I'm sorry for my english math terms, I'm trying to translate them from Italian, hopefully everything is correct.

On my textbook I have this exercise (Powers with irrational exponents chapter):


And I'm asked to say for what values of $a$ this power is meaningful so this is what I did:

I know that $a^2 -1$ must be greater than $0$ because the exponent $-\sqrt{2}$ is a negative irrational number





Now I know this is wrong but I really have no idea how to get to the correct result which is:

$a<-1\lor a>1$

I've searched on the book, online but never received any satfisying answer.

  • $\begingroup$ If you draw a graph of $y=x^2 - 1$ it should be "obvious" that $a^2 - 1 > 0$ when $a < -1$ or $a > 1$. To prove mathematically that this is the correct answer is just a little bit more work. $\endgroup$ – David K Nov 11 '16 at 19:44

You have to be very careful when taking square roots because $\sqrt{a^2}$ really means $|a|$. Your confusion arises when you claim that $a > \pm \sqrt{1}$ because, while this is true for equalities, it isn't necessarily true for inequalities.

In this case you have two cases; either $a$ is positive or $a$ is negative. If $a$ is positive then we can conclude that $a > 1$. However, if $a$ is negative then we have to be careful about our signs: $-a > 1 \implies a < -1$.

A safer way would be to factor the original expression as a difference of squares:

\begin{align} a^2 - 1 &> 0 \\ (a-1)(a+1) &> 0, \end{align}

and then you have $3$ regions to check.


After $a^2 > 1$ you can conclude with $a<-1 \lor a>1$. [Think about which numbers $a$, after squaring, become $>1$.] Be sure you understand why going to $a>\pm \sqrt{1}$ is wrong.

  • $\begingroup$ I don't understand why it's wrong actually. EDIT: Also the correct result makes sense but is there a procedure to follow to get there? $\endgroup$ – Elia Perantoni Nov 11 '16 at 19:39

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