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diagonal elements are all different. However, L1 will not satisfy the conditions of Theorem 12.12, whatever substitutions we make. In L2, replace 0, 1, α, α2 by 0, 1, 2, 3, respectively, and in L3 replace 0, 1, α, α2 by 3, 2, 0, 1, respectively, to obtain the squares L2 and L3 in Table 12.14. Combine these to obtain the square M with entries in base 4. Add 1 to each entry and convert to base 10 to obtain the magic square M in Table 12.14.
EXERCISES
12.1.Construct a latin square of order 7 on {a, b, c, d, e, f, g}.
12.2.Construct four mutually orthogonal latin squares of order 5.
12.3.Construct four mutually orthogonal latin squares of order 8.
12.4.Construct two mutually orthogonal latin squares of order 9.
12.5.Prove that there are at most (n − 1) mutually orthogonal latin squares of order n. (You can always relabel each square so that the first rows are the same.)
12.6.Let L = (lij ) be a latin square of order l on {1, 2, . . . , l} and M = (mij ) be a latin square of order m on {1, 2, . . . , m}. Describe how to construct a latin square of order lm on {1, 2, . . . , l} × {1, 2, . . . , m} from L and M.
12.7.Is the latin square of Table 12.15 the multiplication table for a group of order 6 with identity A?
TABLE 12.15
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12.8.A chemical company wants to test a chemical reaction using seven different levels of catalyst, a, b, c, d, e, f , g. In the manufacturing process, the raw material comes from the previous stage in batches, and the catalyst must be added immediately. If there are seven reactors, A, B, C, D, E, F , G, in which the catalytic reaction can take place, show how to design the experiment using seven batches of raw material so as to minimize the effect of the different batches and of the different reactors.
12.9.A supermarket wishes to test the effect of putting cereal on four shelves at different heights. Show how to design such an experiment lasting four weeks and using four brands of cereal.
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12 LATIN SQUARES |
12.10.A manufacturer has five types of toothpaste. He would like to test these on five subjects by giving each subject a different type each week for five weeks. Each type of toothpaste is identified by a different color—red, blue, green, white, or purple—and the manufacturer changes the color code each week to reduce the psychological effect of the color. Show how to design this experiment.
12.11.Quality control would like to find the best type of music to play to its assembly line workers in order to reduce the number of faulty products. As an experiment, a different type of music is played on four days in a week, and on the fifth day no music at all is played. Design such an experiment to last five weeks that will reduce the effect of the different days of the week.
12.12.The relation of parallelism, //, on the set of lines of an affine plane is defined by l//m if and only if l = m or l ∩ m = Ø. Prove that // is an equivalence relation.
12.13.Let P be the set of points and L be the set of lines of a finite affine plane.
(a)Show that the number of points on a line l equals the number of lines in any parallelism class not containing l.
(b)Deduce that all the lines contain the same number of points.
(c)If each line contains n points, show that the plane contains n2 points and n2 + n lines, each point lying on n + 1 lines. Show also that there are n + 1 parallelism classes.
12.14.Find all the lines in the affine plane of order 3 whose point set is Z23.
Exercises 12.15 to 12.17 refer to the affine plane of order 9 obtained from GF(9 ) = Z3 (α), where α2 + 1 = 0 .
12.15.Find the line through (2α, 1) that is parallel to the line y = αx + 2 + α.
12.16.Find the point of intersection of the lines y = x + α and y = (α + 1)x + 2α.
12.17.Find the equation of the line through (0, 2α) and (2, α + 1).
12.18.Prove that a magic square of order 3 must have 5 at its center.
12.19.Prove that 1 cannot be a corner element of a magic square of order 3.
12.20.How many different magic squares of order 3 are there?
12.21.How many essentially different magic squares of order 3 are there, that is, magic squares that cannot be obtained from each other by a symmetry of the square?
12.22.Is there a magic square of order 2?
12.23.Find two magic squares of order 4 different from the square in Example 12.18.
12.24.Find a magic square of order 5.
12.25.Find a magic square of order 8.
12.26.Can you construct a magic square with the present year in the last two squares of the bottom row?
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Figure 13.2. |
Constructing products and quotients. |
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If a, b K, we can mark off lengths a and b on a line to construct lengths a + b and a − b.
If a, b, c K, mark off a segment OA of length a on one line and mark off segments OB and OC of length b and c on another line through O as shown in Figure 13.2. Draw a line through B parallel to CA and let it meet OA in X. Triangles OAC and OXB are similar, and if OX = x, then x/a = b/c and so x = ab/c.
