0
$\begingroup$

What is wrong with this proof?

$-3 = \sqrt[3]{-27} = {(-27)}^{\frac 13} = {(-27)}^{\frac 26} = \sqrt[6]{{(-27)}^2} = \sqrt[6]{{27}^2} = {(27)}^{\frac 26} = {(27)}^{\frac 13} = \sqrt[3]{27} = 3$

This is obviously false since $-3 \neq 3$.

But still I can't figure out which equation is the wrong one and why that is.

Thanks in advance for anyone who will help.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ The third equality is completely wrong... $\endgroup$ – Yanko Oct 21 '17 at 13:32
  • 1
    $\begingroup$ $(-27)^{1/3} = (-27)^{2/6} \neq ((-27)^2)^{1/6}$ $\endgroup$ – samjoe Oct 21 '17 at 13:36
  • $\begingroup$ @yanko I know it's wrong. I asked this question to understand why. $\endgroup$ – Daniele Oct 21 '17 at 13:48
  • 1
    $\begingroup$ It's obviously wrong because you reach a contradiction.. what made you think that would be true in the first place? $\endgroup$ – Yanko Oct 21 '17 at 14:58
5
$\begingroup$

As suggested in comments, the $(-27)^{2/6} \neq ((-27)^2)^{1/6}$, because $a^{bc}=(a^b)^c$ does not generally hold for $a<0$.

You can find more related info in @mrf's answer here: Is $(-1)^{ab} = (-1)^{ba}$ true? => $(-1)^{ab} = ((-1)^a )^b$ is true?.

|cite|improve this answer
$\endgroup$
3
$\begingroup$

The mistake here:

$$(-27)^{\frac{2}{6}}=(-27)^{\frac{1}{3}}.$$ It's wrong because $$(-27)^{\frac{2}{6}}=|-27|^{\frac{1}{3}}.$$ Also if we write $a^{\frac{1}{3}}$ then it's better to think that $a>0$.

|cite|improve this answer
$\endgroup$
-1
$\begingroup$

you could aswell prove this:

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

Clearly the first equality is wrong (and your's third equality is wrong for similar reason).

|cite|improve this answer
$\endgroup$

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.