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Why y is a function of x ?

  1. Sep 27, 2011 #1
    I have some questions , first : in college Algebra course I didn't used to prove that y is a function of x for example must I prove that polynomials are function (I did by mathematical induction but I see that this is useless) , now I don't know why I am thinking different I want to prove somethings I am not used to even think to prove., is that a good thing to think by this way .??? I think I am stupid in someway than I was during reading the college algebra text-book .
    second : how must I understand if and only if part in fundamental graphing principle of equations and functions.

    The Fundamental Graphing Principle
    The graph of an equation is the set of points which satisfy the equation. That is, a point (x,y) is on the graph of an equation if and only if x and y satisfy the equation.

    I used to not to concentrate at these points , but now after I began to read in other branches of mathematics , I began to think by another ways I don't know why .
    Here I understand that that if (x,y) is on the graph then x and y will satisfy the equation until now we are assuming that it is possible that there exists some x and y which satisfy the eq. but (x,y) is not on the graph . the second implication is saying that if x and y satisfying the equation then (x,y) lies on the graph , which also says that it is possible to have (x,y) on the graph but x and y doesn't satisfy the equation here the first implication says that x and y satisfy the equation so all x and y which satisfy the equation corresponds to all (x,y) on the graph. so in this way I understand why the equation of circle for example describe only the set of all points on the circle and so with ellipse , hyperbola , parabola , lines , ......

    I think this way I think is a stupid so please help me.

    thanks .
  2. jcsd
  3. Sep 27, 2011 #2


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    I think it's a good thing that you're challenging yourself to prove things the books take for granted. Some of those things are non-trivial. The statement "Every polynomial is a function" is very easy to prove, if you understand the definition of a function. So you might want to look at that. Suppose that p is a polynomial. It's sufficient to prove that (a) for each real number x, p(x) is a real number, and (b) for all real numbers x and y, if x=y, then p(x)=p(y).

    The only way to define "the graph of an equation" (assuming that the equation is of the form f(x,y)=0) that makes sense to me is to say that the graph is the set of all (x,y) such that f(x,y)=0. This means that (x,y) is a member of the graph if and only if f(x,y)=0.
  4. Sep 27, 2011 #3
    y is not always a function of x. Think of a circle, x^2 + y^2 = 1. In this case you can NOT express y as a function of x. Do you see why not? This is an instructive example. Sometimes y is a function of x and sometimes it isn't.
  5. Sep 27, 2011 #4


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    The use of "if and only if" is used when the two clauses are linked as a definition. In general meaning, "if and only if", for clauses q and p, would be BOTH of these:
    If q then p, AND if p then q.

    Functions: The Vertical Line Test. If you can find a vertical line and place it every where on a graph and it intersects the graph no more than at one point, then the graph is that of a function. If any vertical line intersects a graph at more than one point, then the graph is not a function.
  6. Sep 29, 2011 #5
    yes , I proved first that Every polynomial with real coefficients and real variables will be real , that is we know polynomial is sum of terms ax^n which can be easily proved considering x^n as a repeating multiplication (so it is real since x is real number , and multiplication of real numbers is enclosed over real set ) similarly we say ax^n is real number , and then we can say that a polynomial is always real because adding real number is real .then after that to prove that every polynomial is function , there are many ways to do that (I think about ) , one of these I have , is to prove that ax^n is a function , that is I assumed , [itex]b_1 = ac^n[/itex] and [itex]b_2 = ac^n[/itex] , which can be proved that [itex]b_1 = b_2[/itex] by many methods one of which i like is to prove by mathematical induction,
    and then define the polynomial to be some of these function , and use the theorem of that the some of functions of x is a function of x , if x is element of domain of each.

