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Theorem. A rectangle R with side lengths 1 and x, where x is irra- tional, cannot be “tiled" by finitely many squares (so that the squares have disjoint interiors and cover all of R).

Theorem. A rectangle R with side lengths 1 and x, where x is irra- tional, cannot be “tiled" by finitely many squares (so that the squares have disjoint interiors and cover all of R).

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter2: Parallel Lines
Section2.5: Convex Polygons
Problem 41E
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Please explain this proof step by step I really don’t understand what the textbook says.. It’d be cool if you could draw a graph or something
We claim that if the 1 x x reclangle R is tiled by the squares Q1, Q2,..., Qn, then v(R) = D v(Qi). This leads to a contra- diction, since v(R) = f(1)f(x) = -1, while v(Q;) = f(s;)² > 0 for all i. To check the claim just made, we extend the edges of all squares Qi of the hypothetical tiling across the whole of R, as is indicated in the picture: This partitions R into small rectangles, and using the linearity of f, it is easy to see that v(R) equals to the sum of v(B) over all these small rectangles B. Similarly v(Qi) equals the sum of v(B) over all the small rectangles lying inside Qi. Thus, v(R) = E, v(Qi).
Transcribed Image Text:We claim that if the 1 x x reclangle R is tiled by the squares Q1, Q2,..., Qn, then v(R) = D v(Qi). This leads to a contra- diction, since v(R) = f(1)f(x) = -1, while v(Q;) = f(s;)² > 0 for all i. To check the claim just made, we extend the edges of all squares Qi of the hypothetical tiling across the whole of R, as is indicated in the picture: This partitions R into small rectangles, and using the linearity of f, it is easy to see that v(R) equals to the sum of v(B) over all these small rectangles B. Similarly v(Qi) equals the sum of v(B) over all the small rectangles lying inside Qi. Thus, v(R) = E, v(Qi).
Theorem. A rectangle R with side lengths 1 and x, where x is irra- tional, cannot be “tiled" by finitely many squares (so that the squares have disjoint interiors and cover all of R). Proof. For contradiction, let us assume that a tiling exists, consisting of squares Q1, Q2, ...,Qn, and let s; be the side length of Qi. We need to consider the set R of all real numbers as a vector space over the field Q of rationals. This is a rather strange, infinite- dimensional vector space, but a very useful one. Let V CR be the linear subspace generated by the numbers r and s1, 82, ..., sn, in other words, the set of all rational linear combi- nations of these numbers. We define a linear mapping f: V → R such that f(1) = 1 and f(x) = -1 (and otherwise arbitrarily). This is possible, because 1 and æ are linearly independent over Q. Indeed, there is a basis (b1, b2,..., b) of V with b, = 1 and b2 f(b1) linearly on V. %3D = x, and we can set, e.g., f(bk) = 0, and extend f %3D 1, f(b2) = -1, f(b3) = %3D ... For each rectangle A with edges a and b, where a, b e V, we define a number v(A) := f(a)f(b).
Transcribed Image Text:Theorem. A rectangle R with side lengths 1 and x, where x is irra- tional, cannot be “tiled" by finitely many squares (so that the squares have disjoint interiors and cover all of R). Proof. For contradiction, let us assume that a tiling exists, consisting of squares Q1, Q2, ...,Qn, and let s; be the side length of Qi. We need to consider the set R of all real numbers as a vector space over the field Q of rationals. This is a rather strange, infinite- dimensional vector space, but a very useful one. Let V CR be the linear subspace generated by the numbers r and s1, 82, ..., sn, in other words, the set of all rational linear combi- nations of these numbers. We define a linear mapping f: V → R such that f(1) = 1 and f(x) = -1 (and otherwise arbitrarily). This is possible, because 1 and æ are linearly independent over Q. Indeed, there is a basis (b1, b2,..., b) of V with b, = 1 and b2 f(b1) linearly on V. %3D = x, and we can set, e.g., f(bk) = 0, and extend f %3D 1, f(b2) = -1, f(b3) = %3D ... For each rectangle A with edges a and b, where a, b e V, we define a number v(A) := f(a)f(b).
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