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

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Recently on a computational engine called wolframalpha, by accident I put this equation for a solution( I wanted a soln of another eqn). But this mistake is not so anymore, it showed me that almost by every mean mathematics can be manipulated.

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.

Also, I searched this on certain sites and they explained with graphs of complex numbers containing parabola , hyperbola etc.  I have not learnt calculus yet.

If anyone can explain this without calculus elaborately , I will be grateful

Best wishes

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17 minutes ago, IndianScientist said:

Recently on a computational engine called wolframalpha, by accident I put this equation for a solution( I wanted a soln of another eqn). But this mistake is not so anymore, it showed me that almost by every mean mathematics can be manipulated.

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.

Also, I searched this on certain sites and they explained with graphs of complex numbers containing parabola , hyperbola etc.  I have not learnt calculus yet.

If anyone can explain this without calculus elaborately , I will be grateful

Best wishes

All that means is there is no real number solution to the equation.

You need to use complex numbers.

You are presumably aware of the nesting of number systems or types  complex numbers contain all real numbers which contain all rational numbers (fractions) which contain all integers (whole numbers).

Mostly we work in the real number system ( symbol R)

For this is paired with solutions to polynomial equations, the simplest have integer solutions, next simplest have rational number solutions, then real number solutions finally complex number solutions.

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Posted (edited)

To add a little to this

The equation

(x-25)(x+25) = 0  has integer roots,  x = +5 and x = -5

The equation

(10x-25)(10x+25) = 0 has no integer roots but has rational roots  x = +25/10 and x = -25/10

The equation

(x-√2.5)(x+√2.5) = 0 has neither integer roots nor rational roots but has irrational real roots x = +√2.5  and x  = -√2.5

Edited by studiot
insert missing square root signs
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2 hours ago, IndianScientist said:

Recently on a computational engine called wolframalpha, by accident I put this equation for a solution( I wanted a soln of another eqn). But this mistake is not so anymore, it showed me that almost by every mean mathematics can be manipulated.

I tried WolframaAlpha and the result seems mathematically correct:

i is the symbol used to denote the principal square root of -1, also called the imaginary unit. (from WolframAlpha's description)

You can also search for "i" on WolframAlpha read more about the definition. Here is a direct link:

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Posted (edited)

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.

Edited by mathematic
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I don't think  this is an imaginary numbers problem.
If the square root of x is minus one, then x = 1 is a solution.

Root (1)  = -1

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Posted (edited)
1 hour ago, John Cuthber said:

I don't think  this is an imaginary numbers problem.
If the square root of x is minus one, then x = 1 is a solution.

Root (1)  = -1

I agree. $$x^2 = -1$$ would be a different matter. But $$\sqrt{x} = -1$$ has one solution, which is $$1$$.

Edited by joigus
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1 hour ago, John Cuthber said:

I don't think  this is an imaginary numbers problem.
If the square root of x is minus one, then x = 1 is a solution.

Root (1)  = -1

On re reading the OP, you could be right,  I may have misread and jumped to the wrong conclusion as to the question.

If that is the case my apologies to IndianScientist and thank you John for the correction, +1

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Posted (edited)
1 hour ago, John Cuthber said:

Root (1)  = -1

I don't believe this is true. sqrt(1) = 1 by definition, assuming by Root(1) you mean $\sqrt{1}$. The square root of a positive real number is the positive of the two numbers whose square is the number.

So if someone asks, find $x$ such that $x^2 = 1$, the answer is {1, -1}. But if someone asks what is $\sqrt{1}$, the answer is 1.

There is no solution to the question in the title. What is true is that $- \sqrt{1} = -1$.

Edited by wtf
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8 hours ago, wtf said:

The square root of a positive real number is the positive

By one convention, and by the other convention, both roots are considered as equally valid.
For example, if you are solving a quadratic, you need to take account of both roots.

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Posted (edited)
10 hours ago, 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.

9 hours ago, wtf said:

I don't believe this is true. sqrt(1) = 1 by definition, assuming by Root(1) you mean 1 . The square root of a positive real number is the positive of the two numbers whose square is the number.

48 minutes ago, John Cuthber said:

By one convention, and by the other convention, both roots are considered as equally valid.

I suggest there is some confusion here between numbers, functions and equations  in Mathematics.

Remembering that I already overestimated IndianScientist's mathematical sophistication I await his reply to avoid confusing him.

Edited by studiot
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In general, $$x^2= a^2$$ has two solutions,  x= a and x=-a.

But if $$x= \sqrt{a^2}$$ then x= |a|.   In particular, there is NO x such that $$\sqrt{x}= -1$$.

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Posted (edited)

$x^{2} =-1$ ( the existence of $\sqrt{-1}$   ) or the creation of complex numbers in other words.

is an acceptance.

if you can (I do not know whether you can) solve quadratic equations, this may help you.

and the graphs may (also) be helpful

Edited by ahmet
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Posted (edited)
32 minutes ago, ahmet said:

x2=1 ( the existence of 1

Me being the picky type, let me point out that $\sqrt{-1}$ is not good notation and is technically not correct.

In the case of a nonnegative real number $x$, we can define $\sqrt x$ as the positive of the two values whose square is $x$.

However in the complex numbers there is no concept of positive or negative. That is, we can't algebraically distinguish between $i$ and $-i$. So we define $i$ as a complex number such that $i^2 = -1$. We pick one of the two possible values and call it $i$, and we call the other one $-i$.

