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

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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

Thanks

Thanks

0

reply

Report

#2

(Original post by

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

Thanks

**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

If it's up to then obviously as you can't root a -ve in the reals!

If it's up to then obviously .

0

reply

(Original post by

What set are you working up to??

If it's up to then obviously as you can't root a -ve in the reals!

If it's up to then obviously .

**RDKGames**)What set are you working up to??

If it's up to then obviously as you can't root a -ve in the reals!

If it's up to then obviously .

0

reply

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.

Thanks

Thanks

0

reply

Report

#5

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 so we expect . 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 then to properly define which function we're talking about you should write:

Which is well-defined function.

However if we take the codomain to be then,

is not well-defined and thus not a function, since there is no element in the codomain for

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 so we expect . 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 then to properly define which function we're talking about you should write:

Which is well-defined function.

However if we take the codomain to be then,

is not well-defined and thus not a function, since there is no element in the codomain for

0

reply

(Original post by

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 so we expect . 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 then to properly define which function we're talking about you should write:

Which is well-defined function.

However if we take the codomain to be then,

is not well-defined and thus not a function, since there is no element in the codomain for

**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 so we expect . 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 then to properly define which function we're talking about you should write:

Which is well-defined function.

However if we take the codomain to be then,

is not well-defined and thus not a function, since there is no element in the codomain for

thanks

0

reply

Report

#7

(Original post by

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

**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

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top