    Here , please help me because I teach my-self Mathematics without an instructor so I am afraid , I don't understand well .as you say I see your definition of the equation is very useful an the best , but I can't use it because I am Not good until now at function of two variables , but for now I will use the fundamental graphing principle I have written.
    I want to know now if I am understanding it well , because I use it always to understand the derivation of equations of Geometric loci . here we take the simplest I think an please see if I am understanding the derivation of point slope form well and why it represents only all points on the line determined .

    first from definition of slope we have every line has an unique slope , and thus if we suppose aline l of slope m , and containing [itex]( x_0,y_0 )[/itex] , and let (x,y) be any other point on the line then we have
    (x,y) is on the line l if and only if [itex]m = \frac{y-y_0}{x-x_0}[/itex]

    her I began to argue with myself asking if the the equation determines all the points on the line only , I then begin to say that if x and y doesn't satisfy the equation then the statement say that (x,y) is not on the line l , then logically all points (x,y) on the line if x and y satisfy the equation but is there any x and y which satisfy the equation but (x,y) is not on the line , the second implication says that if (x,y) is not the line then x and y doesn't satisfy the equation and this answers our question and then we have the equation exactly represent all the points on the line l only .

    Another approach is to say that the graph of equation is the set of all points (x,y) where x and y satisfy the equation and according to "(x,y) is on the line l if and only if [itex]m = \frac{y-y_0}{x-x_0}[/itex]" we have the graph of equation is the set of all points (x,y) such that (x,y) is on the line .

    Another question concerns about implications into general , that is if p and q are two statements and " [itex]p \Rightarrow q[/itex]",is that means that we can't write " [itex]p \Rightarrow q[/itex]" if we are not sure that [itex]\neg q \Rightarrow \neg p[/itex]

    for example if x^2 = 4 , then x =2 , but is this statement is true without adding " or x=-2 "
    and also if x not equal 2 then it is not true that x^2 not equal 4 , so must we are careful when we write if ..... , then ..... statements .

    Thanks for helping me .
    Last edited: Sep 29, 2011
  7. Sep 29, 2011 #6
    I know this wasn't the point I am talking about .
    this was only a title of the topic because I didn't know what to write my only questions what I had write .
    and thank you for helping me .
    Last edited: Sep 29, 2011
  8. Sep 30, 2011 #7

    Thank you for helping me , but I don't men to prove graphically I want an Analytic proof.
  9. Sep 30, 2011 #8


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    You did not say how formal this task must be, but you just express what the graphical approach would mean. Know the meaning of a function. A function is a relationship between a set of numbers, one set being an input, the other set being an output, so that every output has no more than one input. (The more analytical people know what I mean and may give their own statements to correct me.)

    Why is x=y2 not a function? Nevermind that it "looks" like x is a function of y. The independant variable of x has more than one output except at the vertex. It is a relation but not a function.
  10. Oct 1, 2011 #9


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    I would say that axn is a number, not a function. It's a number in the range of the function f defined by f(x)=axn for all x. This is a minor detail, but I find that a lot of confusion can be avoided by being careful with these things. I don't understand what you're trying to do when you set [itex]b_1 = ac^n[/itex] and [itex]b_2 = ac^n[/itex] and then talk about proving that [itex]b_1=b_2[/itex]. You certainly don't have to use induction for that. Clearly x=y and z=y implies x=z. This follows from the fact that an equality sign means that what you have on the left is the same thing as what you have on the right.

    The term "polynomial" can be defined e.g. like this: A function p:ℝ→ℝ is said to be a polynomial if there's a positive integer n and real numbers a0,...,an such that [itex]p(x)=\sum_{k=0}^n a_k x^k[/itex] for alll real numbers x.

    If you insist on proving that a polynomial is a function, then you first have to define the term in a way that ensures that it's not immediately obvious that a polynomial is a function. Do you have such a definition in mind? It's obviously going to be impossible to prove that a polynomial is a function without using both the definition of "polynomial" and the definition of "function". So what are your definitions?

    It's wrong to say "if and only if" here, because [itex]m = \frac{y-y_0}{x-x_0}[/itex] is undefined when x=x0. So there's exactly one point (x,y) that's on the line and doesn't satisfy the equation. It's [itex](x_0,y_0)[/itex]. All points that satisfy the equation are on the line.

