1 = -1, which part of this proof is wrong?

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    Imaginary number Proof
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Discussion Overview

The discussion revolves around a mathematical proof that claims to show 1 equals -1 through the manipulation of square roots and complex numbers. Participants are examining the validity of the proof and identifying potential errors in the reasoning presented.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the proof is incorrect, noting that 1 cannot equal -1.
  • One participant suggests starting from the equation \(i^2 + 1 = 0\) as an alternative approach.
  • Another participant identifies a specific error in the proof, stating that the assumption \(\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}\) holds for real numbers but not for complex numbers.
  • Several posts reiterate the steps of the proof, emphasizing the flawed reasoning without introducing new insights.
  • Some participants express frustration over the repetition of previously identified errors in the proof.

Areas of Agreement / Disagreement

Participants generally agree that the proof is incorrect, but there is no consensus on the specific nature of the error beyond the assumption about square roots of complex numbers.

Contextual Notes

The discussion highlights limitations in the application of square root properties to complex numbers, and the potential for misunderstanding in mathematical proofs involving complex arithmetic.

Byeonggon Lee
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Of course 1 isn't same as -1.
This proof must be wrong but I can't find which part of this proof is wrong.
Could you help me with this problem?
(1)$$1 = \sqrt{1}$$
(2)$$= \sqrt{(-1)(-1)}$$
(3)$$= \sqrt{(-1)} \cdot i$$
(4)$$= i \cdot i$$
$$=-1$$
 
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Would you like better to start with i^2+1=0 ?
 
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In the transition between the second line, = \sqrt{(-1)(-1)}, and the third line, = \sqrt{-1}\cdot i you are assuming that \sqrt{ab}= \sqrt{a}\cdot \sqrt{b}. That is true for real numbers but not for general complex numbers.
 
##1=\sqrt{1}##
##=\sqrt{(-1)(-1)}##
##=\sqrt{-1}\cdot\sqrt{-1}##
##=i^2##
##=-1##
 
Deepak suwalka said:
##1=\sqrt{1}##
##=\sqrt{(-1)(-1)}##
##=\sqrt{-1}\cdot\sqrt{-1}##
##=i^2##
##=-1##
You are just repeating what others have pointed out to be wrong.
 
In a thread that is 7 months old.
 
And for an OP who has not been here since posting the question.
 
LCKurtz said:
In a thread that is 7 months old.
sorry
 
  • #10
Deepak suwalka said:
sorry

It's ok.
 

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