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Homework Help: Spivak Error? Definition: Absolute Value

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

    In Spivak's Calculus 4e, he defines absolute value as:

    [tex]|a|= a,\qquad a\ge 0 \qquad \text{ and } \qquad -a,\qquad a \le 0 [/tex]

    Did he really mean to include the '[itex]\le [/itex]' and not just '<' ?

    I know it does not affect the answer, but I didn't think that you could include the '=' condition for both.

    Maybe I am wrong, and of course I am stressing over the trivial. But that's what I do :wink:

    Any thoughts?
  2. jcsd
  3. Nov 14, 2009 #2


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    could it be valid because, say a = 0, then
    a = a + (-a)
    a = 0 + (-a)
    a = (-a)
  4. Nov 14, 2009 #3


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    You're right to think about it -- this is a question of whether the function is well-defined.

    The relation he defines is clear and unambiguous. It's clearly a total relation in the sense that every element in the domain relates to at least one thing.

    So the only question is if each element on the domain relates to at most one thing (which is easy to check).
  5. Nov 14, 2009 #4


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    An example of something that is not well-defined is the following "function" from the rationals to the rationals:
    f(p/q) = p + q​
    where p ranges over all integers, and q ranges over all nonzero integers.

    Of course, not all definitions of this type are ill-defined: for example, the following function is perfectly well-defined:
    [tex]f\left( \frac{p}{q} \right) = \frac{p^2 + 6 p q - 13 q^2}{3p^2 - 15 p q + 4 q^2}[/tex]​
  6. Nov 15, 2009 #5

    I don't know Hurkly. Nobody else (that I have seen) defines it like this? I have always seen it as:

    |a| = a for all a [itex]\ge[/itex] 0 and -a for all a < 0. (1)

    If you write that f(x)= |a| = a for all a [itex]\ge[/itex] 0 and -a for all a [itex]\le[/itex] 0 (2),

    you are implying that f(0) = 0 and -0.

    Now of course I know that 0 = -0 ....but that is not the point. The point is that (1) seems perfectly well-defined to me, so why bother writing it as (2) ?

    What does the 'both less-than equal to/greater than equal to' qualifier in (2) add to the definition that is not already included in (1)?
  7. Nov 15, 2009 #6


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    If nothing else, it has a chance to provoke thought or a discussion like this one. :wink:

    I write definitions like that too, and I do it mainly for artistic reasons -- it looks more natural to me.

    But if I wanted to rationalize my art, I would point out the following two facts:
    . This version doesn't leave the reader to question whether the expression |x| = -x remains valid at x=0
    . This version makes it even more obvious that it defines a continuous function
  8. Nov 15, 2009 #7
    I don't know.....I am not going to lie: I still don't like it. You're the math ninja, so I will take yours and Spivak's word for it, but in my heart, I still won't like it.

    Hurkly, you said:
    But here is my point: Since a = 0 falls under the |x| = x for x>=0 then the reader should not even be asking this question. Do you see what I am saying?

    x=0 implies |x| = 0 ... no reason to go any further.

    And I will reiterate what i said in the prior post: To me, the way Spivak has defined |x| is NOT a function. I know I am going to get bashed for this, but her is why: If we write out his definition explicitly as a piecewise function with 4 conditions we have:

    |x| =

    i. x if x > 0,
    ii. x if x = 0,
    iii. -x if x < 0,
    iv. -x if x = 0.

    from ii & iv we are assigning both 'x' and '-x' to f(0).

    I realize that f(x=0)=0=-0, but it still does not sit well with me.

  9. Nov 15, 2009 #8
    But what does it matter? It's like saying I can't define x+1 as x + 2 -1 because the expression looks different even though they are equal at all points where it matters. In much of elementary mathematics we are used to give functions as compositions of nice functions, or piecewisely as compositions of nice functions, but a function is just a certain mathematical object and not always easily expressible. When we say: define f : Z -> Z by f(x) =x what we really mean is:
    There exists a unique function g : Z -> Z satisfying g(x) = x. Let f be this function.
    We don't state it this way because we have developed a number of ways to specify a function where we are sure that the function exists and is uniquely determined so this is implicit. If we use a piecewise-definition where several of the conditions can hold simultaneously we have violated the established rules and must explicitly check that all values agree.

