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solution1 - Verify that the set of complex numbers of the...

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Verify that the set of complex numbers of the form x + y 2 , where x and y are rational, is a subfield of the field of complex numbers. Evidently, this set contains 0 and 1 . It is also easy to check that it is closed under the addition and multiplication. Then most of the axioms of the field follows directly from the corresponding properties of complex numbers. The nontrivial part of this exercise is to prove that for any element z = x + y 2 , where x and y are rational, its inverse also has this form. Consider an element z 0 = x - y 2 . Then the product zz 0 = x 2 - 2 y 2 is rational. Hence, the number z 0 x 2 - 2 y 2 has the desired form. On the other hand, its product with z is 1 . So the inverse is given by the formula x x 2 - 2 y 2 - y x 2 - 2 y 2 2 . Are the following two systems of linear equations equivalent? If so, express each equation in each system as a linear combination of the equa- tions in the other system. - x 1 + x 2 + 4 x 3 = 0 x 1 + 3 x 2 + 8 x 3 = 0 1 2 x 1 + x 2 + 5 2 x 3 = 0 x 1 - x 3 = 0 x 2 +3 x 3 = 0 For instance, let us express the first equation of the first system as the linear combination of the equations of the second system. Use the method of undetermined coefficients: - x 1 + x 2 + 4 x 3 = a ( x 1 - x 3 ) + b ( x 2 + x 3 ) , where a, b are the numbers that we want to find. Comparing coefficients with x 1 in both sides, one gets - 1 = a, and comparing coefficients with x 2 , one gets 1 = b. Then it is easy to check that the coefficients with x 3 in both sides coincide. Hence, the first equation of the first system is the linear combination of the equations of the second system with coefficients - 1 and 1 . To complete the solution one just need to apply the same method and express the other equations of the first system via the equations of the second one, and vice versa. Let F be a set which contains exactly two elements, 0 and 1 . Define an addition and multiplication by the tables: + 0 1 0 0 1 1 1 0 · 0 1 0 0 0 1 0 1 1

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Verify that the set F , together with these two operations, is a field. The solution was explained in the first recitation. Prove that each subfield of the complex numbers contains ev- ery rational number. Let F be a subfield of C . Since F is a field it must contain a distinguished element ˜ 0 which plays the role of the additive identity. But this element is unique (regarded as a complex number) therefore ˜ 0 = 0 , where 0 is the usual 0 of the complex numbers. Therefore 0 F . Analogously we may con- clude that 1 F . But if 1 F then we must have that 1 + 1 = 2 F (notice that 1 + 1 = 2
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