# How do you decide whether the relation x=y^2 defines a function?

Mar 5, 2018

This is a function of x and y. Can be wriiten as $f \left(x\right) = {y}^{2}$

#### Explanation:

A function is a relatioship between two variables broadly.

Mar 6, 2018

$\text{The answer is:"\qquad "the relation" \qquad x \ = \ y^2 \qquad "is not a function.}$

$\text{Please see below for a demonstration, and explanation, of this.}$

#### Explanation:

$\text{We are given the relation:} \setminus q \quad \setminus q \quad x \setminus = \setminus {y}^{2.}$

$\text{We are asked to decide if it defines a function.}$

$\text{If no matter what the value of the first variable," \ x, "there is}$
$\text{precisely one value of the second variable," \ y, "connected}$
$\text{to it inside the relationship -- then it will be a function. If this}$
$\text{breaks down for even one value of the first variable, it will fail}$
$\text{to be a function. That is to say, if for some value of the first}$
$\text{variable, there are two or more values (or no values) of the}$
$\text{second variable connected to it inside the relationship, then it}$
$\text{will not be a function.}$

$\text{Note -- in general, there is no procedure to decide if an}$
$\text{arbitrarily given relation is functional [ -- is a function or not].}$
$\text{The truth is, in general, there are no such procedures. Our}$
$\text{case, thankfully, turns out to be simple enough to make the}$
$\text{decision, let's say, using good instincts!! }$

$\text{We have:} \setminus q \quad \setminus q \quad x \setminus = \setminus {y}^{2.}$

$\text{We ask, in our mind, for a given value of" \ \ x, "how many values}$
$\text{of" \ \ y \ \ "are connected to it in the relationship -- one, or more}$
$\text{than one ?}$

$\text{That is to say, for a given value of" \ x, "how many solutions} \setminus \setminus y \setminus \setminus$
$\text{are there to the relation:" \ x \ = \ y^2 \ \ "? -- one, or more than one ?}$

$\text{For example, for" \ \ x \ \ "taking the value" \ 1, "how many solutions} \setminus \setminus y$
$\text{are there to the resulting relation:" \qquad \qquad \underbrace{1}_{x} \ = \ y^2 \ \ "?}$
$\text{ -- one, or more than one -- "?}$

$\text{This is, thankfully (!), easy to decide !! We proceed, looking}$
$\text{at the solutions of:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 1 \setminus = \setminus {y}^{2.}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {y}^{2} \setminus = \setminus 1.$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad y \setminus = \setminus \setminus \pm \sqrt{1} .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad y \setminus = \setminus - 1 , 1.$

$\text{So, for" \ \ x \ \ "taking the value" \ 1, "there are two values for} \setminus \setminus y \setminus \setminus$
$\text{connected to it in the given relation:" \ -1, 1 . \ \ "So, more than}$
$\text{one value for" \ \ y, \ "for this value of" \ \ x. \ \ "This ends the decision}$
$\text{right here.}$

$\text{We can stop immediately now -- and conclude that the given}$
$\text{relation is not a function.}$

$\text{This is our result:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{the relation" \qquad x \ = \ y^2 \qquad "is not a function.}$

$\text{I want to make a perhaps valuable note, to keep perspective.}$

$\text{If in the above work, we had picked the value of" \ \ 0 \ \ "for} \setminus \setminus x \setminus \setminus$
$\text{to take in the relation, and then looked to see how many}$
$\text{solutions" \ \ y \ \ "there are to the resulting relation:} \setminus \setminus 0 \setminus = \setminus {y}^{2} ,$
$\text{we would have looked at the solutions of:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 0 \setminus = \setminus {y}^{2.}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {y}^{2} \setminus = \setminus 0.$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad y \setminus = \setminus 0 , \setminus \quad \text{only} .$

$\text{And we would have concluded that, for" \ \ x \ \ "taking the value} \setminus 0 ,$
$\text{there is exactly one value" \ \ y \ \ "connected to it in the given}$
$\text{relation:" \ \ 0. \ \ "Exactly one value for" \ \ y, \ "connected to this}$
$\text{value of} \setminus \setminus x .$

$\text{What does this tell us about whether the given relation is a}$
$\text{function ? NOTHING !!}$

$\text{Because there is exactly one value for" \ \ y \ \ "for this value of} \setminus \setminus x ,$
$\text{we cannot exclude the relation from being a function, as we did}$
$\text{above using the value of" \ \ 1 \ \ "for} \setminus \setminus x .$

$\text{We also cannot say from this that the relation is a function,}$
$\text{either. Why ? The work here told us what happened with the}$
$\text{values for" \ \ y \ \ "connected with the value" \ \ 0 \ \ "for" \ \ x \ \ "-- exactly one}$
$\text{value for" \ \ y. \ \ "But it told us nothing about the values for" \ \ y \ \ }$
$\text{connected with any other value for" \ \ x. \ \ "Other values for}$
$\setminus \setminus x \setminus \setminus \text{might have exactly one value for" \ \ y \ \ "connected to it, }$
$\text{might have more than one value for" \ \ y \ \ "connected to it, or}$
$\text{might have no values for" \ \ y \ \ "connected to it. We cannot know}$
$\text{unless we go back and check values for" \ \ x, "other than" \ \ 0.}$

$\text{What other values for" \ \ x, "should we check -- other than" \ \ 0 \ \ "?}$

$\text{The truth is, in general, there is no way to determine what}$
$\text{other values for" \ \ x \ \ "(if there are any) we should check. We}$
$\text{were lucky we picked the value" \ \ 1 \ \ "for" \ \ x \ \ "above -- which}$
$\text{allowed us to make a decision on this relation. For certain}$
$\text{types of relations, there are ways to determine other values}$
$\text{to check. In general, there is no such procedure for finding}$
$\text{such luck -- just hope, and good instincts !!}$