    There is nothing wrong with Spivak's definition. What he pretty much says is:
    Lemma: There exists a unique function N : R -> R such that N(x) = x for x >= 0 and N(x) = -x for x <= 0.
    Proof: Omitted due to its simplicity.
    Definition: Let |x| denote this function.

    Math is not a formal language. A mathematical argument is a way to pass on an idea, and as long as that idea is clear we don't care whether the argument stays within some arbitrary framework.

    An alternative way to define the absolute value function would be as (I'm not saying this is a good way, just trying to show that there is no fixed way to specify it):
    [tex]\{(x,y) \in \mathbb{R}^2 | (y-x)(y+x)=0, y \geq 0\}[/tex]
    Here again we need to check that this is really a function, but if we examine it we find that it is.

    By the way MathWorld use the same definition as Spivak:

    My guess is that this is just a reflex from the years of high school math with arbitrary rules forced upon us (I'm assuming you haven't taken much math beyond high school since you're doing Spivak).
  10. Nov 15, 2009 #9
    Really? It seems like every book that I have been turning to lately seems extremely formal.

    A reasonable assumption, but I have actually taken all math requirements for an Engineering curriculum. But I have gotten fed up with the 'Applied Calculus' that we engineers (and presumably a lot of physicists) are spoon fed. So I am giving myself a good once over with Spivak. I originally thought I was going to be a math major before I decided on Engineering, so I enjoy re-teaching myself the maths.


    I will take all that you have said into consideration and I like the way you explained it, but I hope that somebody out there can see why I dislike it.

    Since it is zero that we are talking about, it doesn't matter. But, let's say for a second that it was not. If I tried to define a function as:

    N(x) = x for all 'x' including 'b' and N(x)=-x for 'b' as well.

    This would not fly. It is only by virtue of the fact that 'b' in this case is zero that this definition works.

    So it seems like instead of just following the 'template,' he is just defining this way because, in this particular case, he can....and for no other good reason.

  11. Nov 15, 2009 #10
    I'm sure we have all had similar experiences of being presented a new approach to a problem where we could have used an old one. We don't really see why the new approach is better, and consider our standard approach the easiest, but at some point we come to realize the advantages of the new approach. EDIT: The worst cases of this is when like in this case no simple example shows the real value in the technique.

    Yes formally there is no good reason. But in terms of aesthetics there is a difference. Spivak's def. is symmetric while only having two cases. It's easy to see that it's continuous as Hurkyl mentioned. Really the difference is quite minimal and I'm sure no one would mind if you adopted one of the alternative definitions. This is merely a matter of preference.

    In this case it's pretty trivial, but in other cases it may be significantly simpler to specify a relation first and then confirm that it's actually a function.

    If you need a non-trivial example:
    I'm having a hard time thinking of a simple, but non-trivial example so I'm just going to post what first springs to mind which is a common technique in group theory.
    Let [itex]\mathbb{C}^\times[/itex] denote the non-zero complex numbers.
    Suppose we are given a function [itex]f : \mathbb{C}^\times \to \mathbb{C}^\times[/itex] with the property that for all [itex]x,y \in \mathbb{C}^\times[/itex] we have f(x)f(y)=f(xy), f(1)=f(-1)=f(i)=f(-i)=1. Now let [itex]N = \{1,-1,i,-i\}[/itex]. For all [itex]z \in \mathbb{C}^\times[/itex] let [itex]zN = \{z,-z,zi,-zi\}[/itex]. Now let F denote the collection of all sets of the form zN for some [itex]z\in \mathbb{C}^\times[/itex], that is:
    [tex]F = \{zN | z \in \mathbb{C} \}[/tex]
    Now I claim that there exists a unique map [itex]g : F \to \mathbb{C}^\times[/itex] such that g(zN) = f(z). The easiest way is to first define the relation:
    [tex]g(zN) = f(z)[/tex]
    and then confirm that it's a well-defined function since if zN = wN then there exists some [itex]n \in N[/itex] such that z=wn and thus we have:
    [tex]g(zN) = f(z) = f(z)f(n) = f(zn) = f(w) = g(wN)[/tex]
    which shows that g is well-defined. Thus we have shown the existence of such a function, and it's clear that any other function must be equal to this one (we could carry out the details, but this is not important in this example).
    Last edited: Nov 15, 2009
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