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    surely

    \sqrt -1 \sqrt -1 should be 1 not -1? because the negatives cancel?
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    (Original post by timebent)
    surely

    \sqrt -1 \sqrt -1 should be 1 not -1? because the negatives cancel?
    sqrt(-1) is not a defined quantity in terms other than i. Thus, the algebraic rules of surd multiplication do not apply here.
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    (Original post by oShahpo)
    sqrt(-1) is not a defined quantity in terms other than i. Thus, the algebraic rules of surd multiplication do not apply here.
    so different ball game so for example \sqrt -1 where the -1 is could be subbed in for something else right? so let -1 be x and the root of x squared is still x so where x is -1 the answer is -1 right? or am i wrong?
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    (Original post by timebent)
    surely

    \sqrt -1 \sqrt -1 should be 1 not -1? because the negatives cancel?
    sqrt(x)^2=x right?
    x=-1 in this case.
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    (Original post by timebent)
    surely

    \sqrt -1 \sqrt -1 should be 1 not -1? because the negatives cancel?
    Let x=-1
    Sqrtx*Sqrtx=(Sqrtx)^2=x=-1

    Hope that helps

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    (Original post by timebent)
    so different ball game so for example \sqrt -1 where the -1 is could be subbed in for something else right? so let -1 be x and the root of x squared is still x so where x is -1 the answer is -1 right? or am i wrong?
    Wrong, because if you use sqrt(x) then you must also specify the condition x>= 0.
    Sqrt(-1) means i, that is the only definition we have of this quantity in mathematics.
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    (Original post by timebent)
    so different ball game so for example \sqrt -1 where the -1 is could be subbed in for something else right? so let -1 be x and the root of x squared is still x so where x is -1 the answer is -1 right? or am i wrong?
    Correct.

    The square root of -1, squared, is equal to -1.
    More simply put, i^2 = -1

    In reference to your initial post, i is neither positive nor negative, as it does not lie on the real number line.
    It's coefficients can be positive or negative, but never i itself.
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    You cannot do it that way. We don't write  \sqrt{-1} like that for good reason. We use the imaginary (complex with real part 0) number  i . So for what you have put e would actually write it as  i \times i = -1 by definition. So  \sqrt{-16} = 4i . Using it this way is consistent.
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    (Original post by oShahpo)
    Wrong
    (Original post by JLegion)
    Correct.

    rip my brain
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    (Original post by oShahpo)
    Wrong, because if you use sqrt(x) then you must also specify the condition x>= 0.
    Sqrt(-1) means i, that is the only definition we have of this quantity in mathematics.
    The OP was asking (or seemed to for me) if the square root of -1 multiplied by itself is -1, which is true.
    i^2 = -1

    I think the wording of the post was open to misinterpretation.
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    (Original post by JLegion)
    The square root of -1, squared, is equal to -1.
    More simply put, i^2 = -1
    (Original post by JLegion)
    The OP was asking (or seemed to for me) if the square root of -1 multiplied by itself is -1, which is true.
    i^2 = -1
    No. Not "the square root of -1", there isn't a single square root, there are two square roots of -1, following that, it doesn't make sense to say "the square root of -1 squared is -1".
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    (Original post by timebent)
    \sqrt -1 \sqrt -1 should be 1 not -1? because the negatives cancel?
    To address, your issue, I know you're thinking \sqrt{-1}\sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1} = 1 by using the rule \sqrt{a}\sqrt{b} = \sqrt{ab}. However, something that your teachers neglected to mention is that that rule, that nice little operator nicety we take for granted only works over the real field. Indeed, the true formula should read

    \displaystyle 

\begin{equation*}\sqrt{a}\sqrt{b  } = \sqrt{ab} \quad \quad \quad a, b \geq 0 \end{equation*}

    So, taking a=b=-1 renders that formula obsolete and your approach fails. This is due to a more deeply rooted issue with having to give up certain 'privileges' of the sort when working in the algebraic closure of \mathbb{R}, as a extension to this, you lose commutativity when you move to the quaternion system, but I digress.
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    (Original post by B_9710)
    So  \sqrt{-16} = 4i . Using it this way is consistent.
    am I being dense or should this be 16i?


