Recall Corollary 4 from this post:
Corollary 4. If is a subgroup of , then the following are equivalent:
(i) is well-ordered;
(ii) is not dense;
(iii) is cyclic.
Let be a non-zero real number and take to be the subgroup of generated by and . This is a cyclic group if and only if
for some and
, i.e. is rational.
Now by the above corollary, is dense iff it is not well-ordered, i.e. iff
So we have the following criterion for irrationality:
Criterion 1. is irrational if and only if holds.
Note that is dense in iff is dense in . So we deduce:
Criterion 2. is irrational .
We demonstrate how this may be useful by proving that certain types of numbers are irrational.
Proposition 1. Let be a sequence of non-zero integers such that
Then is irrational.
Proof. We have
Multiplying by if necessary, we can take the expression in brackets on the right to be positive and it tends to as . Moreover, if
which cannot happen infinitely often as the left hand side tends to zero. So the conclusion follows by Criterion 2.
Proposition 2. If is a sequence of non-zero integers such that and , then
exists and is irrational.
Proof. It suffices to show that and above hold.
Convergence follows easily from the ratio test, so holds. Now
Some special cases of Proposition 2 are particularly interesting:
Corollary 1. is irrational for all positive integers .
Proof. Take . .
Corollary 2. is irrational.
Proof. Take in Corollary 1.
Corollary 3. and are irrational.
Proof. Take and for , for sine, for cosine.
Corollary 4. and are irrational, where is the modified Bessel function of the first kind.
Proof. Taking in Corollary 1 shows that is irrational. Taking shows that is irrational.
Corollary 5. is irrational.
Proof. If it is rational, then so is , and so is
Taking in the above shows that this is false.
Corollary 6. Let be an integer and the Fibonacci sequence. Then
(i) is irrational.
(ii) is irrational.
Proof. (i) Take and use .
(ii) Take .