My Math Forum Minus one equals one ( mathematical paradox)

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 November 14th, 2014, 05:29 AM #1 Newbie   Joined: Nov 2014 From: Safi, Morocco Posts: 4 Thanks: 0 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$ $\displaystyle = 1$
 November 14th, 2014, 05:53 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms The fourth line is wrong. You need to be more careful when dealing with functions which have multiple roots. Thanks from CaptainBlack and topsquark
 November 14th, 2014, 06:03 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,142 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions Yep. What CRG said. $\displaystyle (1)^{1/2} = \pm 1$ Thanks from topsquark
 November 14th, 2014, 06:41 AM #4 Newbie   Joined: Nov 2014 From: Safi, Morocco Posts: 4 Thanks: 0 All right guys thanks ! Now I get it.
 November 14th, 2014, 08:35 AM #5 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 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 $\displaystyle a^2 - ab:$ $\displaystyle 2=1$
 November 15th, 2014, 02:14 AM #6 Newbie   Joined: Nov 2014 From: Safi, Morocco Posts: 4 Thanks: 0 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.
November 15th, 2014, 06:39 AM   #7
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Quote:
 Originally Posted by Sonoshee 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.

 November 15th, 2014, 12:01 PM #8 Newbie   Joined: Nov 2014 From: Safi, Morocco Posts: 4 Thanks: 0 Thanks for pointing out, my apologies I didn't double check my reply.
 November 15th, 2014, 12:16 PM #9 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 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.
November 15th, 2014, 09:27 PM   #10
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Quote:
 Originally Posted by Benit13 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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