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EIU MAT 3271 - Proof that the Real Affine Plane is a Model of Affine Geometry

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Proof that the Real Affine Plane is a Model of Affine GeometrySeptember 25, 2009The real affine plane is the interpretation of incidence geometry given by Cartesian coordinates:points are, by definition, pairs of real numbers; lines are solution sets of linear equations; a point isincident with a line if it is a solution to the equation that defines the line.The word “affine” in the name for the model comes from the fact that, in addition to satisfyingAxioms I-1, I-2, and I-3, it satisfies the Euclidean Parallel Property. (Recall that the EPP states that,given a line l and a point P not on l, there is a unique line through P that is parallel to l.) Geometrybased on these four axioms - that is, incidence geometry with the addition of the Euclidean ParallelProp e rty - is called affine geometry; a model of affine geometry is called an affine plane. (The meaningof the word “real” in the name of the model should be obvious!)Let us consider two triples (u, v, w) and (u0, v0, w0) to be equivalent if the equations with thesecoefficients (for x = x1, y = y1, and 1 = x0= y0, respectively) have the same solution sets. (It shouldbe obvious that the relation defined in this way is an equivalence relation!) Recall that two e quationswith coefficients (u, v, w) and (u0, v0, w0) have the same solutions if (u0, v0, w0) = (λu, λv, λw) for someλ 6= 0. (In fact, the converse also holds, but we don’t nee d it for our proofs.)Proposition 1 (Axiom I-1 of incidence geometry). If P = (x1, y1) and Q = (x2, y2) are distinct points,then there is a unique line {(x, y) : ux + vy + w = 0} containing both P and Q.Proof. Case 1. x16= x2. Since we must both show that a line exists and show that at most one lineexists, we divide the proof into two claims. We will prove first that at most one line exists. (Can yousee why we do this first?)Claim 1. There is at most one line incident with both P and Q. Proving the claim is equivalentto showing that, up to equivalence, there is at most one ordered triple of real numbe rs (u, v, w), u 6=0 ∨ v 6= 0, such that ux1+ vy1+ w = 0 ∧ ux2+ vy2+ w = 0.Assume that ux1+ v y1+ w = 0 ∧ ux2+ v y2+ w = 0 and that u 6= 0 ∨ v 6= 0. Since subtraction is awell-defined operation, subtracting the second equation from the first we obtain:ux1+ vy1+ w = 0− ux2+ vy2+ w = 0u(x1− x2) + v(y1− y2) = 0Since x16= x2, we have that x1− x26= 0; therefore, x1− x2has a multiplicative inverse and we maysolve for u, obtainingu =−v(y1− y2)x1− x2.1It follows that v 6= 0, since if it were the case that v = 0, we would have u = 0 as well, and it isassumed that u and v are not both 0. Up to equivalence, we may assume v = 1; henceu =−(y1− y2)x1− x2and the first equation becomes−x1(y1− y2)x1− x2+ y1+ w = 0.But now, solving for w yieldsw = −y1+x1(y1− y2)x1− x2=x2y1− x1y1+ x1y1− x1y2x1− x2=x2y1− x1y2x1− x2.Thus, any ordered triple of real numbers (u, v, w) with u 6= 0 ∨ v 6= 0 and satisfying ux1+ vy1+ w =0 ∧ ux2+ vy2+ w = 0 is equivalent to−(y1− y2)x1− x2, 1,x2y1− x1y2x1− x2.Claim 2. There is a line through both P = (x1, y1) = (x, y1) and Q = (x2, y2) = (x, y2). Considerthe line{(x, y) :−(y1− y2)x1− x2x + y +x2y1− x1y2x1− x2= 0}.By plugging in their coordinates, one checks that points P and Q both lie on this line. (We knew thishad to be the case, since the converses of all the implications used to prove Claim 1 are true: all thesteps used to solve for u, v, and w are reversible, showing that the values obtained do indeed make theequations true.)Case 2. y16= y2. Similar. (Exercise.)Remark. As an alternative proof, we could take as our second case that x1= x2. Of course, thisassumption implies y16= y2, but using the fact that x1= x2as well yields a nice and instructive proofwith a simpler calculation:Claim 1. There is at most one line incident with both P and Q. Assume that ux1+ vy1+ w =0 ∧ ux2+ vy2+ w = 0 and that u 6= 0 ∨ v 6= 0. Since subtraction is a well-defined operation, subtractingthe second equation from the first we obtainux1+ vy1+ w = 0− ux2+ vy2+ w = 0v(y1− y2) = 0It clearly follows that v = 0. Since it is not the case that both u and v are 0, u 6= 0. Up toequivalence, we may assume u = 1. But now, solving for w yields w = −x1= −x2. Thus, any orderedtriple of real numbers (u, v, w) with u 6= 0 ∨ v 6= 0 and satisfying ux1+ vy1+ w = 0 ∧ ux2+ vy2+ w = 0is equivalent to (1, 0, −x1).2Claim 2. There is a line through both P = (x1, y1) = (x, y1) and Q = (x2, y2) = (x, y2). Considerthe line {(x, y) : x − x1= 0}. Points P and Q both clearly lie on this line.Note that the formula derive from our original Case 2 (left as an exercise) reduces to this one ifx1= x2. Note also that if y1= y2in Case 1, a similar simple formula results. (These are, of course, thecases that the line is vertical or horizontal, respectively.)Remark. It was natural to prove the uniqueness of the line before the existence of the line because theproof of uniqueness revealed what the standard equation of the line had to be.Proposition 2 (Axiom I-2 of incidence geometry). Every line is incident with at least two points.Proof. Let l be a line, that is, the solution set to an equation ux + vy + w = 0 with u, v, w ∈ R andu 6= 0 or v 6= 0.Case 1. u 6= 0. Without loss of generality, we may assume u = 1. We easily check that (−w, 0) and(v − w, −1) are solutions. (Note that there are, in fact, infinitely many solutions since, for any value ofy ∈ R, we can solve for x ∈ R.)Case 2. v 6= 0. Similar. (Exercise.)Proposition 3 (Axiom I-3 of incidence geometry). There exist three noncollinear points.Proof. Translated into the language of coordinate geometry, this means there are three ordered pairsthat do not all satisfy the same linear equation. Consider (0, 0), (0, 1), and (1, 0). I leave it as an exerciseto show that these are not collinear. (Since we have proven Axiom I-1 for our interpretation, you canuse the fact that the line through two distinct points is unique. Notice that the standard equation ofthe line through (0, 0) and (0, 1) is just x = 0. It should be pretty obvious that (1, 0) does not satisfythis equation!)Now let us turn to the proof of the Euclidean Parallel Property.Proposition 4 (EPP). Given a line l and a point P not on l, there …


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EIU MAT 3271 - Proof that the Real Affine Plane is a Model of Affine Geometry

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