Proof that 1 = -1
I hope this hasn't been posted before
1 = sqrt(1)
=> 1 = sqrt( (-1) * (-1) )
=> 1 = sqrt(-1) * sqrt(-1)
=> 1 = (sqrt(-1))^2
=> 1 = -1
What!!!!
1 = sqrt(1)
=> 1 = sqrt( (-1) * (-1) )
=> 1 = sqrt(-1) * sqrt(-1)
=> 1 = (sqrt(-1))^2
=> 1 = -1
What!!!!
Labels: mathemagic, SharedPuzzle
posted by Chris at
12:41 PM
29 Comments:
Let there be a=b
a = b |*b
ab = b^2 |-a^2
ab - a^2 = b^2 - a^2 |binomial
a(b-a) = (b+a)(b-a) |/(b-a)
a = b + a |a=b
b = 2b |/b
==> 1 = 2 crazy shiiit!!!! ^^
On the 4th line it reads:
"1 = (sqrt(-1))^2"
Sqrt(-1) cannot be done, as you cannot square root a negative number, so this proof is not a proof, sorry!
Hi tha biatcha.
On 5th line you divided both sides by 0 - totally not allowed. Too easy.
sqrt(-1) can be done.
Mathematicians of say 400 years ago were greatly troubled by sqrt(-1). They called it an impossible number. And so denoted it "i" (without the quotes). Zero and the negative numbers also used to be regarded as gibberish once upon a time.
You are not likely to have come across i before degree level tech or maths.
The problem is more subtle than that. And definitley more subtle than tha biatcha's thingy ;-)
Meant to have said do a Google with "imaginary numbers" if you want to learn more info.
-1^2 = +1^2
To prove it, take a one foot square post, and place it in a one square foot hole.
(The negative area is the hole).
Don't yell at me. It just a smart as the question.
Ragknot, you just made my head hurt and my eyes swivel.
just playing around with funny math stuff chris...
and dividing by zero is pretty cool nowadays ^^
+oo=1/0=1/(-0)=-(1/0)=-oo
If you go to the polar side of the complex plane, and ask an angle, maybe phi gently, he will tell you that sqrt(-1*-1)=!sqrt(-1)*sqrt(-1)
but you really shouldn't yell at him, he has some creepy friends like e^i*pi which hates to get squarerooted by some human.
Oh go on, try my one. I think it's fairly good.
Perhaps I should have actually asked for the resolution to the problem 1 = -1 problem.
tha b:::::::H, was your complex poem meant to have read sqrt(-1* -1) != sqrt(-1)*sqrt(-1)? (You wrote =!).
If so, explain why you say that.
I normally take sqrt(a*b) = sqrt(a)*sqrt(b) as being correct.
OK, Chris, if you are operating with complex numbers, you should do it correctly to avoid paradoxes. Each number has abs. value and phase. All +1, -1, i have abs. value 1, so we are dealing with the phase only.
+1=e^{2\pi i},
-1=e^{\pi i},
i=e^{\pi i/2}.
Of course you may add 2\pi phase everywhere, but it doesn't matter. Then
+1=e^{2\pi i}=(e^{\pi i/2})^2
*(e^{\pi i/2})^2. So you can take square root now and get
e^{\pi i}=e^{\pi i/2}e^{\pi i/2},
i.e. -1=-1.
Note. If you add 2\pi phase initially you will result with +1=+1.
For b::::::H. Do you know, that you can't div. by zero? You can just say, that
+OO=lim_{a->+0}1/a, or
-OO=lim_{a->-0}1/a.
Such paradoxes as your one was solved 250 years ago to teach us to deal with infinitely big and small numbers correctly.
Sorry, I'm to clever for you, I'll no more spoil your pleasure.
Okay chris, you take the blog here quite seriously, don't you?
Yes I meant Sqrt[(-1)*(-1)] not equal to Sqrt[-1]*Sqrt[-1]. What you do for your a and b is when they are > 0 (I hope).
Why is that like that? It's not that sqrt(-1) is defined, rather we define a number(or a unit if you prefer) "i" which is defined by:
i^2 = -1
and when you take the root you double the solution set giving you a plus and minus solution.
So Sqrt[(-1)*(-1)]=1 and Sqrt[-1]*Sqrt[-1]= (+,-)*i^2= (+,-)(-1)
and only the second (--1) solves our eq.
That's my explanation =)
That's more like it. The problem is too good to let it be ignored.
I loved the first few answers. I should have used a LOL or two.
Last anonymous and tha b::::::H, you are in the right territory.
But I'm seeking an explanation of where the given math went wrong,
not a proof that 1 != -1 or anything else like that.
Last anonymous, your notation is a bit non-standard. Just use pi,
not \pi. e.g. e^(pi i) or e^(pi*i). It is obvious that tha b;:::::H
knows that you can't divide by 0 etc. His first example was better.
The explanation doesn't need exponential representations.
It is unlikely that a decent mathematician would have been foolish
enough to proceed the way I did. (That isn't a riddle).
Oops, with the same success you may say sqrt (1)=-1 and have no problems.
