1873_solutions

It cannot be written as the union of two nonempty

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Unformatted text preview: that 2a 2  3b 2  2bc 3  c 2 . Therefore, unless bc  0 we have 2 3  2a 3b 2 c 2 2bc which contradicts the fact that the number 3 is irrational. Therefore at least one of the number b and c must be zero. In the event that c  0, the equation a 2  b 3 c becomes a 2 b 3 and, unless a  0, the latter equation gives us 2 b a 3 which contradicts the fact that 2 3 is irrational. So in the case c  0 we also have a  0 and we see at once that b  0 as well. In the event that b  0, the equation a 2  b 3 c becomes a 2 c and, unless a  0, the latter equation gives us c 2a which contradicts the irrationality of 2 . So, once again, a  0 and we see at once that c  0 as well. c. Prove that if m, n, p and q are integers then it is impossible to have 3p 2m  n q and deduce that if  is any real number and if H  n n Z then the subgroup H  Z cannot contain both of the numbers 2 and 3 . Solution: The equation 2 m n  3 p q implies that q 2 n 3 np  mq which, by part b, tells us that 0  q  n  mq np which is clearly impossible since n and q appear denominators of the fractions in the equation 3p m  n q. Now, to obtain a contradiction, suppose that the subgroup H  Z contains both of the numbers 2 and 3 . Choose integers m and n such that 2 2  m  n and choose integers p and q such that 127 3  p  q. Since 2 is irrational, we know that 2 m and so n m 2 n  0; and we know similarly that q 3p q 0. Thus which we know to be impossible. d. Suppose that G is a subgroup of R other than 0 , that p  inf x G x  0 and that the number p is positive. Prove that the set G is closed. Solution: We know from an earlier exercise that G  np n Z. e. Prove that if G is a subgroup of R other than 0 and that G has no least positive member then G  R. Solution: This fact was established in an earlier exercise. f. Suppose that  is an irrational number, that H  n n Z and that G  H  Z. Prove that although the sets H and Z are closed subgroups of R and although the set G is also a subgroup of R, the set G is not closed. Solution: Since G cannot contain both of the numbers 2 and 3 we know that G R. To show that G is not closed we shall make the observation that G  R and, for this purpose, all we have to show is that if p  inf x G x  0 then p  0. Suppose that p is defined in this way and, to obtain a contradiction, suppose that p  0. We know that G  np n Z and, using the fact that both of the numbers 1 and  belong to G, we choose integers m and n such that 1  mp and   np. From the fact that p  1/m we see that p is rational but from the fact that p  /n we see that p must be irrational. Thus we have arrived at the promised contradiction. Exercises on Limit Points 1. Prove that in the metric space R we have L Z  . Given any number x, the interval x 1, x  1 can contain at most two integers. We know that a neighborhood of a limit point of a set must always contain infinitely many members of that set and so we conclude that no number x can be a limit point of...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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