Characterizations of C(X) among its subalgebras by R. B. Burckel

By R. B. Burckel

Publication by way of Burckel, R. B.

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100 , .. 100 . 200   x1 So, x1 − 34 x3 = −100, and x2 + 73 x3 = 200. Now, we can write our solution vectors in terms of x3 :  x2  = x3  4  x − 100 3 3  − 7 x3 + 200 . Since all of our values must be non-negative, x1 must be greater than or equal to zero, or 3 x3 4 x − 100 ≥ 0, which means that x3 ≥ 75. 3 3 Also, x3 must be greater than or equal to zero, meaning that − 73 x3 + 200 ≥ 0 or x3 ≤ 600 7 . Since x3 must be an integer, this forces x3 ≤ 85. Thus, we are looking for solutions where 75 ≤ x3 ≤ 85.

B.. b2 ..   1   . c   c2 1 c−a (c − a)(c − b)     .  This system is consistent if and only if c = a or c = b. Thus the vector is a linear combination if c = a or c = b. 63 This is the line parallel to w which goes through the end point of the vector v. 64 This is the line segment connecting the head of the vector v to the head of the vector v + w. 65 This is the full parallelogram spanned by the two vectors v and w. 66 Write b = 1 − a and av + bw = av + (1 − a)w = w + a(v − w) to see that this is the line segment connecting the head of the vector v to the head of the vector w.

3  . We find that our solutions are bound by 0 ≤ x3 ≤ 100. However, Thus, our solutions are of the form  1500−15x 19 x3 3 3 3 3 3 since both 400−4x = 4 100−x and 1500−15x = 15 100−x must be non-negative integers, the quantity 100−x must 19 19 19 19 19 be a non-negative integer, k, so that x3 = 100 − 19k. The condition x3 ≥ 0 now leaves us with the possibilities k = 0, 1, 2, 3, 4, 5. 31 400 19 1500 19  . Chapter 1              20 16 12 8 4 0 ducks Thus, we find our solutions for  sparrows  :  0  .

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