If $q^n$ is irrational for all $n>1$, then $q$ is irrational.

Question

Theorem. Let $q \in \mathbb{R}$ an arbitrary given number. If $q^n$ is irrational for all $n>1$ integer, then $q$ is irrational.

My Questions. What is a the name of this statement and what is the shortest proof for it? Thanks.

Answer 1

The rationals $\,\Bbb Q\,$ are closed under multiplication, therefore, by induction

$\qquad\qquad\qquad\qquad\qquad\ \ \forall i\!:\ q_i \in \Bbb Q\ \Rightarrow\ q_1\cdots q_n\in \Bbb Q$

${\rm Contrapositively}\qquad\ \ \ q_1\cdots q_n \not\in \Bbb Q\ \Rightarrow\ q_i\not\in\Bbb Q,\ \text{for some}\ i$

$\text{The constant case}\,\ q_i = q\ \ {\rm is}\ \ q^n \not\in \Bbb Q\ \Rightarrow\ q\not\in \Bbb Q,\ $ which yields your claim.

So the result is a contrapositive of the closure of $\,\Bbb Q\,$ under multiplication. Generally, there are no widely-used names for the contrapositive forms of these ubiquitous closure properties. See here for many other such "complementary" views of closure properties of algebraic operations.

Answer 2

Your claim: $$\not\exists a\in \mathbb{N}, b\in \mathbb{N} : bq^n = a ~\forall n > 1$$

(or alternatively)

$$\forall n > 1\not\exists a\in \mathbb{N}, b\in \mathbb{N} : bq^n = a$$

A proof:

Using your statement as hypothesis, let's consider $q^2$. You say that $q^2$ is irrational. By contradiction, if $q$ is integer or rational, then:

$$q = \frac{a_q}{b_q},$$

with $b_q \neq 0$.

In this case $q^2 = \frac{a_q^2}{b_q^2}$ must be integer or rational. In fact, both numerator and denominator are integer. This contradicts your hypothesis and then $q$ is irrational too.

Answer 3

Isn't this trivial? Proof by contradiction.

Suppose $q$ is rational. Then $q^n$ is also rational for all $n>1$. Since the assumption is that they are irrational, the assumption is false and $q$ must be irrational.

Don't know if there is a name for this.

Answer 4

You can prove this via contradiction pretty quickly. FTSOC suppose that $q$ is not irrational. Then $q$ is rational. Any finite product of rational numbers is rational, so $q^{n}$ must be rational. This contradicts the assumption that $q^{n}$ is irrational for all $n>1$, so $q$ has to be irrational.