My Math Forum 1=-1 Incorrect Proof

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 April 3rd, 2017, 07:45 AM #1 Member   Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14 1=-1 Incorrect Proof 1=1 1²=1² Since, (-1)²=1=1², we have 1²=(-1)² So, taking square root on both sides gives 1=-1. Obviously, this is wrong, but where is the mistake? I suppose it has to do with the square root of (-1)², but why is that an issue? (-1)² is greater than 0, so afaik, its square root shouldn't cause any trouble. Thanks for the help.
April 3rd, 2017, 07:59 AM   #2
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Quote:
 Originally Posted by sKebess 1=1 1²=1² Since, (-1)²=1=1², we have 1²=(-1)² So, taking square root on both sides gives 1=-1. Obviously, this is wrong, but where is the mistake? I suppose it has to do with the square root of (-1)², but why is that an issue? (-1)² is greater than 0, so afaik, its square root shouldn't cause any trouble. Thanks for the help.
The square root of $\displaystyle (-1)^2$ is $\displaystyle \pm 1$. The mistake is building an equation valid if the mapping is many-to-one and then failing to acknowledge that the relationship on the inverse operation is one-to-many, not one-to-one.

April 3rd, 2017, 08:18 AM   #3
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Quote:
 Originally Posted by Benit13 The square root of $\displaystyle (-1)^2$ is $\displaystyle \pm 1$. The mistake is building an equation valid if the mapping is many-to-one and then failing to acknowledge that the relationship on the inverse operation is one-to-many, not one-to-one.
Hi Benit

So, if I follow, then it's because my last equation is assuming a one-to-one relationship when in fact it was a "many-to-many" relationship?

So, instead, the correct implication should be
1²=(-1)² => +1=+1 ?

I'm not sure I clearly get this, so please bear with me if what I'm saying is nonsense...

 April 10th, 2017, 02:40 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,822 Thanks: 750 Strictly speaking the square root function returns the positive root. That is, $\sqrt{4}= 2$, not $\pm 2$. That's the reason we need to write the $\pm$ when solving the equation $x^2= c$ ($x= \pm\sqrt{c}$). If we wrote only $x= \sqrt{c}$ we get only the positive root. $\sqrt{1^2}= \sqrt{(-1)^2$ is the same as $\sqrt{1}= \sqrt{1}$ and results in $1= 1$. Thanks from Benit13 and Joppy Last edited by Country Boy; April 10th, 2017 at 02:44 PM.
 April 10th, 2017, 05:59 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,641 Thanks: 959 Math Focus: Elementary mathematics and beyond $$\sqrt{a^2}=|a|$$ Thanks from agentredlum
 April 12th, 2017, 03:59 AM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 As greg1313 has pointed out, perhaps we should not be too quick to 'strike out' the exponents on both sides of the equation ... $1^2 = (-1)^2 \ \ \ \ \ \ \$<-- TRUE $\sqrt{1^2} = \sqrt{(-1)^2} \ \ \ \ \ \ \ \$ <-- TRUE Now be careful ... $|1| = |-1| \ \ \ \ \ \ \ \ \ \ \ \ \ \$ <-- TRUE $1 = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ <--TRUE There is no paradox. Last edited by skipjack; April 12th, 2017 at 04:26 AM.

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