Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What exactly is meant by that you cannot take the square root of a negative number? Is $-1$ not equivalent to $1$ w.r.t distance from origin, or $-1$ equivalent to $2$ w.r.t $+1$ on a line diagram. Why the square roots cannot approach to origin from both positive and negative sides? Is square root not a reducing operator that converge things? If the product of two negative numbers can yield a positive number then why the square root of a negative number cannot make a positive number? I think $\sqrt{-1}=1$ if theres exist a $0$; or $\sqrt(-1)=1.4142...$ from +1 side.

The sqrt of 0 is 0 and sqrt of 1 is 1, where does the sqrt operation sits ? , at 1 or at 0? I think it sits at 1 because sqrt operation is derived from multiplication/division not from addition/subtraction.

Correct my concepts please.

Edit: i mean that the nth root of n number converge/nears to 1, which gives the impression that all the roots should fall into 1 which acts as a base point for the roots whether they are negative or positive. Therefor -1 has corresponding number 2 on +axis.

share|improve this question

closed as not constructive by Amzoti, Pedro Tamaroff, Julian Kuelshammer, Thomas Andrews, Hagen von Eitzen May 28 '13 at 21:14

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

8  
-1 for "bullshit". Edit: The user iostream007 has erased it for you, but I recommend that you avoid pointless vulgarity, as well as the arrogance to think that something you don't understand is "bullshit". –  Zev Chonoles May 28 '13 at 20:15
2  
$\large {-1}$ only for '' "bullshit'' –  Maisam Hedyelloo May 28 '13 at 20:18
2  
@Waqar $\sqrt{-1}$ it is not a bullshit its solution invent a new branch in mathematics –  iostream007 May 28 '13 at 20:21
1  
@Waqar $1$ is not square root of $-1$ because $1\cdot 1=1\ne -1$. Neither is $\sqrt 2$. –  Berci May 28 '13 at 20:24
3  
Ok, back to the basics here: What is a square root, Waqar? –  imranfat May 28 '13 at 20:37

2 Answers 2

The square of a positive number is positive.
The square of a negative number is positive.

We say that $a$ is a square root of $b$ if $a\cdot a=b$.

If such $a$ exists among the real numbers, then by the above $b\ge 0$.


This observation can be translated as 'the equation $x^2=-1$ has no solution in $\Bbb R$'. This is just a similar situation that Pythagoras or some of his mates met some thousands years ago, when they realized that $x^2=2$ has no solution among the fractions (i.e., in $\Bbb Q$), that is, that $\sqrt2$ is irrational.

But, in mathematics we have some wonderful weapons. As long as the new element doesn't lead to logical contradiction, we are (kind of) free to include a new (ideal) element with prerequisted properties.

In that manner, one can extend $\Bbb Q$ with a fictive element $x$ such that $x^2=2$, and similarly, we can extend $\Bbb R$ with a fictive element $i$ such that $i^2=-1$.

share|improve this answer
    
i mean that the nth root of n number converge/nears to 1, which gives the impression that all the roots should fall into 1 which acts as a base point for the roots whether they are negative or positive. Therefor -1 has corresponding number 2 on +axis. –  Waqar Ahmad May 28 '13 at 21:16
    
After having introduced $i$ (that is, among complex numbers - which is the algebraic completion of the reals) we will have $n$ roots of $x^n=-1$. All of their absolute values indeed converge to $1$, as $n\to\infty$ (actually, all of their absolute values is $1$) but we can select a sequence of $n$th roots such that it doesn't converge. –  Berci May 28 '13 at 22:12

If you choose to avoid the definitions that the rest of mathematics already agrees to use, then you can define $1+3=5$. Not particularly useful.

A square root of a number $x$ is a number $y$ such that $y\cdot y = y^2 = x$. When restricting $y$ to be a real number, this can only happen if $x\geq 0$.

However, if you define square root to mean whatever it is you think it means, you are not going to get far in mathematics, because mathematics is a form of language, and if you decide to use words in a language to mean what you want them to mean against all other person's understandings of their meaning, then radiology campsite marmalade.

share|improve this answer
    
i m not saying in 1+3=5 sense. I am saying that theres no such thing as negative numbers at all . I m saying that they are proving the sqrt of negative numbers with symbols(i, etc) only, which are infact real i think. –  Waqar Ahmad May 28 '13 at 20:55
1  
Well, if you are going to say there is no such thing as negative numbers at all, then why not say that? Neither your topic nor the body makes it clear that your objection is to negative numbers. Indeed, by talking about $-1$, and whether it has a square root, you are implicitly are talking about a negative number. It's a relatively elementary mathematical step to define the set of all integers from the set of natural numbers, thus defining negative numbers. But it might be a step beyond where you are at just now. –  Thomas Andrews May 28 '13 at 20:57
    
i mean that the nth root of n number converge/nears to 1, which gives the impression that all the roots should fall into 1 which acts as a base point for the roots whether they are negative or positive. Therefor -1 has corresponding number 2 on +axis. –  Waqar Ahmad May 28 '13 at 21:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.