- #1

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$$2=2*1=2*\sqrt{1}=2*\sqrt{(-1*-1)}=2*i*i=2*i^2=-2$$

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- Thread starter fahraynk
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- #1

- 184

- 5

$$2=2*1=2*\sqrt{1}=2*\sqrt{(-1*-1)}=2*i*i=2*i^2=-2$$

- #2

fresh_42

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No, because it is wrong. You applied rules outside their area of validation. See:

$$2=2*1=2*\sqrt{1}=2*\sqrt{(-1*-1)}=2*i*i=2*i^2=-2$$

https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

- #3

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First of all square root does not come out of nowhere.

If y^2 = 1 then y=±(1)^(1/2). That means y is +1 or - 1.

Next i is a special complex unit in the form of (0,1) where as 1 is a real number. A REAL NUMBER CANNOT BE REPRESENTED AS PURELY COMPLEX.

<Edited>

A real number system is a subset of Complex number system but the converse isn't true.

I assume that this was a random post from a popular social media. The most of them are baseless, only written to attain popularity. Don't waste time on them.

If y^2 = 1 then y=±(1)^(1/2). That means y is +1 or - 1.

Next i is a special complex unit in the form of (0,1) where as 1 is a real number. A REAL NUMBER CANNOT BE REPRESENTED AS PURELY COMPLEX.

<Edited>

A real number system is a subset of Complex number system but the converse isn't true.

I assume that this was a random post from a popular social media. The most of them are baseless, only written to attain popularity. Don't waste time on them.

Last edited:

- #4

jbriggs444

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##-\sqrt{3}## is indeed equal to ##\sqrt{3} \times i^2##-3=sqrt(3) x i^2 FALSE

Edit: I'd failed to notice that the claimed inequivalence could rest in part on the obvious fact that 3 <> ##\sqrt{3}##. I've repaired that oversight and hope that I've now rendered the intended claim properly.

What is not true is that ##\sqrt{-1} \times \sqrt{-1}## is equal to ##\sqrt{-1 \times -1}##.

Last edited:

- #5

mathman

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- #6

Stephen Tashi

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square roots are two valued functions.

Yikes! How many threads are there where sudents are lectured that ##y = \sqrt{x}## is a function ? (i.e. not a "multi-valued" function).

"Square root" is an example of ambiguous terminology in mathematics. "##y## is equal to

In complex analysis no one blinks at speaking of "the n-th roots of unity". or even "multi-valued" functions.

- #7

mathman

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You can argue about terminology, but the fact remains 4 has two square roots, 2 and -2. This is the source of many silly proofs, such as in the original post!!Yikes! How many threads are there where sudents are lectured that ##y = \sqrt{x}## is a function ? (i.e. not a "multi-valued" function).

"Square root" is an example of ambiguous terminology in mathematics. "##y## is equal tothe square rootof ##x##" has one definition as a function. If ##x^2 = y## then ##x## isa square rootof ##y## has a different definition, which describes apropertyof ##x##.

In complex analysis no one blinks at speaking of "the n-th roots of unity". or even "multi-valued" functions.

- #8

Mark44

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Yes, no one disputes that, but by common agreement, the symbol ##\sqrt 4## evaluates to a single number, + 2.You can argue about terminology, but the fact remains 4 has two square roots, 2 and -2.

The OP's question has been answered, and there are multiple threads here about this "paradox" and similar ones, so I'm closing this thread.

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