1
$\begingroup$

There was a long discussion in a forum I visit in where a calculus teacher was being critical of Stewarts Calculous for making a distinction between points and vectors. He argued that no such distinctions are made in Eastern Europe or Germany and argued that mainstream US Mathematics has been perverted by bad textbooks.

I was just wondering what this guy was talking about. What is the differences between how vectors and points are defined in and outside the US.

Here is one of his arguments,

There is no such terminology as "geometric point". The terminology "geometric vector" is simply confusing and inaccurate and should be avoided. The "displacement vector" referred to by Stewart is not a vector. It models a force applied to a point. In Bulgarian, that is called an "arrow" by physicists (to avoid confusing it with a "vector"). That object is very different from vector, and the point of application matters. When adding arrows based at different points, the resulting arrow has an application point different from the bases of the two original arrows.

The "geometric point" terminology stems from misunderstanding of the terminology of Euclidean geometry. Such misunderstanding often occurs to lower level educators such as Stewart. ... However, what he calls "vector" in english is called "displacement vector", and in Bulgarian it's called "arrow". A "displacement vector" is not a vector. Displacement vectors do not satisfy the axioms of a vector space. Vectors can be defined as equivalence classes (set-theoretically) of displacement vectors.

$\endgroup$
5
  • 1
    $\begingroup$ I think the point is that you can't really add points or do many operations with them. Points are just coordinates with no extra structure. In general, points are elements of Cartesian products of sets and sets need not carry any sort of notion of addition or any other meaningful operation. Vectors however are equipped with a notion of addition (among other things) by definition. $\endgroup$
    – Cameron Williams
    Oct 12, 2014 at 2:23
  • 1
    $\begingroup$ Here is an extremely relevant MSE post from a while back: math.stackexchange.com/questions/645672/… $\endgroup$
    – Cameron Williams
    Oct 12, 2014 at 2:26
  • $\begingroup$ While mainstream math may suffer from bad texts in the US, I disagree with that dude's assessment of points and vectors. Conflating the two would certainly limit a student. $\endgroup$
    – rschwieb
    Oct 12, 2014 at 2:26
  • $\begingroup$ I'd be interested in reading the thread you're referring to. Can you link it in the OP? $\endgroup$
    – Kaj Hansen
    Oct 12, 2014 at 2:26
  • $\begingroup$ I added some of his arguments to the question. I won't link the discussion because of it's poor overall quality. $\endgroup$
    – MVTC
    Oct 12, 2014 at 2:54

1 Answer 1

Reset to default
1
$\begingroup$

Not all vectors are based at the same point. Consider the vector based at (0,0) and ending at (0,-1) and the vector based at (3,2) and ending at (3,1). They have the same magnitude and direction, but they are not strictly speaking the same vector. The teacher would probably denote the first vector as the point (0,-1), but how would the teacher denote the second vector? And how would the teacher express the idea of a vector field, where there are many vectors at the same time not based at the same point?

For some purposes two vectors that are parallel translates of one another can be regarded as the same, but not for all purposes. If the teacher thinks the vector corresponding to the force of gravity on a copy of Stewart's calculus book is the same no matter where the book is held, then drop one copy of the book on the floor and one copy on the teacher's foot. I suspect the teacher will quickly agree that those two vectors are not the same even though they have the same magnitude and direction.

$\endgroup$
2
  • $\begingroup$ I added his argument corresponding to this issue in the question. $\endgroup$
    – MVTC
    Oct 12, 2014 at 2:47
  • $\begingroup$ Note that the "controversial" identification is the standard $T_p({\bf R}^n)\cong {\bf R}^n$ (tangent space at a point). $\endgroup$
    – tomasz
    Oct 12, 2014 at 2:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .