What is the Flawed Proof That 1 Equals -1?

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SUMMARY

The forum discussion centers on flawed mathematical proofs that incorrectly assert 1 equals -1. Participants share various examples, including manipulations involving complex numbers and square roots, such as the erroneous step where \(\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}\) leads to the conclusion that -1 equals 1. The consensus is that all presented proofs contain fundamental errors, particularly in the handling of complex numbers and the properties of square roots. The discussion emphasizes the importance of recognizing these flaws to avoid misconceptions in mathematical reasoning.

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_Mayday_
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Hey Everyone,

A while back I found this cool little proof that showed 1 = -1. Now I am fully aware there was a little cheat in there somewhere, but I have lost the little proof. Has anyone come across it, or have anything similar? I just think it's cool, even though in one of the steps there is a mistake. I know it starts with like rooting one, and then putting 1 = (-1)(-1) etc.

Cheers.

_Mayday_
 
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I got another version:

\frac{-1}{1}=-1 and \frac{1}{-1}=-1

so:

\frac{-1}{1}=\frac{1}{-1}

if \sqrt{-1}=i

then

\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}

so:

\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}

\frac{1}{2}(\frac{i}{1})=\frac{1}{2}(\frac{1}{i})

becomes

\frac{i}{2}=\frac{1}{2i}

\frac{i}{2} + \frac{3}{2i} = \frac{1}{2i} + \frac{3}{2i}

i(\frac{i}{2} + \frac{3}{2i}) = i(\frac{1}{2i} + \frac{3}{2i})

\frac{-1}{2}+\frac{3}{2}=\frac{1}{2}+\frac{3}{2}

\frac{2}{2} = \frac{4}{2}

1 = 2
 
YOu got a problem right here: \frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}} Since this gives i=1/i.


\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}

Just because a=b doesn't mean that \sqrt a = \sqrt b.
 
Last edited:
robert Ihnot said:
\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}

Just because a=b doesn't mean that \sqrt a = \sqrt b.

yes it does (as long as we've agreed on some convention so that \sqrt{x} is a function, which we have)

And that line is correct. The problem is that in the complex numbers \sqrt{\frac{a}{b}} = \sqrt{\frac{c}{d}} does not imply that \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{c}}{\sqrt{d}}. This is true for the positive real numbers, but not for complex numbers in general.

But the poster was just asking for "proofs" that 1 = -1. Of course they are all flawed. But to the OP: There are a lot of "proofs" of this, so any more description, if you could remember any part of it, would be useful.
 
Last edited:
Hey Mayday,

You pretty much have a proof in the one that Dirk_mec1 posted.

Once you get to this step:

\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}

you now have

i/1 = 1/i

multiply both sides by i

you have (i^2)/1 = i/i --> -1/1 = 1/1 --> -1 = 1.
 
Thank you! Wait till my class see this stuff!
 
Another one that freaks people out

a=b
aa=ab
aa-bb=ab-bb
(a+b)(a-b)=b(a-b)
divide by a-b
a+b=b
since a=b then
2b=b
2=1
Naturally this is completely fake, the error in this logic is that when you divide by a-b you are dividing by zero. If you want to can keep repeating this and get like 1=4 and stuff. Kinda freaks people out but make sure you explain it to them in the end =P.
 

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