    [itex]p\Rightarrow q[/itex] and [itex]\lnot q\Rightarrow\lnot p[/itex] are equivalent statements. The former is true if and only if the latter is true. This can be proved by writing down the truth tables.

    Yes, [itex]x^2=4\Rightarrow x=2[/itex] is a true statement and so is [itex]x^2=4\Rightarrow (x=2\text{ or } x=-2)[/itex] and [itex]x^2=4\Leftrightarrow (x=2\text{ or } x=-2)[/itex]
  11. Oct 1, 2011 #10
    My definition of polynomial is the same you have written . I know What you meant by when x=y and y=z , then x=z ( transitive property ), so sorry because I Proved that [itex] b_1 = b_2[/itex] by induction . But is this enough to say that a polynomial as a function.
    Also I have two definitions of functions one I learned in college algebra and the other I have known from a book about set theory

    A relation in Which each y-coordinate is matched with only one y-coordinate is said to describe y as a function of x

    Another one : A binary relation F is called a function if [itex]aFb_1[/itex] and [itex]aFb_2[/itex] implies [itex]b_1 = b_2[/itex] for any [itex] b_1 ,b_2 ,a[/itex].In other words abinary relation F is a function if and onlt if for every a from dom F there is exactly one b such that aFb. the value of F at a is denoted F(a) (F(a) is not defined when a is not in the dom F).

    Another one I don't usually use is a function n f is a rule that assigns every x in a set D into only one value named f(x) in a set B .

    Second , I know that , I used if and only if many times so I removed the stipulation of except (x_0 , y_0 ) because see it is not important as it will be recovered in the equation of the form [itex]y - y_0 = m (x-x_0 )[/itex] right , so I avoided writing
    (x,y) on the line l where (x,y) not equal (x_0,y_0 ) if and only if [itex] m = \frac{y-y_0}{x-x_0}[/itex]

    I know I am not going precise But I thought that you will understand that I removed the stipulation Deliberately . What I wanted only to see if I understand well the reason that the equation of the line express only the line wanted .

    Third , I don't know why [itex]x^2=4\Rightarrow x=2[/itex] is true if it is , then if x not equal 2 then x^2 will not equal 4 ,and I see that is wrong because x may be -2 if it is not equal 2 (Right ? )

    Thank you very much .
    Last edited: Oct 1, 2011
  12. Oct 1, 2011 #11


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    Oops. No, [itex]x^2=4\Rightarrow x=2[/itex] is not a true statement. I don't know how I ended up typing that. I don't have time for a long reply now, but I'll explain these things in more detail when I get back (a few hours from now).
  13. Oct 1, 2011 #12
    yes , I thought I was going crazy , and have no background about logic when you said it is true . So now I realized why in solving equations we must be careful in using "implies".
  14. Oct 1, 2011 #13


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    Regarding the implications...

    The sentence [itex]x^2=4\Rightarrow x=2[/itex] doesn't have a truth value (it's neither true nor false), but each assignment of a value to the variable x associates a truth value with the sentence. That truth value is "false" if and only if the value assigned to x is -2. If you don't see why it's "true" when the value assigned to x is 1, consider the truth table that defines the [itex]\Rightarrow[/itex] symbol.

    When we write [itex]x^2=4\Rightarrow x=2[/itex] or say "if [itex]x^2=4[/itex], then [itex]x=2[/itex]", the sentence we have in mind is usually not [itex]x^2=4\Rightarrow x=2[/itex], but [tex]\forall x\in\mathbb R~~(x^2=4\Rightarrow x=2).[/tex] Even that is kind of imprecise. It's actually an abbreviation of the more precise [tex]\forall x~~ (x\in\mathbb R\Rightarrow(x^2=4\Rightarrow x=2)).[/tex] This sentence is false, because its negation is true. There are several ways to write the negation, for example: [tex]\exists x~~ (x\in\mathbb R\ \land\ (x^2=4\ \land\ x\neq2))[/tex] This sentence is true because [itex]-2\in\mathbb R[/itex], [itex](-2)^2=4[/itex] and [itex]-2\neq 2[/itex].

