Homework Assignment 1

pg.22 prob.3

Show that the system with points (0,0), (1,0), (0,1), and (1,1) and lines given by the equations x=0, x=1, y=0, y=1 and x=y is not an affine plane.

The points (1,0) and (0,1) are not contained in any of the five lines, so axiom 1 is violated.

pg.22 prob.7

Prove that if a line in an affine plane is parallel to one of two intersecting lines, it intersects the other.

Let l and m be two lines of an affine plane which intersect at point P. Let k be parallel to one of these lines, say l. BWOC suppose that k is parallel to m. Since k is parallel to m, k does not contain P. Now, both of l and m pass through P and are parallel to k. This contradicts axiom 2, so k and m can not be parallel.

pg.28 prob.3

Let l~m stand for the statement "l is parallel or equal to m". Prove that ~ is an equivalence relation on the set L of lines of an affine plane.

  1. l~l for every line l (reflexitivity) since "l is equal to l" is always true.
  2. If l~m, then m~l (symmetry). If l = m, the statement is obvious. If l is parallel to m, the statement is also obvious.
  3. If l~m and m~k, then l~k (transitivity). If any two of these lines are equal the statement is obvious, so we may assume that the three lines are distinct. Suppose then that l is parallel to m and m is parallel to k. If l and k intersect, then m would be parallel to l and therefore (by prob.7 pg. 22 above) would have to intersect k. This contradiction shows that l must be parallel to k.

pg.28 prob.7

Prove that if a line in an affine plane intersects one of two parallel lines, it intersects the other.

Let l and m be parallel lines. Let k be a line which intersects l at a point P. If k is parallel to m, then through P there would pass two lines (l and k) both parallel to m, contradicting axiom 2. Thus, k must intersect m.

pg.36 prob.4

Suppose that A and B are orthogonal Latin squares of order n and that J and K are constant n × n matrices with entries j and k, respectively. Prove that A + J and B + K are orthogonal Latin squares (again the addition is done mod n).

Assume that A+J and B + K are not orthogonal. That means that there are two positions (i,m) and (s,t) where (A+J)im = (A+J)st and (B+K)im = (B+K)st. This means, aim + j = ast + j and bim + k = bst + k. Hence, aim = ast and bim = bst, which would contradict the fact that A and B are orthogonal.