About powers of irrational numbers

Square of an illogical number can be a sensible number as an example $\sqrt{2}$ is illogical yet its square is 2 which is sensible.

Yet exists an illogical number square origin of which is a sensible number?

Is it secure to think, as a whole, that $n^{th}$ - origin of illogical will constantly offer irrational numbers?

2019-05-09 11:33:50
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Answers: 2

It's real specifically due to the fact that the rationals $\mathbb Q$ consist of a multiplicative subsemigroup of the reals $\mathbb R$,
i.e. the part of rationals is shut under the reproduction procedure of $\mathbb R$. Your declaration emerges by taking the contrapositive of this declaration - which moves it right into an equal declaration in the enhance set $\mathbb R \backslash \mathbb Q$ of illogical reals.

Hence $\rm\quad\quad\quad r_1,\ldots,r_n \in \mathbb Q \;\Rightarrow\; r_1 \cdots r_n \in \mathbb Q$

Contra+$\rm\quad\; r_1 r_2\cdots r_n \not\in \mathbb Q \;\Rightarrow\; r_1\not\in \mathbb Q \;\:$ or $\rm\;\cdots\;$ or $\rm\;r_n\not\in\mathbb Q$.

Your instance $\rm\;\;\; r^n\not\in \mathbb Q \;\Rightarrow\; r\not\in \mathbb Q \;$ is the unique constant instance $\rm r_i = r$

Obviously the very same holds true if we change $\rm\mathbb Q\subset \mathbb R$ by any kind of subsemigroup chain $\rm G\subset H$

The contrapositive kind is necessary in algebra given that it identifies prime perfects in semigroups, rings, etc

2019-05-11 01:06:23

Obviously, if p is sensible, after that p 2 has to additionally be sensible (unimportant to confirm).

$$ p \in \mathbb Q \Rightarrow p^2 \in \mathbb Q. $$

Take the contraposition, we see that if x is illogical, after that √ x have to additionally be illogical.

$$ p^2 \notin \mathbb Q \Rightarrow p \notin \mathbb Q. $$

By adverse power I think you suggest (1/n) - th power (it is noticeable that $(\sqrt2)^{-2} = \frac12\in\mathbb Q$). It holds true by the declaration above-- simply change 2 by n .

2019-05-10 09:05:59