What is wrong with this proof? -3 = 3

Question

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.

@yanko I know it's wrong. I asked this question to understand why. — Daniele, Oct 21 '17 at 13:48
It's obviously wrong because you reach a contradiction.. what made you think that would be true in the first place? — Yanko, Oct 21 '17 at 14:58
You can also check some other related posts, such as: Which step in this process allows me to erroneously conclude that $i = 1$, Which step in this process allows me to erroneously conclude that $i = 1$, Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$? and When does $\sqrt{a b} = \sqrt{a} \sqrt{b}$? — Martin Sleziak, Sep 11 '20 at 20:40

Sil · Accepted Answer · 2017-10-21T13:54:44.997

4

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?.

edited Oct 21 '17 at 13:54

answered Oct 21 '17 at 13:49

Sil

16,612

Michael Rozenberg · Answer 2 · 2017-10-21T13:47:47.123

3

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 $a>0$.

edited Oct 21 '17 at 13:47

answered Oct 21 '17 at 13:37

Michael Rozenberg

194,933

score -1 · Answer 3 · answered Oct 21 '17 at 13:33

-1

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).

answered Oct 21 '17 at 13:33

Yanko

13,758

What is wrong with this proof? -3 = 3

3 Answers3