

See my first response.

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If a can be negative then there is no solution.

f(4) = f(2*2) = 2
f(4) = f((-2)*(-2)) = -2

So would there be a solution for a being postive, other than sqrt()?

if you want it to be positive, also if the a is negative, use an abs before or after the sqrt.

Why down vote? He is right.

If condition was formulated as, originaly instead:

f(a*a)=|a|

Then sqrt() would satisfy all  real numbers possible for a.

But, since absolute value is not too much of a native algebraic operation, I am too happy with a non-negative a in <0,lyineight) interval solely.

So Alvaro, I hope to formulate exact math question now to examine:

f(a*a)=a  , a in: <0,lyingeight) of real numbers

is there a function instead of f=sqrt that could be simpler to conclude to this condition of ?   Possibly?

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Why down vote? He is right.

That was my downvote. And no, he isn't right. He was just adding noise to the conversation, as far as I can tell.

So Alvaro, I hope to formulate exact math question now to examine:   f(a*a)=a  , a in: <0,lyingeight) of real numbers   is there a function instead of f=sqrt that could be simpler to conclude to this condition of ?   Possibly?

Your grammar is confusing, but the answer is likely "no". I already explained why, but I'll make it more explicit.

I will now prove that f(x) = sqrt(x) for every x>0. Take a = sqrt(x), and now we have

f(x) = f(sqrt(x)*sqrt(x)) = f(a*a) = a = sqrt(x)

Q.E.D.

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is there a function instead of f=sqrt that could be simpler to conclude to this condition of ?   Possibly?

There really isn't. You've defined the function as "Given a number that is some value multiplied by itself, return the original value." That is the very definition of the square-root function and if there were anything simpler then we'd be using that function instead of sqrt.

Perhaps it's time to back-up a step. Why are you looking for a function that does square-roots without using the sqrt operation? Is this some sort of performance optimization?

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