is f(x)=rt(x) a function for the domain x any real number?

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hiyatt
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is f(x)=rt(x) a function for the domain x any real number? So can you extend the range to complex numbers?

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RDKGames
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(Original post by hiyatt)
is f(x)=rt(x) a function for the domain x any real number? So can you extend the range to complex numbers?

Thanks
What set are you working up to??

If it's up to \mathbb{R} then obviously x \geq 0 as you can't root a -ve in the reals!

If it's up to \mathbb{C} then obviously x \in \mathbb{C}.
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hiyatt
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(Original post by RDKGames)
What set are you working up to??

If it's up to \mathbb{R} then obviously x \geq 0 as you can't root a -ve in the reals!

If it's up to \mathbb{C} then obviously x \in \mathbb{C}.
don't really get what you mean by what set
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hiyatt
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can you have a function which just goes from real to complex numbers? like with sqrt(x) x any real number. i don't what set are you working up to means.

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Ryanzmw
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A function consists of three parts:
1) The domain (the set the function maps from)
2) The codomain (the set the function maps to)
3) The defining relation between the two sets.

Here I'm assuming you're saying the domain is  \mathbb{R} so we expect  f(x) = \sqrt{x} . Which is all fine and dandy except we're missing the 2nd part of the definition. Why is this important? Well a function must be well-defined, it maps every element of the domain to a unique element of the codomain.

If we take the codomain to be  \mathbb{C} then to properly define which function we're talking about you should write:
 f: \mathbb{R} \mapsto \mathbb{C}

$ f(x) = \sqrt{x} $

Which is well-defined function.

However if we take the codomain to be  \mathbb{R} then,
 f: \mathbb{R} \mapsto \mathbb{R}

$ f(x) = \sqrt{x} $
is not well-defined and thus not a function, since there is no element in the codomain for  f(-1)
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hiyatt
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(Original post by Ryanzmw)
A function consists of three parts:
1) The domain (the set the function maps from)
2) The codomain (the set the function maps to)
3) The defining relation between the two sets.

Here I'm assuming you're saying the domain is  \mathbb{R} so we expect  f(x) = \sqrt{x} . Which is all fine and dandy except we're missing the 2nd part of the definition. Why is this important? Well a function must be well-defined, it maps every element of the domain to a unique element of the codomain.

If we take the codomain to be  \mathbb{C} then to properly define which function we're talking about you should write:
 f: \mathbb{R} \mapsto \mathbb{C}

$ f(x) = \sqrt{x} $

Which is well-defined function.

However if we take the codomain to be  \mathbb{R} then,
 f: \mathbb{R} \mapsto \mathbb{R}

$ f(x) = \sqrt{x} $
is not well-defined and thus not a function, since there is no element in the codomain for  f(-1)
For the first function would you not have to split the domain into two parts one for x>0 with normal radical notation and then one for x<0 but as normal radical notation is ambiguous don't you have to specify the principle root of x where x is negative as irt(-x) where x is negative
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Ryanzmw
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(Original post by hiyatt)
For the first function would you not have to split the domain into two parts one for x>0 with normal radical notation and then one for x<0 but as normal radical notation is ambiguous don't you have to specify the principle root of x where x is negative as irt(-x) where x is negative
thanks
Edit: Ah yeah I see your point because we have to make a choice of branch to define  \sqrt{x} on.
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