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November 14th, 2014, 06:29 AM   #1
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Question Minus one equals one ( mathematical paradox)

Hello everyone,
I'm not sure whether it's appropriate to call this a "paradox" but anyways.
I want you to have a look at this and explain what step isn't right about it :

$\displaystyle -1 $
$\displaystyle = (-1)^1 $
$\displaystyle = (-1)^2 $$\displaystyle ^/$$\displaystyle ^2$
$\displaystyle = ((-1)^2)$$\displaystyle ^1$$\displaystyle ^/$$\displaystyle ^2$
$\displaystyle = (1)^1$$\displaystyle ^/$$\displaystyle ^2$
$\displaystyle = 1 $
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November 14th, 2014, 06:53 AM   #2
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The fourth line is wrong. You need to be more careful when dealing with functions which have multiple roots.
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November 14th, 2014, 07:03 AM   #3
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Yep. What CRG said.

$\displaystyle (1)^{1/2} = \pm 1$
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November 14th, 2014, 07:41 AM   #4
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All right guys thanks ! Now I get it.
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November 14th, 2014, 09:35 AM   #5
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Mathematical fallacies can be fun! Here's another to "prove" that 2=1

Let a=b

$\displaystyle ∴ a^2 = ab$

Add $\displaystyle a^2$ to each side:

$\displaystyle 2a^2 = a^2 + ab$

Subtract 2ab from each side:

$\displaystyle 2a^2 -2ab = a^2 + ab - 2ab$

"Take 2 out of the left side" and simplify the right side:

$\displaystyle 2(a^2-ab) = a^2 - ab$

Multiply the right side by 1:

$\displaystyle 2(a^2-ab) = 1(a^2 - ab)$

Divide each side by $\displaystyle a^2 - ab:$

$\displaystyle 2=1$
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November 15th, 2014, 03:14 AM   #6
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Thanks for sharing Timios !
But the error is present in the last line where you divide by $\displaystyle a^2 - ab$.
This difference is equal to zero and thus multiplication isn't allowed.
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November 15th, 2014, 07:39 AM   #7
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Quote:
Originally Posted by Sonoshee View Post
Thanks for sharing Timios !
But the error is present in the last line where you divide by $\displaystyle a^2 - ab$.
This difference is equal to zero and thus multiplication isn't allowed.
This difference is equal to zero and thus division isn't allowed.
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November 15th, 2014, 01:01 PM   #8
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Thanks for pointing out, my apologies I didn't double check my reply.
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November 15th, 2014, 01:16 PM   #9
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You got it Sonoshee! I knew you meant "division".

Board Tutor, division by zero is undefined. There's no one allowing or disallowing it. Anyone is free to go ahead and try it.
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November 15th, 2014, 10:27 PM   #10
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Quote:
Originally Posted by Benit13 View Post
Yep. What CRG said.

$\displaystyle (1)^{1/2} = \pm 1$
Hello!
No!I say that $\displaystyle (1)^{\frac{1}{2}} =\sqrt{1}=1$.
If $\displaystyle (1)^{\frac{1}{2}} = \pm 1$ , then how do $\displaystyle -(1)^{\frac{1}{2}}=-\sqrt{1}$?
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