Why is $\sqrt x\ne\frac1{x^2}$?

Question

Which step is illegal?

$\sqrt x=x^{\frac12}$
$\frac12=2^{-1}$
$x^{\frac12}= x^{2^{-1}}$
$a^{b^c}=a^{bc}$
$x^{2^{-1}}=x^{-2}$
$a^{-b}=\frac1{a^b}$
$x^{-2}=\frac1{x^2}$

Am I correct in assuming that it is step 3, because $a^{b^c}$ is not the same as $a^d$ where $d=b^c$ but rather $a^d=a^{(b^c)}$?

Not to be rude, but I can barely follow the steps as written... — The Count, Mar 18 '17 at 00:59
Step 3 looks ambiguous as you point out, but step 4 is the real error. You are correct that you can't associate in the way that step 4 suggests. — theREALyumdub, Mar 18 '17 at 01:03
Just want to point out that $a^{b^c}$ is $a^d$. If you want $(a^b)^c$, you have to write it like that. That's at least the convention I know. — martin.koeberl, Mar 18 '17 at 01:04

score 2 · Answer 1 · answered Mar 18 '17 at 01:07

2

The illegal step is Step 4. Exponentiation is not associative, as is suggested in one of the comments you got, i.e.

$$a^{bc} = ({a^b})^c \neq a^{b^c}$$

By the way, I don't know what you would like to show with steps 6 and 7. Given that your step 4 is wrong anyway, the correctness of latter steps does not really matter...

answered Mar 18 '17 at 01:07

bfrguci

123

Even though step 4 is the error it is step seven that made me realize that there was a mistake – Adam Lee Mar 18 '17 at 07:01

score 1 · Answer 2 · answered Mar 18 '17 at 01:07

1

In step 3, you have $x^{(2^{-1})}$, but when you apply step 4 to step 3 to get step 5, you treat it as $(x^2)^{-1}$, because step 4 really is $(a^b)^c = a^{bc}$

answered Mar 18 '17 at 01:07

Bram28

100,612
6
70
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score 1 · Accepted Answer · answered Mar 18 '17 at 01:27

1

For sake of natation:

$(a^b)^c=a^{bc} $

But $a^{(b^c)}\ne a^{bc} $

So $x^{\frac 12}=x^{(2^{-1})}\ne (x^2)^{-1}=\frac 1 {x^2} $.

Others explained but I hope notation makes it clear.

answered Mar 18 '17 at 01:27

fleablood

124,253

Why is $\sqrt x\ne\frac1{x^2}$?

3 Answers3