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hw Copy.pdf

# hw Copy.pdf - Some Hints and Solutions to Homework Problems...

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Some Hints and Solutions to Homework Problems Homework 1. 1. Prove Euclid I.6 : if two angles in a tiangle are equal to each other then the triangle is isosceles. Euclid argues this way. Let a triangle ABC has the angle BAC equal the angle BCA but | AB | > | BC | . Take D on AC such that | AD | = | BC | . Then the triangles ADC and CBA have equal areas by the side-angle-side proposition, which is absurd since ADC is strictly inside CBA . 2. Suppose that the triangles ABC and DEF have | AB | = | DE | , | AC | = | DF | and the angle ABC equal the angle DEF . Does this imply that the triangles are congruent? No. Fix the angle ABC and the length | AB | . Circles centered at A may intersect the ray BC at two different points C and C 0 . Then the triangles ABC and ABC 0 are not congruent since one of them is strictly inside the other. 1.10. T a R = R T - a . Respectively, ( T a R ) ( T b R ) = ( T a R ) ( R T - b ) = T a ( R R ) T - b = T a T - b = T a - b . 1.18. g · g = e is equivalent to g - 1 = g . If this is true for any g G , then for any a, b G we have ab = ( ab ) - 1 = b - 1 a - 1 = ba . 2.14. If a linear transfromation A is an isometry then A t A = I . The formula ( u + v ) 2 = u 2 +2 uv + v 2 applied to the dot-product of vectors shows how to express the dot-product in terms of sums and lengths of vectors: h u, v i = 1 2 ( h u + v, u + v i - h u, u i - h v, v i ) for any u, v . Since A preserves lengths of all vectors and A ( u + v ) = Au + Av (due to lineariry of A ), we conclude that A preserves dot-products: h Au, Av i = h u, v i for any u and v . Therefore for any u, v 0 = h Au, Av i - h u, v i = h A t Au, v i - h Iu, v i = h ( A t A - I ) u, v i . Thus for any u the vector ( A t A - I ) u = 0 since it is orthogonal to any v . This means that the matrix A t A - I = 0. 1

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Homework 2. 1.15. g = ag implies e = gg - 1 = ( ag ) g - 1 = a ( gg - 1 ) = ae = a . 1.16. gh = gk implies g - 1 ( gh ) = g - 1 ( gk ) implies ( g - 1 g ) h = ( g - 1 g ) k implies eh = ek implies h = k . 1.17. e = fg implies g - 1 = eg - 1 = ( fg ) g - 1 = f ( gg - 1 ) = fe = f . 2.20a. For any vector x we have T u ( x ) = x + u . Thus ρ ( T u ( x )) = ρ ( x + u ) = ρ ( x ) + ρ ( u ) = T ρ ( u ) ( ρ ( x )) . 2.26b. Hint: Any isometry on the plane is the composition of a translation with a rotation or reflection. A translation can be written as the composition of two reflections in two parallel lines perpendicular to the direction of the translation. Thus compositions of a translation with a reflection can be represented as the composition of three reflections. The composition of a translation T u with a rotation ρ is in fact the rotation through the same angle about the fixed point defined by the equation ρ ( x ) + u = x . Homework 3. 1. Prove that a group homomorphism f from a group G to a group H maps the identity element of G to the identity element of H and maps inverse elements to inverse elements. Indeed, f ( a ) e H = f ( a ) = f ( ae G ) = f ( a ) f ( e G ) implies e H = f ( e G ). Similarly, e H = f ( e G ) = f ( aa - 1 ) = f ( a ) f ( a - 1 ) implies f ( a ) - 1 = f ( a - 1 ). 3.7. The subgroup C 4 of rotations in the group D 4 of symmetries of the square is a cyclic group of order 4. It is not isomorphic to V 4 since any cyclic subgroup in V 4 has at most two elements.
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