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April 3rd, 2017, 07:45 AM   #1
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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.
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April 3rd, 2017, 07:59 AM   #2
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Quote:
Originally Posted by sKebess View Post
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.
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April 3rd, 2017, 08:18 AM   #3
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Quote:
Originally Posted by Benit13 View Post
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...
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April 10th, 2017, 02:40 PM   #4
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Strictly speaking the square root function returns the positive root. That is, , not . That's the reason we need to write the when solving the equation (). If we wrote only we get only the positive root.

is the same as and results in .
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Last edited by Country Boy; April 10th, 2017 at 02:44 PM.
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April 10th, 2017, 05:59 PM   #5
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$$\sqrt{a^2}=|a|$$
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April 12th, 2017, 03:59 AM   #6
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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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