2+2=5

You introduce an error on this line:

√(1)/√(-1) = √(-1)/√(1)

2+2=5 or sometimes 2+2=3 depending on which one is best for the party's interests.

I am reminded of this gem. [grin]

It's been a little while, but isn't taking the square root of a negative 'against the law'...

EDIT: Anyways you could've started with this line:

sqrt(1) = sqrt(1)1       = -1

Line 5: 1/i <> i
also for all integers a,b: (a/b)^1/2 <> a^(1/2) / b ^(1/2)

By abott and costello of who's on first base fame:

Proof that 7 x 13 = 28, from the movie In the Navy starring Abbott and Costello

Proof:           Check #1:                 Check #2:                     13                            13                 x 7                           ---               -----                       7 /28                   21                          - 7                   + 7                          ---               -----                          21                 28                         - 21                                            ----                                               0      Check #3:                      13                 13                 13                 13                 13                 13               + 13             ------                 28  (3+3+3+3+3+3+3) + (1+1+1+1+1+1+1)    In Check #1, multiply 7 x 3 to get 21.  Then multiply 7 x 1 to get 7.      Then add 21 and 7 to get 28.    In Check #2, we check the multiplication problem by dividing one factor into the product.    Divide 7 into 2 but it doesn't go.    So divide 7 into 8 to get 1.  Subtract 7 from 28 to get 21.    Now divide 7 into 21 to get 3.      So the answer is 13.    In Check #3, another way to check a multiplication problem is by addition.    To check if 7 x 13 = 28, we add seven 13s together.

Just a question, is the whole point of this thread to show incorrect math? Or can you not see the errors that are being made 0.0?

Quote:

Original post by Deranged Just a question, is the whole point of this thread to show incorrect math? Or can you not see the errors that are being made 0.0?

heeheeeheekeke, mine, with its flagrant errors is a joke, OP's whose "subtilty" is questionable is a trick?

Quote:

Original post by Deranged It's been a little while, but isn't taking the square root of a negative 'against the law'... EDIT: Anyways you could've started with this line: sqrt(1) = sqrt(1)1 = -1

The problem with the "proof" I linked to is that is misapplies the identity √(a) × √(b) = √(a × b). This identity does not necessarily hold on negative real arguments, yet the author assumes it does.

Edit: assuming it was my post you were referring to, which appears, upon re-reading your post, is not the case...

I guess I do not know how to read that proof. When you assert that 2 = 3 is True, I would have to disagree, even if 1 does equal 1. I have not had a proofs class, maybe that is why I do not see how this could prove anything.

Quote:

Original post by Deranged It's been a little while, but isn't taking the square root of a negative 'against the law'...

Well, that depends on what you mean. The square root of a negative is not defined for real numbers, but it is for complex numbers (or, more specifically, imaginary numbers).

Quote:

Original post by nilkn Quote: Original post by Deranged It's been a little while, but isn't taking the square root of a negative 'against the law'... EDIT: Anyways you could've started with this line: sqrt(1) = sqrt(1)1 = -1 The problem with the "proof" I linked to is that is misapplies the identity √(a) × √(b) = √(a × b). This identity does not necessarily hold on negative real arguments, yet the author assumes it does. Edit: assuming it was my post you were referring to, which appears, upon re-reading your post, is not the case...

I am confused - according to what the author is saying, sqrt(a * b) = sqrt(a) * sqrt(b) ALWAYS. So how is it disproved then?

Quote:

Original post by dustamulet Quote: Original post by nilkn Quote: Original post by Deranged It's been a little while, but isn't taking the square root of a negative 'against the law'... EDIT: Anyways you could've started with this line: sqrt(1) = sqrt(1)1 = -1 The problem with the "proof" I linked to is that is misapplies the identity √(a) × √(b) = √(a × b). This identity does not necessarily hold on negative real arguments, yet the author assumes it does. Edit: assuming it was my post you were referring to, which appears, upon re-reading your post, is not the case... I am confused - according to what the author is saying, sqrt(a * b) = sqrt(a) * sqrt(b) ALWAYS. So how is it disproved then?

sqrt(a * b) = sqrt(a) * sqrt(b) is only true where a and b are positive real numbers.

Quote:

Original post by Roboguy Quote: Original post by dustamulet Quote: Original post by nilkn Quote: Original post by Deranged It's been a little while, but isn't taking the square root of a negative 'against the law'... EDIT: Anyways you could've started with this line: sqrt(1) = sqrt(1)1 = -1 The problem with the "proof" I linked to is that is misapplies the identity √(a) × √(b) = √(a × b). This identity does not necessarily hold on negative real arguments, yet the author assumes it does. Edit: assuming it was my post you were referring to, which appears, upon re-reading your post, is not the case... I am confused - according to what the author is saying, sqrt(a * b) = sqrt(a) * sqrt(b) ALWAYS. So how is it disproved then? sqrt(a * b) = sqrt(a) * sqrt(b) is only true where a and b are positive real numbers.

Yes, but if you take into account what he has said, sqrt(a * b) = sqrt(a) * sqrt(b) for negative numbers too??

Quote:

Original post by dustamulet Quote: Original post by Roboguy Quote: Original post by dustamulet Quote: Original post by nilkn Quote: Original post by Deranged It's been a little while, but isn't taking the square root of a negative 'against the law'... EDIT: Anyways you could've started with this line: sqrt(1) = sqrt(1)1 = -1 The problem with the "proof" I linked to is that is misapplies the identity √(a) × √(b) = √(a × b). This identity does not necessarily hold on negative real arguments, yet the author assumes it does. Edit: assuming it was my post you were referring to, which appears, upon re-reading your post, is not the case... I am confused - according to what the author is saying, sqrt(a * b) = sqrt(a) * sqrt(b) ALWAYS. So how is it disproved then? sqrt(a * b) = sqrt(a) * sqrt(b) is only true where a and b are positive real numbers. Yes, but if you take into account what he has said, sqrt(a * b) = sqrt(a) * sqrt(b) for negative numbers too??

Take what into account? What are you referring to?

WARNING: this site may offend...
nilkn posted a link to http://cubicao.tk/stupidevil1.html

2 + 2 = 5 for sufficiently large values of 2...

My math teacher once gave proof that 1 = 5. She did that using a geometry, drawing some elaborate figure, defining a few angles in it and the calculating the rest of the angles in the figure from that using the F and Z rules. One of the angles would be 1 degree or 5 degree, depending on where you started. It stumped the class for quite a while. Sadly I can't remember the proof anymore.

Quote:

Original post by Deranged you could've started with this line: sqrt(1) = sqrt(1)1 = -1

No, because roots are defined as being positive. A "function" that allows two f(x) for a given x is not a function.

sqrt((-5)*(-5))=sqrt(25)=5
sqrt(-5)*sqrt(-5) = i*sqrt(5)*i*sqrt(5) = i*i*25 = -25

Actually, whenever sqrts are involved, you must be extra careful to remember that equation x²=a for nonzero a has two different roots, x₁=+sqrt(a) and x₂=-sqrt(a)
e.g. both -1 and 1 works in equation x²=1 .
All sqare root "fallacies" is based on at some point assuming that two roots of x²=a is equal, or rather implicitly assuming that x²=a has one root that is same.

It isn't really proof, just there is mistake in the chain that may go unnoticed because chain is long.