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# Math Homework Solutions

2.1.1. Show that there is an uncountable number of Cauchy sequences of rational numbers equivalent to
any given Cauchy sequence of rational numbers.

Proof. Let α be any Cauchy sequence of rational numbers, and suppose that

[ α ] { x l x is a Cauchy sequence and x is equivalent to α }

is countable. We would then have an enumeration of [ α ] , and thus
We define β to be the sequence such that for n ∈ N

Clearly β is equivalent to α, since for each n, β(n) differs from α(n) by at most  . However, we also
have for all n since they differ in the nth place. Thus β is not in the enumeration of [ α ], and
we have thus obtained a contradiction, so [ α ] is uncountable.

2.1.2. Show that every real number can be given by a Cauchy sequence of rationals ,  where none
of the rational numbers is an integer.

Proof. Let c be an arbitrary real number. If c is an integer, then the sequence is a
Cauchy sequence of rational numbers that converges to c, none of which are integers.

Suppose that c is not an integer and let k be an integer such that k - 1 < c < k. We construct a
sequence by arbitrarily choosing a rational number from the interval for each .
Clearly none of the are integers, since c < < k for all n, and is equivalent to c, as desired.

2.1.3. What kinds of real numbers are representable by Cauchy sequences of integers.

Only integers are representable as Cauchy sequences of integers. To see this, let be any
Cauchy sequence of integers. Then there exists such that j, k ≥ m implies . If
and were distinct integers, then would be at least 1, and thus we must have .
It follows that there exists such that k ≥ m implies , and therefore x is equivalent
to the constant sequence

2.1.4. Suppose and are two sequences of rational numbers. Define the shuffled sequence
to be Prove that the shuffled sequence is a Cauchy sequence if and only if
and are equivalent Cauchy sequences.

We will make use of the fact that a sequence is Cauchy if and only if it is equivalent to some Cauchy
sequence to prove both implications.

Claim: If and are equivalent Cauchy sequences, then the shuffle sequence
is a Cauchy sequence.

Proof. For any , there exists such that implies .
Similarly, since x is equivalent to y, there exists such that implies .

Let , and choose k ≥ m. If k is even, then k = 2k' for some integer
k'. It follows that . It then follows that

by the triangle inequality and since k and k' are greater than both and .

Otherwise, if k is odd, then k = 2k' − 1 for some integer k'. Similarly, , and thus

since both k and k' are greater than .

Thus, in either case , so x and z are equivalent, and z is Cauchy.

Claim: If is a Cauchy sequence, then and are
equivalent Cauchy sequences.

Proof. Since z is Cauchy, for every   there exists m such that j, k ≥ m implies
. In particular, if we let j = 2k − 1, then , and we have

for all k ≥ m. Thus, z and x are equivalent, so x is Cauchy. A similar proof shows that z and
y are equivalent, so y is Cauchy. Finally, since x is equivalent to z and z is equivalent to y,
x is equivalent to y, by the transitivity of equivalence. Thus if z is Cauchy then x and y are
both Cauchy and equivalent to each other.

2.1.8. Can a Cauchy sequence of positive rational numbers be equivalent to a Cauchy sequence of negative
rational numbers?

Consider the sequence for and for . Clearly, for any , we have
, and thus these two sequences are equivalent.

2.1.9. Show that if is a Cauchy sequence of rational numbers there exists a positive integer N
such that for all j.

Proof. Let x be any Cauchy sequence, and suppose that for every integer N there exists a j such
that . Since x is Cauchy, there exists m such that for all k ≥ m,. Let N' be
any integer greater than , and obtain j such that . If j m, then ,
which is impossible since so . Thus, j < m. Clearly, there
c an only be finitely many positive j that satisfy this inequality, and thus we can take the maximum of
. The axiom of Archimedes guarantees that there is an integer N that is larger than
this maximum, and thus for all j. This is a contradiction, so there must exist some integer
N such that for all j.