I think I broke math

So I've been pondering this question for a while now and I cant see anything wrong. Please tell me where I go wrong in the following equations:

If...

$\sqrt{ab^2}=b\sqrt{a}$

Then

$\sqrt{5}=\sqrt{5(-1)^2}=(-1)\sqrt5$

Therefore

$\sqrt5=-\sqrt5$

How is this possible. (There has to be some simple detail I missed)

#Algebra #Numbers #Proofs #Easymoney #How

Easy Math Editor

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Comments

Remember that $\sqrt{x^{2}} \not = x \Rightarrow \sqrt{x^{2}} = |x|$

Jordi Bosch - 6 years, 9 months ago

The contradiction you get is precisely why mathematicians decided to say $\sqrt{x^{2}} = |x|$

Jordi Bosch - 6 years, 9 months ago

I'll work with this case if necessary $\sqrt{ab^{2}} = b\sqrt{a}$ if b > 0 and $\sqrt{ab^{2}} = -b\sqrt{a}$ if b < 0. Since in this case $b = -1 \rightarrow \sqrt{5}= \sqrt{5*(-1)^{2}} = -(-1)\sqrt{5} = \sqrt{5}$

Jordi Bosch - 6 years, 9 months ago

@Jordi Bosch – .Ahh, that makes sense. Thanks for the explanation

Trevor Arashiro - 6 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$