    OK, then there's no need to prove that all polynomials are functions. They are functions by definition.

    This is the one I use. Note that it says that a function and its graph are the same thing.

    This one appears in some form in most "introduction to calculus" textbooks, but you shouldn't think of it as a definition. Think of it as a suggestion about how to think about functions in order to understand them on an intuitive level.

    No matter what we choose to think of as the foundation of mathematics, there are always a few terms and symbols that are left undefined. It's perfectly OK to leave "function" undefined. If we do, we still need to explain how to think about functions, and we would do that with the statement "a function is a rule that...". It's not a definition but an elucidation (explanation) of an undefined term.

    Most mathematicians choose to think of ZFC set theory as the foundation of mathematics. Then the only things left undefined are the term "set", and the symbol [itex]\in[/itex]. In other words, we don't try to explain what a set "really" is, or what it "really" means for a set to be a member of a set.
  15. Oct 2, 2011 #14
    Yes That is I want to make sure about implication , so we must be careful in using if ... , then .......... , in solving equations .In fact I try to do my best to use if and only if in solving many equations in order to avoid the time of checking the solution .

    OK , it can be seen from the definition of polynomial that it is a function , I understand that as if [itex]b_1 = \sum_{k=0}^n c_k a^k [/itex] and [itex]b_2 = \sum_{k=0}^n c_k a^k[/itex], so since the right hand side of both equation is the same unique real number , then [itex]b_1 = b_2[/itex].

    What about my understanding to the graph of equations ???

  16. Oct 2, 2011 #15


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    Can you be more specific?
  17. Oct 5, 2011 #16
    This my last question is about Related rates ,

    for an easy example : a ladder 10ft long rests against the wall . if the bottom of the ladder slides away from the wall at a rate 1 ft /s , how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall ?

    solution : le x be the distance from the wall to the bottom of the ladder , y be the distance from the top of the ladder to the ground .

    we have [itex]\frac{dx}{dt} = 10 ft/s[/itex] .

    we use Pythagorean theorem :

    [itex]x^2 + y^2 =100[/itex]

    Now her is the problem , how can we differentiate both side with respect to t (time) , without knowing if y is defined implicitly as a differentiable function of t or not , i.e how we can know that dy/dt exist .

  18. Oct 5, 2011 #17


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    From the statement of the problem, you can infer that both x and y vary over time, which means that they are both functions of time, t. You can also assume that the motion of the ladder is smooth enough so that x(t) and y(t) are continuous, and that both functions are differentiable.
  19. Oct 6, 2011 #18


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    The proof goes like this:

    Let t0 be the real number such that [itex]x(t_0)=6[/itex]. By assumption, y satisfies [itex]y(t)=\sqrt{100-x^2(t)}[/itex] for all t in some open interval that contains t0. Denote the following functions by f,g and h respectively. [tex]
    & s\mapsto s^2\\
    & s\mapsto 100-s\\
    & s\mapsto \sqrt{s}
    \end{align}[/tex] Since
    [itex]y=h\circ g\circ f\circ x,[/itex]
    x is differentiable at t0,
    f is differentiable at x(t0),
    g is differentiable at f(x(t0)),
    h is differentiable at g(f(x(t0))).​
    the chain rule tells us that y is differentiable at t0.
  20. Oct 6, 2011 #19
    yes , I know this proof but some it becomes very hard to solve the equation , for general how we can know that the equation is define y implicitly as a differentiable function of x
    , is it good to me to take advanced course in Calculus to know how .

  21. Oct 6, 2011 #20


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    Check out the implicit function theorem (the indented statement at the end of the section "statement of the theorem").
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