A question out of curiosity....

Can a number have 2 square roots? If you say no, surely 4 has 2 squared and minus 2 squared bothe as square roots? Do minus numbers get involved in square roots? And surely a minus number can't have a square root?

It's not exactly a 13+ question (which is what I'm doing), it's just out of curiosity.

Re: A question out of curiosity....

Re: A question out of curiosity....

But when you are restricting to **real numbers** it is standard to **define** the square root on b to be the **positive** real number a such that . In that sense a number has at most one square root. 2 is **the** square root of 4. (The reason we say that the two roots of are is precisely because itself is only the positive root.) We do this to make the square root function "single valued"- a very useful property.

Unfortunately, if we require functions of **complex numbers** be "single valued" we would lose most of our most useful functions! So for functions of complex variables we drop that requirement, making Plato's answer correct.

Re: A question out of curiosity....

..after you find the positive root (called the principal square root) multiply it by -1 to get the negative root.

example: square root of 9 = 3 (3 is the principal square root) ; 3 x -1 = -3 (-3 is the negative root) Thus the square root of 9 has two answers: 3 and -3

Now, for the square root of a negative number, find the square root of the same positive number and multiply by the letter "i" to find your answer. Where "i" will mean your answer is an imaginary number.

example: square root of -9 = (square root of 9)x(i) =3i ; the negative root will be 3i x -1 = -3i Thus the square root of -9 has two answers: 3i and -3i

Now, for the square root of an imaginary number I usually use a square root calculator that can handle imaginary numbers or complex numbers. http://www.squarerootcalcultor.co can help.

example: square root of 9i = 2.12 + 2.12i (which is a complex number) ; the negative root is -2.12 - 2.12i Thus the square root of 9i has two answers: 2.12 + 2.12i and -2.12 - 2.12i

I hope that helps.

Re: A question out of curiosity....

Re: A question out of curiosity....

The symbol (read "*the* square root of x") is defined as a function. That is, for each x, there is only one . However, for any positive real number x, there are exactly 2 real numbers r_{1},r_{2} such that r_{1}^{2} = x and r_{2}^{2} = x. (In this case r_{1} = -r_{2}.)

If we include complex numbers we can extend this concept and assert: for any complex number x, there are exactly n complex numbers r_{1},r_{2},...,r_{n} such that r_{i}^{n} = x, 1 ≤ i ≤ n. The roots r_{i} may not be distinct, but always come in pairs of complex conjugates. (If n is odd, there will be at least one real r_{i}, and the rest are the pairs of complex conjugates. All pairs of real numbers are complex conjugates.)

Example:

2^{4} = 16,

(-2)^{4} = 16,

(2*i*)^{4} = 16,

(-2*i*)^{4} = 16,

Note: we define *i* = *i*^{2} = -1 (*i*^{2})^{2} = *i*^{4} = 1

(*i* is not a real number.)

Re: A question out of curiosity....

defining i = √(-1) is somewhat problematic. the reason being, we don't know how to distinguish between i and -i (algebraically a+ib and a-ib act identitically: the map

z-->z* is a field automorphism).

the best we can say is i is "a" square root of -1 (in other words, orientation in the complex plane is chosen arbitrarily). behold the following "proof":

1 = √(1) = √((-1)(-1)) = √(-1)√(-1) = (i)(i) = -1,

which highlights the risky business of dealing with complex square roots. what happens is: multiplication in the complex plane can "rotate by more than a whole circle", the exponential function is no longer 1-1. so things like principal values and branches become very important (if we regard the complex number 1 = 1+0i as having angle 2π, it's "principal square root" is -1, if we regard it as having angle 0, it's principal square root is 1, the "standard convention").

perhaps the least confusing way to answer the OP's question is this:

for any real number a:

if a > 0, there are two real numbers that satisfy x^{2} = a. we call the positive one √a, and the negative one -√a.

if a = 0, there is only one number satisfying x^{2} = 0, namely: 0.

if a < 0, there are two complex numbers satisfying x^{2} = a, either one of which can be called the/a square root (there's no sense of "positiveness" in the complex plane).

Re: A question out of curiosity....

Quote:

Originally Posted by

**Deveno** defining i = √(-1) is somewhat problematic. the reason being, we don't know how to distinguish between i and -i (algebraically a+ib and a-ib act identitically: the map

z-->z* is a field automorphism).

the best we can say is i is "a" square root of -1 (in other words, orientation in the complex plane is chosen arbitrarily). behold the following "proof":

1 = √(1) = √((-1)(-1)) = √(-1)√(-1) = (i)(i) = -1,

which highlights the risky business of dealing with complex square roots. what happens is: multiplication in the complex plane can "rotate by more than a whole circle", the exponential function is no longer 1-1. so things like principal values and branches become very important (if we regard the complex number 1 = 1+0i as having angle 2π, it's "principal square root" is -1, if we regard it as having angle 0, it's principal square root is 1, the "standard convention").

perhaps the least confusing way to answer the OP's question is this:

for any real number a:

if a > 0, there are two real numbers that satisfy x^{2} = a. we call the positive one √a, and the negative one -√a.

if a = 0, there is only one number satisfying x^{2} = 0, namely: 0.

if a < 0, there are two complex numbers satisfying x^{2} = a, either one of which can be called the/a square root (there's no sense of "positiveness" in the complex plane).

No, that's not the best we can say nor do we have to choose 'arbitrarily'. The standard way to define the complex numbers is to the set of complex numbers as the set of **pairs** or real numbers (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). We can then interpret the real number 'a' as the pair (a, 0) and **define**, without ambiguity, i= (0, 1). Then we can write (a, b)= (a, 0)+ (0, b)= a+ bi and show that [tex]i^2= (0, 1)^2= (0(0)- 1(1), 0(1)+1(0))= (-1, 0)= -1.

Re: A question out of curiosity....