Expert: Scott A Wilson - 5/27/2010

Question
Why is the domain in a function always all real numbers?

Answer
The domain of a function is all the values of x that can be put in the function.
The domain of all polynomials, trig function, and powers is all real numbers.

However, if f(x) = √x, then the domain is only those x that are non-negative.

If f(x) = 1/(x-2), the domain is all x values except for the value x = 2.

If f(x) = sec(x), the domain is all x except for the values where cos(x) = 0,
for sec(x) = 1/cos(x).  Thus, that is all x except for nπ, where n is any integer from -∞ to ∞.

So when dealing with all real numbers, the domain of the function may not be all real numbers.

Complex Numbers - most likely don't have them yet, but if you're into math, read on...
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I don't know if you've got there yet, but more advanced math deals with all number in the x-y plane where x is the reals and y is for the imaginary numbers, that is, the multiplier on √-1.
If complex numbers are dealt with, the √-9 = 3i, where i = √-1.  For this, the squareroot has
a range that includes positive and negative numbers.

This came about just like negative numbers.  See, back in Roman times, they had no concept of negative numbers.  In fact, I don't even think they had the concept of 0.  In this way, given
x=3-5, x just didn't exist.  In more recent centuries, they didn't have a concept that negative numbers existed, so √-4 was just non-existant.  It really is, if only reals are dealt with.
However, is √-1 is defined as i, the √-4 = 2i.

For example, the √-1 = i, and the √i = (√2 + (√2)i)/2.  Computing [(√2 + (√2)i)/2]² give
(2 + 2i + 2i²)/2.  Now since i²=-1, this is (2 + 2i - 2)/2 = 2i/2 = i.

When dealing with complex numbers, the value is taken as an x-y value and then converted to polar coordinates, which involve r (the distance from the origin) and Θ, the angle with the
x axis.  r = √(x²+y²) and Θ = arctan(y/x).  To compute a complex number in polar coordinates,
the √(r,Θ) is (√r,Θ/2).  Thus, √-1 is ... well, in polar coordinates, that is r=1 and Θ=π,
where Θ is in radians.  To compute the squareroot would be to say that r=√1=1 and Θ/2=π/2.
Since π/2 radians = 90° and r=1, that is the value straight up, which is i.  To compute the squareroot again, the √1 is still 1, so r is 1, but Θ=π/4, which is 45².  Thus, the length of the hypoteneuse of a triangle with this point has length 1 and the angle is 45°.  This makes
both of the sides have length √2/2.

However, even with complex numbers, if the function is f(x) = 1/(x-3), f(3) is still ∞,
so it is undefined at this point.

Now trig functions are defined in the complex plain to me by looking at the Taylor's expansion.
For example, sin(x) = sum(n=0 to ∞){[(-1)^n][x^(2n)]/(2n)!}.  The ! is used to mean factorial.
Note that 1! = 1, 2!= 2*1! = 2, 3! = 3*2!= 6, 4!=4*3! = 24, etc.  In really advanced mathematics (the junior level courses and higher in college in mathematics), a definition has been found to
apply n! to all reals and not just integer, but that involves calculus.

I could write even more, but this is probably more than you want to know already.
If, however, more is desired, this address gives an intro into what can be done:
http://en.wikipedia.org/wiki/Calculus

Highlight it, cut it, and paste it in your search box.