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From my readings on geometry, I understand that a line is a "undefined" or "primitive" term. If that is the case, how do we justify saying a 1st degree polynomial represents a line or is "linear", when we do not have a definition for what a line is in the first place? I am inclined to think that this is also an axiom, something that we take for granted as per our observations. But I am not sure and would like to know if there is a more detailed explanation as to why this is so.

jacob78
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  • My understanding is that in Hilbert's axioms for geometry (which are based upon those of Euclid), a "line" is indeed a primitive term. But this is not so for analytical geometry, i.e. the study of geometry using cartesian coordinates. In the latter case, I think it is typical to give a line a specific definition: for instance, in the case of the cartesian plane, we could define a line as a subset of $\mathbb R^2$ of the form ${(x,y):ax+by=0}$ for some $a,b\in\mathbb R$. – Joe Nov 06 '23 at 00:52
  • To add to Joe's comment, one can show that if points and lines are defined (respectively) as ordered pairs, and solution sets of ordered pairs, as they are in Cartesian coordinates, then (this is oversimplifying a bit since other definitions need to be clarified as well) Hilbert's axioms are satisfied, so all of the theorems you learn in a typical proof based geometry class (which will be provable from Hilbert's axioms) apply to analytic geometry as well. – M W Nov 06 '23 at 00:59
  • @Joe I see what you're saying. The definition of a first degree polynomial is taken as the definition of a line in analytical geometry. Still, I think the core of my question lies in justifying the assumption that the definition of a first degree polynomial, when graphed as a relation in the cartesian plane, represents the geometric object that we call a line and do not have a definition for. – jacob78 Nov 06 '23 at 01:24
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    @jacob78: Well, I think the justification is that if you try to plot the graph of $ax+by=0$, then it looks like a line. I don't think it is possible to give a rigorous justification because before you formally define what a line is in some way (or leave it as a primitive notion as you do in classical geometry), it is necessarily an informal notion. Nevertheless, we possess a great deal of geometric intuition for what a line is, and so when we plot the graph of $ax+by=0$, this confirms that our definition was "correct" in some sense. – Joe Nov 06 '23 at 01:31
  • @jacob78: Alternatively, you could try to turn whatever informal idea you had of a line into a rigorous definition, and then you could prove that it is equivalent to my one. E.g. you could define a line as the graph of a function with a constant gradient (there is a slight hiccup here with lines parallel to the $y$-axis, but otherwise your definition is completely equivalent). – Joe Nov 06 '23 at 01:34
  • There is something that I haven't mentioned that I really ought to have. The more fundamental question is: why should we think that the real number system can be represented by a line in the first place? Thankfully, that question has already been answered. – Joe Nov 06 '23 at 12:20

2 Answers2

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It is a definition in analytic, coordinate geometry, and a theorem in synthetic, axiomatic geometry.

The confusion comes about because in synthetic geometry the fundamental entities are points and lines: a priori there is no notion of number with which to assign coordinates, so we must invent one.

A number $a$ is an ordered triple of collinear points $(O,I,A)$ such that $O$ and $I$ are distinct. We imagine $a$ to be the formal ratio $$a = OA:OI\text{.}$$

Two numbers $a=OA:OI$ and $a'=O'A':O'I'$ are equivalent, $a\equiv a'$, if lines $OO'$, $II'$, $AA'$ are parallel. von Staudt showed that we can define addition and multiplication operations on numbers that obey the usual laws.

Consequently, we can show that, if $OXPY$ is a parallelogram with $OIXA$ and $OJYB$ each collinear sets, and if we define the numbers $x=OX:OI$, $a=OA:OI$, $y=OY:OJ$, $b=OB:OJ$, then $P$ lies on $AB$ iff $$\tfrac{x}{a}+\tfrac{y}{b}\equiv 1 \text{.}$$

K B Dave
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I'll take for granted that we have the usual definition of cartesian axes and of coordinates $(x,y)$ of a point in the plane, together with the usual method to compute the distance of two points from their coordinates.

Note first of all that an equation like $y=a$ (where $a$ is any given real number) represents a line. In fact point $A=(x,a)$ forms a rectangle with points $P=(0,a)$, $O=(0,0)$ and $H=(x,0)$, because $AH=OP$ and $AP=OH$. Hence $A$ lies on the line through $P$ parallel to $x$-axis.

The same reasoning also works for an equation like $x=a$.

It remains to consider the case where the equation can be written as $y=mx+q$. Suppose $m>0$: the points satisfying that equation belong to the locus $L=\{(x,mx+q)\ |\ x\in\mathbf R\}$. Point $Q=(0,q)$ is in $L$ and all points $(x,q)$, as shown above, lie on a line $r$ through $Q$ parallel to $x$-axis.

If $P_1=(x_1,mx_1)$, $P_2=(x_2,mx_2)$ are two points in $L$, let $H_1=(x_1,q)$, $H_2=(x_2,q)$ be their perpendicular projections on $r$. Triangles $QH_1P_1$ and $QH_2P_2$ are similar, because $P_1H_1/H_1Q=P_2H_2/H_2Q=m$, hence $\angle P_1QH_1=\angle P_2QH_2$.

If $x_1$ and $x_2$ have the same sign then $P_1$ and $P_2$ lie on the same half-plane with respect to $r$, otherwise they lie on opposite half-planes. In both cases, from the existence and uniqueness of the ray forming a given angle with another ray, we conclude that $Q$, $P_1$, $P_2$ are aligned and $L$ is a line.

A similar argument can be used if $m<0$.

Intelligenti pauca
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