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Square Root Function

Date: 12/16/98 at 18:37:26
From: Jason Lynch
Subject: Square root problem

Given this problem: 

   sqrt(2x + 3) = -1

solve for x.

I say the answer would be -1, because:

   sqrt(2(-1) + 3) = -1
      sqrt(-2 + 3) = -1
           sqrt(1) = -1

My second year Algebra teacher says otherwise. She says that there 
would be no solution. I fail to see the reasoning, as -1 is one 
possible square root of 1.

Jason Lynch

Date: 12/17/98 at 16:58:23
From: Doctor Rick
Subject: Re: Square root problem

Hi, Jason! You've asked a good question.

What's going on here, I think, is that we have two separate concepts 
that can easily be confused. 

One concept is the process of finding the root of a square; for 
instance, the number x such that x^2 = 4. You know that this has two 
solutions: 2 and -2 are both roots of this equation.

The other concept is the square root FUNCTION. A function takes in one
number and returns another number. The number that it returns must be 
UNIQUE since a function is by definition single-valued. You can't put 
in 4 and get out both 2 and -2.

The square root function is therefore DEFINED so that sqrt(x) returns 
the NON-NEGATIVE root of y^2 = x. Then we say that the two roots of 
x^2 = 4 are +-sqrt(x), plus or minus the [non-negative] square root 
of x.

Admittedly this is a rather arbitrary definition, and your teacher's 
answer feels artificial. But if functions were not single-valued, we'd 
have a mess. And when you use the square root function in real 
situations, it will make sense. For instance, the distance between the 
points (0, 0) and (x, y) is sqrt(x^2 + y^2), and it makes sense that 
the distance is a non-negative number.

Here is something from our Archives that might add a bit to my 

- Doctor Rick, The Math Forum   

Date: 03/02/2011 at 13:12:20
From: Rosemary
Subject: 1:n mappings in relations and functons

Why must functions be single-valued?

In answering this question about the square root function, why does Dr.
Rick say "if functions were not single-valued, we'd have a mess"? What is
wrong with having multiple values, as in square roots or quadratic
equations? Relations can have 1:n mappings. Why not have the square root
function be a two-valued relation?

This is part of a more general question I have about why single-valued
functions are so dominant in pure mathematics, including both analysis and
group theory.  Why not base analysis, including calculus, and group theory
on something richer such as relations?

Date: 03/03/2011 at 03:52:34
From: Doctor Jacques
Subject: Re: 1:n mappings in relations and functons

Hi Rosemary,

The short answer is simply that "function" is the name we give to a
single-valued relation. Everything in mathematics is based on definitions;
we try to choose definitions to make them as useful as possible.

More general relations are indeed also present in all areas of
mathematics. For example, in analytic geometry, we study curves like
conics, which are described by relations that are generally not functions;
and, usually, the last thing you should do is to try to express one
variable as a (multi-valued) function of the other. So general relations
do have their place in mathematics.

On the other hand, some concepts require the use of functions.

Consider, for example, the definition of a continuous function. A function
f : R -> R is continuous at 'a' if, for every eps > 0, there is an open
interval N containing a such that, for every x in N:

  |f(x) - f(a)| < eps

What happens if f is multi-valued? In general, you cannot choose any value
for f(x) and any value for f(a) independently; try it with f(x) = sqrt(x)
and a = 1. To ensure that the definition makes sense, you would have to
divide the values of f(x) into "classes" of related values, and each such
class would be a continuous, single-valued function. As you see, we are
back to single-valued functions again. Things will get even uglier if you
try to define derivatives or functions over the complex numbers.

There are indeed some cases where multi-valued functions would make sense.
This happens, for example, for functions (like sqrt or the inverse
trigonometric functions) that are defined as inverses of functions that
are not injective (one-to-one).

In the case of sqrt, we can define a single-valued function by choosing
the positive square root. This choice is not totally arbitrary, because it
ensures that the relation ...

   sqrt(xy) = sqrt(x)sqrt(y)     [1]

... holds for all non-negative x and y. This would not be true if we had
chosen the negative square root.

However, if we consider the function sqrt over the complex numbers, such a
preferred choice is no longer available. In fact, it can be proved that no
single-valued definition of sqrt over C will make [1] true in all cases.

In such a case, other techniques are used, based, for example, on Riemann


The idea is to define the function over a larger domain that contains two
copies of the complex plane, and to have a single-valued function from
that extended domain to C. For each non-zero z in C, there are two points
z1 and z2 on the extended domain; f(z1) corresponds to one square root and
f(z2) to the other.

To summarize:

  * single-valued functions are needed in many cases.
  * general relations are also used uin many cases.
  * there are techniques to define something like multi-valued
    functions (in some particular cases).

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum   
Associated Topics:
High School Square & Cube Roots
Middle School Square Roots

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