Inverse relationships

[Larry Hanawalt]

Is a function without an inverse possible?

 

[Dr.Peterson]

Is a function without an inverse possible?

Do you mean, a function without an inverse function?

Any function is a relation, and any relation has an inverse (relation), but the inverse is not always a function.

Have you learned about one-to-one functions?

 

[HallsofIvy]

The function $\displaystyle f(x)= x^2$, for x any real number, does not have an inverse function.

 

[Larry Hanawalt]

Do you mean, a function without an inverse function?

Any function is a relation, and any relation has an inverse (relation), but the inverse is not always a function.

Have you learned about one-to-one functions?

Dr. Peterson, here is what I really want to ask.

It's about the convention sqrt(1) = 1 ... if (e^(i*pi))^2 = 1 and [sqrt(e^(i*pi))]*[sqrt(e^(i*pi))] = i^2 = -1 ... then it seems to me that the sqrt(1) = 1 convention contradicts calculation ... (e^(i*pi))^2 = 1 represents a rotation of 180 degrees ... but sqrt(1) = 1 does not allow for the inverse rotation ... [minus times minus] = plus but [sqrt(minus times minus)] is also = plus ... doesn't this make the conventional representation of the complex number field asymmetrical?

A related issue is that abs(-1) = 1 ... my questions are all about conventions, assumptions, axioms, the most basic assumptions about the relationships among 1, 0, zeta(0), -1, sqrt(1) etc

 

[Dr.Peterson]

This has very little, if anything, to do with your original question.

The convention that sqrt(1) = 1 is part of the definition of the square root function. It is a choice of a principal value for the square root, a somewhat arbitrary choice that results in some useful properties. This applies primarily to the function over the reals, not over the complex numbers.

What you show with complex numbers is a separate matter; but it shows that if you try to define a principal value of the square root over the complex numbers leads to some complications. I'm not exactly sure what you mean by "inverse rotation" and "asymmetrical", but basically, yes, there are tricky things about the complex numbers.

I have no idea what you are concerned about with the absolute value, which is just part of the definition. But definitions and conventions are not the same as assumptions.

 

[pka]

It's about the convention sqrt(1) = 1 ... if (e^(i*pi))^2 = 1 and [sqrt(e^(i*pi))]*[sqrt(e^(i*pi))] = i^2 = -1 ... then it seems to me that the sqrt(1) = 1 convention contradicts calculation ... (e^(i*pi))^2 = 1 represents a rotation of 180 degrees ... but sqrt(1) = 1 does not allow for the inverse rotation ... [minus times minus] = plus but [sqrt(minus times minus)] is also = plus ... doesn't this make the conventional representation of the complex number field asymmetrical?
A related issue is that abs(-1) = 1 ... my questions are all about conventions, assumptions, axioms, the most basic assumptions about the relationships among 1, 0, zeta(0), -1, sqrt(1) etc

These are addition comments. There are a group of complex analysts/authors who would limit the use of radical signs to non-negative real numbers. It is discouraged using say $\sqrt{16i}$ rather as you did above use $\left(16\exp\dfrac{i\pi}{2}\right)^{\frac{1}{2}}$
About absolute value as a real valued function. In each of $a~\&~b$ is a number then $|a-b|$ is the distance between $a~\&~b$. Thus absolute value is a metric and as such has all properties of a metric and is a non-negative real number. Note that $|x|=|x-0|$ so $|x|$ is the distance $x$ is from $0$. Also note that is also true for complex numbers. $|-3+4i|=\sqrt{9+16}=5$ or five units from zero.

 

[Larry Hanawalt]

This has very little, if anything, to do with your original question.

The convention that sqrt(1) = 1 is part of the definition of the square root function. It is a choice of a principal value for the square root, a somewhat arbitrary choice that results in some useful properties. This applies primarily to the function over the reals, not over the complex numbers.

What you show with complex numbers is a separate matter; but it shows that if you try to define a principal value of the square root over the complex numbers leads to some complications. I'm not exactly sure what you mean by "inverse rotation" and "asymmetrical", but basically, yes, there are tricky things about the complex numbers.

I have no idea what you are concerned about with the absolute value, which is just part of the definition. But definitions and conventions are not the same as assumptions.

Thanks for staying with me. The word arbitrary is right on the money for the question I'm asking:

If you were on a professional committee exploring the "useful properties" of the "arbitrary" choice between sqrt(1) = 1 and sqrt(1) = -1 would you vote for:

sqrt(-i^2) = sqrt(1) = 1 (current convention) or sqrt(-i^2) = i sqrt(i^2) = i*i = -1 (calculation, not convention)

Which roots would you call "trivial" -- negative or positive?

I am not asking you to vote so much to tell me whether you see this a valid distinction, worthy of exploring the relative advantages of the two options.

Again, thank you for staying with me. I am looking for help with articulating my question is such a way that mathematicians might be willing to give it some thought.

Another way of asking the question: if I could prove employing the calculation [sqrt(-i^2) = i sqrt(i^2) = i*i = -1] that a hypothesis is false -- and this could not be proven employing the convention sqrt(1) = 1, would the proof be accepted by a journal for peer review?

 

[Dr.Peterson]

But as I said, the convention is for real numbers; we don't really have a convention at all for complex numbers.

And for real numbers, the convention is not truly arbitrary. I said it was "somewhat arbitrary", and pointed to the fact that this choice (rather than, say, the negative root) leads to convenient rules such as [MATH]\sqrt{xy} = \sqrt{x}\sqrt{y}[/MATH]. Would you want to abandon that? It would not be true for real numbers if you took the square root of 1 to be -1.

The fact is that when you bring in complex numbers, there is no choice of convention that makes everything work out nicely. Believe me, mathematicians have given this plenty of thought.

I'm not at all sure what you mean by "trivial". Has someone used that word?

 

[Cubist]

if I could prove employing the calculation [sqrt(-i^2) = i sqrt(i^2) = i*i = -1]

The above calculation doesn't seem valid for either "sqrt(x)" or "-sqrt(x)". Did you type it in correctly?

To illustrate let's rewrite it as:-
step A: f(-i^2) = i*f(i^2)
step B: i*f(i^2) = i*i
step C: i*i = -1

CASE 1: Let f(x)=sqrt(x). The following manipulation: f(-x) = i*f(x) only holds for x≥0 therefore step A would not hold
CASE 2: Let f(x)=-sqrt(x). Then f(i^2) = f(-1) = -i therefore step B would not hold

Please check my logic, I might not be correct!

 

[Dr.Peterson]

The above calculation doesn't seem valid for either "sqrt(x)" or "-sqrt(x)". Did you type it in correctly?

To illustrate let's rewrite it as:-
step A: f(-i^2) = i*f(i^2)
step B: i*f(i^2) = i*i
step C: i*i = -1

CASE 1: Let f(x)=sqrt(x). The following manipulation: f(-x) = i*f(x) only holds for x≥0 therefore step A would not hold
CASE 2: Let f(x)=-sqrt(x). Then f(i^2) = f(-1) = -i therefore step B would not hold

Please check my logic, I might not be correct!

Yes, what you say about step A amounts to the fact that, with the usual definition of the principal root, [MATH]\sqrt{xy} = \sqrt{x}\sqrt{y}[/MATH] applies only for positive x and y, where all numbers involved, including the roots, are real. (It happens to hold if both are negative, but that does not fit the conditions of the theorem anyway.) This is what I meant in saying that the definition is applied only over the reals.

I think the OP is thinking that that rule must always apply, and that the fact that things don't work here is due to an error in the definition of the principal root, rather than to the application of a theorem where it doesn't apply.