Results 1 to 9 of 9

Math Help - A question out of curiosity....

  1. #1
    Newbie
    Joined
    Apr 2012
    From
    London
    Posts
    3
    Thanks
    1

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,959
    Thanks
    1783
    Awards
    1

    Re: A question out of curiosity....

    Quote Originally Posted by Questioner View Post
    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.
    Both 2~\&~-2 are square roots of 4.
    Both 4i~\&~-4i are square roots of -16.

    All numbers except zero have two square roots.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853

    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 a^2= b. 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 x^2= a are x= \pm \sqrt{a} is precisely because \sqrt{a} 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2012
    From
    North Carolina
    Posts
    4

    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2012
    From
    North Carolina
    Posts
    4

    Re: A question out of curiosity....

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member Mathhead200's Avatar
    Joined
    Oct 2011
    From
    Washington DC
    Posts
    50
    Thanks
    6

    Re: A question out of curiosity....

    The symbol \sqrt{x} (read "the square root of x") is defined as a function. That is, for each x, there is only one \sqrt{x}. However, for any positive real number x, there are exactly 2 real numbers r1,r2 such that r12 = x and r22 = x. (In this case r1 = -r2.)

    If we include complex numbers we can extend this concept and assert: for any complex number x, there are exactly n complex numbers r1,r2,...,rn such that rin = x, 1 ≤ i ≤ n. The roots ri may not be distinct, but always come in pairs of complex conjugates. (If n is odd, there will be at least one real ri, and the rest are the pairs of complex conjugates. All pairs of real numbers are complex conjugates.)

    Example:
    24 = 16,
    (-2)4 = 16,
    (2i)4 = 16,
    (-2i)4 = 16,

    Note: we define i = \sqrt{-1} \implies i2 = -1 \implies (i2)2 = i4 = 1
    (i is not a real number.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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 x2 = a. we call the positive one √a, and the negative one -√a.

    if a = 0, there is only one number satisfying x2 = 0, namely: 0.

    if a < 0, there are two complex numbers satisfying x2 = a, either one of which can be called the/a square root (there's no sense of "positiveness" in the complex plane).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853

    Re: A question out of curiosity....

    Quote Originally Posted by Deveno View Post
    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 x2 = a. we call the positive one √a, and the negative one -√a.

    if a = 0, there is only one number satisfying x2 = 0, namely: 0.

    if a < 0, there are two complex numbers satisfying x2 = 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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2012
    From
    new york
    Posts
    5
    Thanks
    1

    Re: A question out of curiosity....

    amazing!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Curiosity about surface are
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 4th 2011, 07:35 AM
  2. Curiosity : integration, measure and probability ?
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: February 25th 2009, 03:02 PM
  3. My curiosity needs satisfying
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 25th 2008, 09:21 PM

Search Tags


/mathhelpforum @mathhelpforum