Ooops forgot to say that (a*b)^c = (a^c)*(b^c) for all complex numbers a, b, c. But just be careful how you use it.
sqrt(a) = a^(1/2) (again be careful how you use it).
Idioms like a > 0 are not defined for complex a.
Last Anonymous, you are getting very close.
The actual real problem with your proof is this
if you multiply 2 positive real root numbers like
rtA.rtB = rt(A.B)
where A>0 and B>0, then this is pissible
If even one of them is negative also it is possible
rtA.rtB = rt(A.B)
where A>0 while B<0 OR
where A<0 while B>0
But is they both are negative it is not possible i.e
rtA.rtB not equal to rt(A.B)
when A<0 AND B<0
SO the rt(-1*-1) is not equal to the rt(-1) * rt(-1)
This is a proerty or imaginary numbers some1 mentioned this earlier.
ALSO you can just again square on both sides we can get:
1^2 =(-1)^2
But in this method nearly everything can b proved like anything can be multiplied by 0 on both sides and hence he proof. But 0, infinity and imaginary numbers are obscure parts of maths.
I mean the numbers are imaginary so why r they taken seriously at all...my imaginary friend was never taken seriously when i was 7 yrs. old ><
Hi A
rt(-1 * -1) = rt(1) = +/-1
rt(-1)*rt(-1) = (+/-i)*(+/-i)
= (+/-)i^2
= +/- (-1)
= +/-1
So your assertion is wrong.
The problem is far more pernicious than the divide by 0 etc. trivia. I wouldn't have posted it otherwise.
I'm not sure what you mean by 0, infinity and complex numbers as being obscure. They certainly require careful attention though.
But my sincere thanks for trying.
People, this is not a trivial problem.
Here's my take on it:
"Though I don't think that you're going to like it"
- (Deep Thought, THHGTTG).
The error comes about because sqrt(1) is multi-valued. Conventions
are needed to deal with this. One convention is to define a
principal value for an expression. That the pv is being specified
is often indicated by capitalising the first letter of the function.
Using that, i = Sqrt(-1) and 1 = Sqrt(1) are single-valued and well
defined. In fact they can be regarded as definitions.
Looking at the original statements, we are in trouble on the first
line because sqrt(1) = +/- 1. We can fix that by changing it to
1 = Sqrt(1). Similarly, we can modify the second line to
1 = Sqrt((-1)*(-1)). The next line has a new problem, we implicitly
are now "taking roots" of -1 (twice and independently). We have a
multi-multi-valued expression on the right. Now we can choose to say
this is nonsense and stop, or that we must choose to take the "+"
from one and the "-" from the other and use the principal values to
forestall further ambiguity. It should have been written
1 = -Sqrt(-1)*Sqrt(-1). This choice is in keeping with well know
ploy of throwing away unacceptable results from solving quadratic
equations that describe a physical system. The 4th line is wrong.
I won't argue if you want to claim the 3rd line is unfixable.
Obvioususly, the situation is contrived. It is madness to convert a
single-valued expression into a multi-valued one. You would
calculate the (-1)*(-1) = 1 first, then extract the roots. This all
boils down to, working with complex numbers requires great care.
Despite me bashing your attempts, I thought you had guys and gals?
had a pretty good try.
That was thrilling. Thank you for watching.
"It was a tough assignment," said Deep Thought.
Notes:
Just to cover the point, it is fine to say sqrt(a*b)=sqrt(a)*sqrt(b)
(complex a,b), but you must check that you address the multi-values.
Many would-be mathematical proofs have broken down due to these
difficulties (with not so easy to see resolutions as in our case),
causing the authors severe professional embarassment. This is
especially notable for many Riemann Hypothesis attempts.
Just for interest, the three cube roots of 1 are often written
1, w and w^2. (w is meant to be lower case omega). Note the "S".
This can be used to resolve the ambiguity of cuberoot(1).
w = -1/2 + i*Sqrt(3)/2
w^2 = -1/2 - i*Sqrt(3)/2
Sorry about that. I hadn't realised just how boring the explanation was going to be.
The quick explanation is that the first statement:
1 = sqrt(1) is false.
I reckon tha b::!:::h was the closest.
On rechecking Anonymous Aug 15, 1:17 am, I noticed a crafty move:
both sides of the equation were sqrt'd. That is a correct thing to
do, but it deviated from the original scheme - disqualified (soz).
negative number squared = positive number squared
e.g 8^2=64 -8^2=64 sqrt64=8 OR -8
PQ.
This post has been removed by the author.
the explanation is quite complicated because sqrt(-1) is imaginative and certain things cannot be done with those numbers like with real ones...
I don't understand what you mean. Please provide an example to illustrate your statement.
The square root of 1 is 1 so this proof is false. 1 does not = -1.
Duh! we all knew that the original proof was false.
You haven't demonstrated any connection between your statements and your conclusion. It isn't clear to me that you are even talking abut the posted problem.
Hmmmm, seems to me it is a Proof showing that 1 = -1. Chris, you obviously think your super smart and can solve this problem all by yourself so why don't you do that. After all, solving this riddle (which clearly you can do) will help you succeed in your life and help you get recognized by the people of the world.
Anonymous, your post is a blatant violation of the rules of this site.
OK, one troll fed.
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