By taking c = 1, we can construct ab, and by taking b = 1, we can construct
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Corollary 13.2. K is an extension field of Q. |
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Proof. Since 1 K and sums and differences of constructible numbers are constructible, it follows that Z K. Since quotients of constructible numbers
are constructible, Q K. |
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Proposition 13.3. If k K and k > 0, then |
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k K. |
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Proof. Mark off segments AB and BC of lengths k and 1 on a line. Draw the circle with diameter AC and construct the perpendicular to AC at B as shown
in Figure 13.3. Let it meet the circle at D and E. Then, by a standard theorem |
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in geometry, AB · BC = DB · BE: thus BD = BE = |
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Example 13.4. √2 is constructible. |
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Solution. We apply the construction of Proposition 13.3 twice to construct 2
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and then √2 = √2. |
We can construct any number that can be written in terms of rational numbers, |
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+, −, ·, ÷, and √ |
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signs. For example, the numbers 1 + 4 5, |
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4/5, and |
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√
3 − 7 are all constructible. If k1 is a positive rational number, all the elements
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GEOMETRICAL CONSTRUCTIONS |
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Figure 13.3. |
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Constructing square roots. |
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of the extension field K1 |
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( k1) are constructible. K1 has degree 1 or 2 over |
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Q depending on whether |
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of K1, all the elements of K2 |
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= K1( k2) = |
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[K2 : K1] = 1 or 2, depending on whether or not |
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now show that every constructible number lies in a field obtained by repeating the extensions above.
Theorem 13.5. The number α is constructible if and only if there exists a sequence of real fields K0, K1, . . . , Kn such that α Kn Kn−1 · · · K0 = Q and [Ki : Ki−1] = 2 for 1 i n.
Proof. Suppose that α Kn Kn−√1 · · · K0 = Q, where [Ki : Ki−1] = 2. By Proposition 11.16, Ki = Ki−1( γi−1) for γi−1 Ki−1, and since Ki is real, γi−1 > 0. Therefore, by repeated application of Proposition 13.3, it can be shown that every element of Kn is constructible.
Conversely, suppose that α K; thus α appears as the coordinate of a point constructible from (0, 0) and (1, 0) by a finite number of the operations 1 to 5 preceding Theorem 13.1. We prove the result by induction on m, the number of constructible numbers used in reaching α.
Suppose that Xk = {x1, . . . , xk } is a set of numbers that have already been constructed, that is, have appeared as coordinates of constructible points. When the next number xk+1 is constructed, we show that [Q(Xk+1) : Q(Xk )] = 1 or 2, where Q(Xk+1) = Q(x1, . . . , xk , xk+1).
We first show that if we perform either operation 1 or 2 using previously constructed numbers in Xk , the coefficients in the equation of the line or circle remain in Q(Xk ). If we perform operation 3, the newly constructed numbers remain in Q(Xk ), and if we perform operation 4 or 5, the newly constructed numbers are either in Q(Xk ) or an extension field of degree 2 over Q(Xk ).
Operation 1. The line through (α1, β1) and (α2, β2) is (y − β1)/(β2 − β1) = (x − α1)/(α2 − α1), and if α1, α2, β1, β2 Xk , the coefficients in the equation of this line lie in Q(Xk ).
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Operation 2. The circle with center (α1, β1) and radius equal to the distance from
(α2, β2) to (α3, β3) is (x − α1)2 + (y − β1)2 = (α2 − α3)2 + (β2 − β3)2, and all the coefficients in this equation lie in Q(Xk ).
Operation 3. Let αij , βj Q(Xk ). Then the lines
α11x + α12y = β1
α21x + α22y = β2
meet in the point (x, y), where using Cramer’s rule,
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as long as they are not parallel. Both of these coordinates are in Q(Xk ).
Operation 4. To obtain the points of intersection of a circle and line with coefficients in Q(Xk ), we eliminate y from the equations to obtain an equation of the form
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αx2 + βx + γ = 0 |
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Q(Xk )( |
β2 − 4αγ |
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degree 2 over Q(X ) |
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Operation 5. The intersection of the two circles
x2 + y2 + α1x + β1y + γ1 = 0
x2 + y2 + α2x + β2y + γ2 = 0
is the same as the intersection of one of them with the line
(α1 − α2)x + (β1 − β2)y + (γ1 − γ2) = 0.
This is now the same situation as in operation (4).
Initially, m = 2, X2 = {0, 1}, and Q(X2) = Q. It follows by induction on m, the number of constructible points used, that
α Q(Xm) Q(Xm−1) · · · Q(X3) Q(X2) = Q,