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    (Original post by drandy76)
    am I being dense or should this be 16i?


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    Nah mate. \sqrt{-16} \stackrel{?}{=} \sqrt{16} \sqrt{-1} = 4\sqrt{-1} = 4i. Indeed, (4i)^2 = 16i^2 = -16
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    (Original post by timebent)
    rip my brain
    The answer is correct, but the method is wrong.
    Square root of x^2 is not x but + or - x. The way to deal with this is to understand that sqrt of -1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
    Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = -1 you get sqrt(1) which is 1 or -1. You see the confusion here?
    The reason for this confusion is that sqrt(x) is not defined for negative numbers, if you are using negative numbers you must break it down in terms of i. So in this case it becomes sqrt(x^2) = sqrt(x*x)= sqrt(x) * sqrt(x) = sqrt(-1) ^2 = i^2 = -1 only.
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    (Original post by Zacken)
    Nah mate. \sqrt{-16} \stackrel{?}{=} \sqrt{16} \sqrt{-1} = 4\sqrt{-1} = 4i. Indeed, (4i)^2 = 16i^2 = -16
    Fek, I'm growing rusty


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    (Original post by Zacken)
    To address, your issue, I know you're thinking \sqrt{-1}\sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1} = 1 by using the rule \sqrt{a}\sqrt{b} = \sqrt{ab}. However, something that your teachers neglected to mention is that that rule, that nice little operator nicety we take for granted only works over the real field. Indeed, the true formula should read

    \displaystyle 

\begin{equation*}\sqrt{a}\sqrt{b  } = \sqrt{ab} \quad \quad \quad a, b \geq 0 \end{equation*}

    So, taking a=b=-1 renders that formula obsolete and your approach fails. This is due to a more deeply rooted issue with having to give up certain 'privileges' of the sort when working in the algebraic closure of \mathbb{R}, as a extension to this, you lose commutativity when you move to the quaternion system, but I digress.
    so the bit where it says a,b is greater than or equal to 0 surely you applied up there in your first paragraph where you said "a" and "b" are both -1, your second step of \sqrt -1x-1 should thus equal 1 in the end.

    However it seems that if you use negative numbers you still use the negative regardless

    so in this case then \sqrt -2 \sqrt -2 should then equal \sqrt -4 right?

    so if i'm right

    then what is the outcome of \sqrt 2 \sqrt -2??
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    (Original post by Zacken)
    No. Not "the square root of -1", there isn't a single square root, there are two square roots of -1, following that, it doesn't make sense to say "the square root of -1 squared is -1".
    I'm not sure I understand where you are coming from.
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    (Original post by oShahpo)
    The answer is correct, but the method is wrong.
    Square root of x^2 is not x but + or - x. The way to deal with this is to understand that sqrt of -1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
    Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = -1 you get sqrt(1) which is 1 or -1. You see the confusion here?
    The reason for this confusion is that sqrt(x) is not defined for negative numbers, if you are using negative numbers you must break it down in terms of i. So in this case it becomes sqrt(x^2) = sqrt(x*x)= sqrt(x) * sqrt(x) = sqrt(-1) ^2 = i^2 = -1 only.
    Ah I see, a mistake on my part.

    Sorry.
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    (Original post by oShahpo)
    The answer is correct, but the method is wrong.
    Square root of x^2 is not x but + or - x. The way to deal with this is to understand that sqrt of -1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
    Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = -1 you get sqrt(1) which is 1 or -1. You see the confusion here?
    The reason for this confusion is that sqrt(x) is not defined for negative numbers, if you are using negative numbers you must break it down in terms of i. So in this case it becomes sqrt(x^2) = sqrt(x*x)= sqrt(x) * sqrt(x) = sqrt(-1) ^2 = i^2 = -1 only.
    No, this is wrong. \sqrt{x\cdot x} \neq \sqrt{x}\sqrt{x} when x < 0. Indeed, the definition of i is i^2 = -1, any approach you try to use to derive this will be flawed because it's a definition.
 
 
 
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