# sqrt(x)= -1 , no soln. ?

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8 minutes ago, ahmet said:

sorry but I did not understand what you exactly imply or what the correlation was ,with this thread.

could  you provide more details please?

dear  Ahmet ;

You say ( in the original Q ) :

"

Anyways, the engine showed there exists no soln . for this.

I tried this on few others and still got the same. I can't understand this.

-1 and 1 both give 1 when multiplied by itself. No square of a number can be in negative form.

So sqrt(x)=-1 should have the solution as 1.

Can anyone explain this. "

Whether it be a case of imag number sol OR another one ,  THe  POint  REmains  that :

A complex number's "space" cannot be non-reflecting upon its real number counterpart. A complex number is equal to its complex conjugate if its imaginary part is zero, or equivalently, if the number is real. In other words, real numbers are the only fixed points of conjugation. The product of a complex number with its conjugate is equal to the square of the number's modulus. Two differing varieties of such number spaces , in case the imag turns out to be real or not , still remain to differ from one another in that more-than-one solutions as compared to single-solution eq(s) are to be representative of what we symbolize them to be. Fortunately or unfortunately , there  is not a standard (sort of objective) outside the realm of the symbolization that we , mathematicians , impose upon them . They simply do not lose their intimate connexion with analysis and solutions of variables defined by complex difference eq(s).

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Posted (edited)
7 hours ago, studiot said:

Thanks but that doesn't answer my question.

I looked back at your comments in this thread and perhaps I missed your question, can you please repeat it? The only question I saw was that you asked what is the additive inverse of a + bi; and it's of course -a =bi.

If there was another question I did not see it.

Edited by wtf
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4 minutes ago, wtf said:

I looked back at your comments in this thread and perhaps I missed your question, can you please repeat it? The only question I saw was that you asked what is the additive inverse of a + bi; and it's of course -a =bi.

If there was another question I did not see it.

Yes that was the question and I hope you mean (-a-ib)

🙂

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46 minutes ago, studiot said:

Yes that was the question and I hope you mean (-a-ib)

Yes thanks.

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16 minutes ago, wtf said:

Yes thanks.

Cool.

So whi8lst in the reals numbers, to use actual examples, we have the numbers +13 and -13 which are additive inverses in the real number system, we have no equivalent in the reals of

(13 + 13i) ; (-13 -13i) ; (13 -13i) and (-13 + 13i)

So there are four complex numbers, because the 13 can be positive or negative  one sign convention and that generates the four different permutations.

This is what I meant by saying there are two different conventions involved.

The sign of a and b

The operations of adding or subtracting

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Here is a site which expains the Fundamental Theorem of Algebra in our context.

It is vry easy to follow.

https://www.mathsisfun.com/algebra/fundamental-theorem-algebra.html

It goes up to include the use of complex numbers for quadratics that do not have a real solution.

Any questions  ?

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What about when strong integers are turned into computer bits and bytes ?

What about when complex conjugates do not exactly mirror one another in 2D  and/or  3D  ?

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2 minutes ago, Prof Reza Sanaye said:

What about when strong integers are turned into computer bits and bytes ?

What about when complex conjugates do not exactly mirror one another in 2D  and/or  3D  ?

again I did not understand what '"strong" integer' meant

and have you implied the symmetry via the underlined phrase?

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Posted (edited)
14 minutes ago, ahmet said:

again I did not understand what '"strong" integer' meant

and have you implied the symmetry via the underlined phrase?

Yes / Yes / Very Dear Ahmet ::  I am talking about the symmetry we expect to exist in case(s) of conjugates . . ... Be they integers or complexes ..

I shall be very grateful if you will kindly explain to me********

BTW : strong integers are those whose symmetric additive property extends to graphs where we have axes or curves containing both integers and complexes . . . . . .  .

Edited by Prof Reza Sanaye
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13 minutes ago, Prof Reza Sanaye said:

I am talking about the symmetry we expect to exist in case(s) of conjugates . . ... Be they integers or complexes ..

I shall be very grateful if you will kindly explain to me********

this will be off topic ,but based on  your query;I can suggest you check these topics in complex analysis

--- >> residues and solutions

--- >> C-R equations

--- >> analitic functions and their properties.

also, further:

you can look to:

--- >> conform transforms

16 minutes ago, Prof Reza Sanaye said:

BTW : strong integers are those whose symmetric additive property extends to graphs where we have axes or curves containing both integers and complexes . . . . .

strong integer, I have not heard it until now. (Weak limit does exist , but it seems no correlation with this)

what is more, when you mention a conjuugate you mention just one complex number. But if you are finding more than one conjugate number ,then most probably you mention complex functions.

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3 minutes ago, ahmet said:

this will be off topic ,but based on  your query;I can suggest you check these topics in complex analysis

--- >> residues and solutions

--- >> C-R equations

--- >> analitic functions and their properties.

also, further:

you can look to:

--- >> conform transforms

strong integer, I have not heard it until now. (Weak limit does exist , but it seems no correlation with this)

what is more, when you mention a conjuugate you mention just one complex number. But if you are finding more than one conjugate number ,then most probably you mention complex functions.

HHhhmmmmm  ..............  Many thanx ...........

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On 3/31/2021 at 4:40 PM, mathematic said:

Mathematically sqrt(1) has two solutions +1 and -1.  There is a convention that only +1 is allowed and the program may be set up to follow the convention.

"sqrt(1)" does not have any "solutions" because it is not a problem to be solved!  The equation $x^2= 1$ has two solutions, x= 1 and x= -1.  By definition, $\sqrt{y}$ is the positive number, x, such that $x^2= y$.

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