The notation $\sqrt{-1}$ is technically inaccurate regardless of its ubiquity

That's why when people want to use that notation and are also being precise, they'll say, "Let $\sqrt{-1}$ be a square root of -1, rather than "the" square root of -1. The use of the word "a" signals that the writer understands the point and is using the notation $\sqrt{-1}$ anyway.

Edited by wtf
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Posted (edited)
17 minutes ago, wtf said:

Me being the picky type, let me point out that 1 is not good notation and is technically not correct.

In the case of a nonnegative real number x , we can define x as the positive of the two values whose square is x .

However in the complex numbers there is no concept of positive or negative. That is, we can't algebraically distinguish between i and i . So we define i as a complex number such that i2=1 . We pick one of the two possible values and call it i , and we call the other one i .

The notation 1 is technically wrong regardless of its ubiquity

technical means , the methods used in one, of branch of science, profession or art.

Edited by ahmet
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1 hour ago, ahmet said:

technical means , the methods used in one, of branch of science, profession or art.

Yes. The notation $\sqrt{-1}$ is often used casually, but it's imprecise in the branch of science, profession, or art of mathematics.

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Posted (edited)
13 hours ago, wtf said:

Yes. The notation 1 is often used casually, but it's imprecise in the branch of science, profession, or art of mathematics.

Actually there is another 'explanation'.

This is a another case where not one but two sign conventions  (as in the use of the symbols  + and - )  are in play and both have to be satisfied.

Each has a different technical meaning or operation.

This is similar to situations in Thermodynamics,  Mechanics and Electrics.

Brackets can be employed to highlight this situation.

Edited by studiot
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10 hours ago, studiot said:

Actually there is another 'explanation'.

I'm afraid I didn't see where you said what it is. There are no positive or negative numbers in the complex numbers, that's why you can't unambiguously use the sqrt sign convention that works in the real numbers.

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2 hours ago, wtf said:

I'm afraid I didn't see where you said what it is. There are no positive or negative numbers in the complex numbers, that's why you can't unambiguously use the sqrt sign convention that works in the real numbers.

So what is the additive inverse of the complex number z = (a +ib)  , given that a or b or both could themselves be either positive or negative ?

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Posted (edited)
11 minutes ago, studiot said:

So what is the additive inverse of the complex number z = (a +ib)  , given that a or b or both could themselves be either positive or negative ?

-a -ib. Why are you asking such an elementary question whose answer you perfectly well know? If z is a complex number, -z is its reflection through the origin.

Perhaps you'll find this helpful. There are no positive or negative complex numbers because it's not possible to put a total order on the complex numbers that is compatible with their addition and multiplication.

Or

Edited by wtf
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Posted (edited)
2 hours ago, studiot said:

So what is the additive inverse of the complex number z = (a +ib)  , given that a or b or both could themselves be either positive or negative ?

ps -- I don't think my previous explanations were very good. I found a much better page explaining this matter. If you google "why can't you distinguish i from -i" you get hundreds of totally irrelevant hits no matter how you alter or rephrase the question. Took me a while to find this.

The right answer is that there's an automorphism of $C$ that takes $i$ to $-i$; namely, complex conjugation. In other words the difference between the two amounts to a relabeling with no change of meaning.

By comparison, there is no automorphism of the reals that takes 1 to -1. They are algebraically different.

Edited by wtf
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9 hours ago, wtf said:

ps -- I don't think my previous explanations were very good. I found a much better page explaining this matter. If you google "why can't you distinguish i from -i" you get hundreds of totally irrelevant hits no matter how you alter or rephrase the question. Took me a while to find this.

The right answer is that there's an automorphism of C that takes i to i ; namely, complex conjugation. In other words the difference between the two amounts to a relabeling with no change of meaning.

By comparison, there is no automorphism of the reals that takes 1 to -1. They are algebraically different.

Thanks but that doesn't answer my question.

No one would expect complex numbers to be identical to the reals in every respect or there would be no point having them.

Real numbers have only two big divisions, complex numbers have four, in this sense.
So clearly a single pair of + and - is insufficient to label all the complex numbers.

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Posted (edited)
13 hours ago, studiot said:

So what is the additive inverse of the complex number z = (a +ib)  , given that a or b or both could themselves be either positive or negative ?

I was also thinking this. And gave a potential probability to wtf 's notation , stating or pointing out :

$e^{i \theta } = cos(\theta) + i sin(\theta)$

formula. but as I realised that OP has no basic knowledge in analysis,I did not want to give any instructions in complex analysis.

so, automorphism seems like useless to mention for ensuring the OP understand the issue

Edited by ahmet
locating formula and adding a sentence,one spelling error
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I guess the machine imaginary number "lock" has operated in such a way as to minimally bring our sqrt(s). If we do not accord the solution with some level of imaginary dimensionality , then unit becomes non-unit , which in modern maths does not necessarily offer problematics. The ancients , however , would have been shocked at it.

anyway , auto_difference(s) in eq(s) that are not difference ones , is still available in systems that deal with more-than-one solution.

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1 hour ago, Prof Reza Sanaye said:

I guess the machine imaginary number "lock" has operated in such a way as to minimally bring our sqrt(s). If we do not accord the solution with some level of imaginary dimensionality , then unit becomes non-unit , which in modern maths does not necessarily offer problematics. The ancients , however , would have been shocked at it.

anyway , auto_difference(s) in eq(s) that are not difference ones , is still available in systems that deal with more-than-one solution.

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

could  you provide more details please?

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