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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
2a
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.
 Fall '08
 STAFF
 Math, Calculus

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