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Math Help - Can you tell me if I'm right? :)

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    Can you tell me if I'm right? :)

    My sister and I are currently arguing about something regarding square roots. We'd like someone here to tell use who's right and why?

    Basically, the question is: Is is true that Square root of 16 = 4?

    My reasoning is pretty simple:

    Since Square root of 16 is 16^1/2, using the opposite operation, we could prove that both 4 and -4 are equal to 16^1/2 since (4) = 16 and (-4) = 16.

    My sister's reasoning (well it's not hers since she's too stubborn to write it down...) is pretty much this:

    First of all, she brought that the square root of -16 doesn't exist. I don't see the point in that, but she did... What she's basically saying is that you are finding the square root of (4) which then means that in can only be 4 for some magic... Honestly, I doubt she even understands her own logic...

    When answering, please take into account that I want an answer that is useful for maths a 16 years old would do. Please don't include Ph.D. level reasoning!

    Thank you!
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by Snowmen View Post
    My sister and I are currently arguing about something regarding square roots. We'd like someone here to tell use who's right and why?

    Basically, the question is: Is is true that Square root of 16 = 4?

    My reasoning is pretty simple:

    Since Square root of 16 is 16^1/2, using the opposite operation, we could prove that both 4 and -4 are equal to 16^1/2 since (4) = 16 and (-4) = 16.

    My sister's reasoning (well it's not hers since she's too stubborn to write it down...) is pretty much this:

    First of all, she brought that the square root of -16 doesn't exist. I don't see the point in that, but she did... What she's basically saying is that you are finding the square root of (4) which then means that in can only be 4 for some magic... Honestly, I doubt she even understands her own logic...

    When answering, please take into account that I want an answer that is useful for maths a 16 years old would do. Please don't include Ph.D. level reasoning!

    Thank you!
    By convention \sqrt{16} = 4. This convention holds for any use of numbers. But if we are talking about a variable \sqrt{x^2} = \pm x.

    -Dan
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by topsquark View Post
    By convention \sqrt{16} = 4. This convention holds for any use of numbers. But if we are talking about a variable \sqrt{x^2} = \pm x.

    -Dan
    I agree with the convention thing but say you had a test that asked you whether square root of 16 = 4 is true or false, what would you answer? We're not talking about a variable by the way.
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by topsquark View Post
    By convention \sqrt{16} = 4. This convention holds for any use of numbers. But if we are talking about a variable \sqrt{x^2} = \pm x.-Dan
    Dan, that is simply wrong mathematically. \sqrt{x^2} =|x| is correct.
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    Re: Can you tell me if I'm right? :)

    The definition of "square root" as a function is commonly defined to be specifically the POSITIVE number that when squared gives the original number. Yes it is an extremely asymmetrical and imbalanced definition, but it is agreed upon so that "square root" can be a function. Functions gives only one value to each number.

    So the square root of 16 is 4.

    But if you define it as a multiple-valued function, it can be -4 as well. So it can depend on the context. Honestly the argument is simply a question of notation or definition, much like arguing whether the first positive number is 1 or 0. Its not really a mathematical question. On a different planet, it may very well be that the square root of 16 as a function is -4.

    I agree with the convention thing but say you had a test that asked you whether square root of 16 = 4 is true or false, what would you answer? We're not talking about a variable by the way.
    I would ask them to define "square root".

    Quote Originally Posted by Plato View Post
    Dan, that is simply wrong mathematically. \sqrt{x^2} =|x| is correct.
    But the square root if -1 doesn't equal 1, the absolute value of i.
    Last edited by SworD; September 3rd 2012 at 05:48 PM.
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by SworD View Post
    I would ask them to define "square root".
    That is not the question.
    The square root of 4 is \pm 2.

    BUT \sqrt4=2

    Do not confuse the radical symbol with the words "square root". They are two different concepts.
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    Re: Can you tell me if I'm right? :)

    Ah thanks for the correction. I don't know too much about the precise definitions either. :P In that case it is true that the square root of 16 is both 4 and -4, but 4 is the principal square root.
    Last edited by SworD; September 3rd 2012 at 05:59 PM.
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by SworD View Post
    The definition of "square root" as a function is commonly defined to be specifically the POSITIVE number that when squared gives the original number. Yes it is an extremely asymmetrical and imbalanced definition, but it is agreed upon so that "square root" can be a function. Functions gives only one value to each number.
    So the square root of 16 is 4.
    But if you define it as a multiple-valued function, it can be -4 as well. So it can depend on the context. Honestly the argument is simply a question of notation or definition, much like arguing whether the first positive number is 1 or 0. Its not really a mathematical question. On a different planet, it may very well be that the square root of 16 as a function is -4
    I would ask them to define "square root"
    But the square root if -1 doesn't equal 1, the absolute value of i.
    Any one who is up to date with current mathematical notation knows that \sqrt{x} is used only if x\in\mathbb{R}^+.
    So that \sqrt{-1} has no meaning.

    What has happen by way of model theory, we know that we can add the symbol i such that i^2=-1 to the reals, thus we have the complex number set.
    Last edited by Plato; September 3rd 2012 at 06:25 PM.
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    Re: Can you tell me if I'm right? :)

    o.O That level of rigor is too much for me. But yeah, to the OP, you can ignore my post and just listen to Plato xD
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by Plato View Post
    That is not the question.
    The square root of 4 is \pm 2.

    BUT \sqrt4=2

    Do not confuse the radical symbol with the words "square root". They are two different concepts.
    I should have remembered \sqrt{x} = |x| But this one I didn't know. Thank you for both the correction and the concept here.

    -Dan
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by SworD View Post
    The definition of "square root" as a function is commonly defined to be specifically the POSITIVE number that when squared gives the original number. Yes it is an extremely asymmetrical and imbalanced definition, but it is agreed upon so that "square root" can be a function. Functions gives only one value to each number.

    So the square root of 16 is 4.

    But if you define it as a multiple-valued function, it can be -4 as well. So it can depend on the context. Honestly the argument is simply a question of notation or definition, much like arguing whether the first positive number is 1 or 0. Its not really a mathematical question. On a different planet, it may very well be that the square root of 16 as a function is -4.



    I would ask them to define "square root".



    But the square root if -1 doesn't equal 1, the absolute value of i.
    That wasn't what Plato said. he said \sqrt{x^2}= |x|. [itex]\sqrt{(-1)^2}= \sqrt{1}= 1[/itex].
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    Re: Can you tell me if I'm right? :)

    Quote Originally Posted by HallsofIvy View Post
    That wasn't what Plato said. he said \sqrt{x^2}= |x|. [itex]\sqrt{(-1)^2}= \sqrt{1}= 1[/itex].
    No you misunderstood, I wondered what would happen if you let x equal i, not -1. But Plato answered that anyway.
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