B Domain and imaginary numbers

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Consider the function sqrt(x).

What is the domain of this function? Is it all real positive numbers?

This is what I was taught in high school, but I was also taught that plugging in -1 would give an answer of i.

So if the function takes negative inputs, shouldn’t they be part of the domain?
 

fresh_42

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Consider the function sqrt(x).

What is the domain of this function? Is it all real positive numbers?
Almost. It's all non negative real numbers.
This is what I was taught in high school, but I was also taught that plugging in -1 would give an answer of i.
And now we pay the price for sloppiness! sqrt(x) is a sequence of seven letters over the latin alphabet extended by some special characters. It is not a function. By calling it a function we made several hidden assumptions:
  1. sqrt(x) = ##+\sqrt{x}##
  2. sqrt(x) = ##\left( f\, : \,\mathbb{R}_0^+\longrightarrow \mathbb{R} \,;\,x \stackrel{f}{\longmapsto} +\sqrt{x} \right)##
So if the function takes negative inputs, shouldn’t they be part of the domain?
No, the domain should be part of the definition in the first place! It isn't honest to rely on hidden assumptions and then change horses in the middle of the race.
 
Almost. It's all non negative real numbers.

And now we pay the price for sloppiness! sqrt(x) is a sequence of seven letters over the latin alphabet extended by some special characters. It is not a function. By calling it a function we made several hidden assumptions:
  1. sqrt(x) = ##+\sqrt{x}##
  2. sqrt(x) = ##\left( f\, : \,\mathbb{R}_0^+\longrightarrow \mathbb{R} \,;\,x \stackrel{f}{\longmapsto} +\sqrt{x} \right)##
No, the domain should be part of the definition in the first place! It isn't honest to rely on hidden assumptions and then change horses in the middle of the race.
I’m not super familiar with some of the notation you used for assumption 2. Does that say that the domain of the function is all real numbers equal to or greater than 0?

Why is this assumption implied by the fact that sqrt(x) is a function?

I thought a function was just of mapping of elements from one set to another set. I don’t know how sqrt(x) being a function would exclude negative inputs and/or imaginary outputs.
 

mathman

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##\sqrt{z}## is defined for all complex ##z##. Using the notation above f:C-> C.
 

fresh_42

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I’m not super familiar with some of the notation you used for assumption 2. Does that say that the domain of the function is all real numbers equal to or greater than 0?
Yes.
Why is this assumption implied by the fact that sqrt(x) is a function?
Because you have said ...
Consider the function sqrt(x).
... and human automatically adds everything which isn't said and makes an assumption, which allows the sentence to make sense. It is a convention that real variables are noted by an ##x## and complex variables by a ##z##, same as function values are associated with ##y##. This isn't written anywhere, it is just what most people do. So ##x## is associated with a real variable, unless otherwise stated! If ##x<0## then ##\sqrt{x}## is no real function anymore. This is why ##x\geq 0## is assumed. Again, this is an extrapolation of the many unmentioned aspects of a function: domain, range, mapping definition.
I thought a function was just of mapping of elements from one set to another set ...
... with the property, that no element can be mapped on two different ones, yes. But as you said: two sets are involved, ##\mathbb{R}_0^+ \, , \, \mathbb{R}## or ##\mathbb{C}## or whatever. Since they haven't been specified, there are two possibilities left:
  • reject your entire question due to insufficient data
  • add required data, which again offers two possibilities
    • add the data to yield a solution which is most general
    • add the data under the assumption that context is as usual
I don’t know how sqrt(x) being a function would exclude negative inputs and/or imaginary outputs.
It doesn't. You created a lack of information and now refuse the attempt to make sense of this manko.

This means: the only reason for this thread is reasoning, based on a trap.

Thread closed.
 

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