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Math Help - What is the square root of 4?

  1. #1
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    What is the square root of 4?

    square root of 4= 2
    since 22 = 4
    and (-2)2 = 4

    is the above ture as we were taught that square root of 4=+2 only
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  2. #2
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    Quote Originally Posted by nazz View Post
    square root of 4= 2
    since 22 = 4
    and (-2)2 = 4

    is the above ture as we were taught that square root of 4=+2 only
    It is true that there are two square roots of 4, which are 2 and -2.

    However, often we use "square root" to mean "principal square root" which is the nonnegative value, 2. With a radical sign \sqrt{4} the value is by definition the principal square root. If we say "the square root" this is a pretty clear indication that we mean principal square root. Otherwise we would say "a square root." I think it is a slight abuse of terminology but doesn't really cause communication problems.

    Square root - Wikipedia, the free encyclopedia
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  3. #3
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    \sqrt{4} = 2 ... only

    if x^2 = 4 , then x = \pm 2 ... there is a difference.
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  4. #4
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    Quote Originally Posted by nazz View Post
    square root of 4= 2
    since 2(2) = 4
    and (-2)2 = 4
    is the above ture as we were taught that square root of 4=+2 only
    Think of it this way: there are two square roots of 4, \pm2.

    But the symbol \sqrt{4} stands for only one number, 2.

    -\sqrt{4}=-2.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Think of it this way: there are two square roots of 4, \pm2.

    But the symbol \sqrt{4} stands for only one number, 2.

    -\sqrt{4}=-2.
    So symbol \sqrt{-4} stands for 2i and -2i is not included?
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  6. #6
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    The reason we write the solution to x^2= a as \pm\sqrt{a} is that " \sqrt{a}" itself only stands for the "non-negative number such that x^2= a".

    But your last question raises an ambiguity. The real numbers can be "ordered"- that is we can define "less than" and so "positive", "negative", and "non-negative". Since the imaginary numbers are just a real number times i, we can do the same and say that "ai" is "positive" if and only if a is positive. In that sense we could define "the" square root of -4 as "2i" and not "-2i".

    However, the complex numbers, as a whole, cannot be ordered in such a way that the basic rules for an "ordered field"
    a) For any two numbers a and b, one and only one holds:
    i) a= b
    ii) a< b
    iii) b< a
    b) If a< b then a+ c< b+ c for all c
    c) If a< b then ac< bc for all c such that c< 0.

    Because of that, for general complex numbers we cannot define "positive" or "non-negative" and have no consistent way to distinguish between the two roots of " x^2= i". That is why, generally, we must allow multiple valued functions and take either root of x^2= a as " \sqrt{a}".

    That is why there is an ambiguity here. We can either consider " \sqrt{-a}", for a a positive real number, to be "[tex]i\sqrt{a}[/itex]" and assert only that single value as the square root, or we can consider " \sqrt{-a} to be the square root of a complex number and so take both values as the "square root".

    Strictly speaking the same problem occurs with square roots of positive real numbers, because the real numbers are a subset of the complex numbers. That is, we can think of \sqrt{4} as the square root of a real number and so have value 2 only or we can think of it as the square root of the complex number 4+ 0i and assert that both 2 and -2 are correct. But the real numbers have a "precedence", at least historically, so that we almost never do the latter.
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