0
$\begingroup$

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):

$(a^2-1)^{-\sqrt{2}}$

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

$a^2-1>0$

$a^2>1$

$a>\pm\sqrt1$

$a>\pm1$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.