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    Hi,
    If (3rt(x)-rty)(2rt(x)-5rt(y)) is expanded would the expanded version only equal that one there if x is positive?
    Thanks
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    (Original post by 111davey111)
    Hi,
    If (3rt(x)-rty)(2rt(x)-5rt(y)) is expanded would the expanded version only equal that one there if x is positive?
    Thanks
    They would be equivalent only if both x,y \geq 0
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    (Original post by RDKGames)
    They would be equivalent only if both x,y \geq 0
    Thanks, could you explain why?
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    (Original post by 111davey111)
    Thanks, could you explain why?
    Because LHS has roots of x and y, and if you try to root a negative you will get an error - you cant root -ve numbers without extending into \mathbb{C}
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    (Original post by RDKGames)
    Because LHS has roots of x and y, and if you try to root a negative you will get an error - you cant root -ve numbers without extending into \mathbb{C}
    What if you used complex numbers?
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    (Original post by 111davey111)
    What if you used complex numbers?
    Then there would be no restrictions on the (real) values of x and y, taking \sqrt{x} to be shorthand for i\sqrt{-x} in the case that x is negative, of course.
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    (Original post by 111davey111)
    What if you used complex numbers?
    If you work in \mathbb{C}, then

    (3\sqrt{x}-\sqrt{y})(2\sqrt{x}-5\sqrt{y}) = 6x+5y-17\sqrt{x} \sqrt{y}

    for all x,y \in \mathbb{C} (and hence for all x,y \in \mathbb{R}) [noting that \sqrt{x} \sqrt{y} \not\equiv \sqrt{xy} in \mathbb{C}]
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    (Original post by RDKGames)
    If you work in \mathbb{C}, then

    (3\sqrt{x}-\sqrt{y})(2\sqrt{x}-5\sqrt{y}) = 6x+5y-17\sqrt{x} \sqrt{y}

    for all x,y \in \mathbb{C} (and hence for all x,y \in \mathbb{R}) [noting that \sqrt{x} \sqrt{y} \not\equiv \sqrt{xy} in \mathbb{C}]
    Have to be careful though, you need to define what you mean by 'the' square root of a complex number.
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    (Original post by RDKGames)
    If you work in \mathbb{C}, then

    (3\sqrt{x}-\sqrt{y})(2\sqrt{x}-5\sqrt{y}) = 6x+5y-17\sqrt{x} \sqrt{y}

    for all x,y \in \mathbb{C} (and hence for all x,y \in \mathbb{R}) [noting that \sqrt{x} \sqrt{y} \not\equiv \sqrt{xy} in \mathbb{C}]
    thanks, also another thing i don't get when you expand that bracket you go from 6rt(x^2) to 6x but isn't it equal to 6lxl
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    (Original post by IrrationalRoot)
    Have to be careful though, you need to define what you mean by 'the' square root of a complex number.
    You’re right, overlooked this one last night.

    (Original post by 111davey111)
    thanks, also another thing i don't get when you expand that bracket you go from 6rt(x^2) to 6x but isn't it equal to 6lxl
    They are not equivalent in C. The notion of abs(x) in C is that this is the value of the distance that x is from the origin on the complex plane.
    Certainly, if you choose x=i then you would get i=1 which is not true.

    Anyhow, as mentioned by IrrationalRoot, allowing C would make the relation true for x,y in R at the very least, and in R we have abs(x)=sqrt(x^2), but what we do not have is (sqrt x)^2 = sqrt(x^2) as this is only valid for positive x in reals. (Try x=-1 as a counterexample)
    In your expansion, you end up with sqrt(x)sqrt(x)=(sqrt x)^2 =/= sqrt(x^2) in R so you cannot use the absolute value here.
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    (Original post by RDKGames)
    You’re right, overlooked this one last night.



    They are not equivalent in C. The notion of abs(x) in C is that this is the value of the distance that x is from the origin on the complex plane.
    Certainly, if you choose x=i then you would get i=1 which is not true.

    Anyhow, as mentioned by IrrationalRoot, allowing C would make the relation true for x,y in R at the very least, and in R we have abs(x)=sqrt(x^2), but what we do not have is (sqrt x)^2 = sqrt(x^2) as this is only valid for positive x in reals. (Try x=-1 as a counterexample)
    In your expansion, you end up with sqrt(x)sqrt(x)=(sqrt x)^2 =/= sqrt(x^2) in R so you cannot use the absolute value here.
    so if x is greater than or equal to 0 the expansion would be equal to the LHS and with or without the modulus would be correct either way as you are only dealing with positive x and y and if x and y is any real then you have for example (the first term in expansion) 6rt(x)rt(x) which gives 6rt(x)^2 not 6rt(x^2). and this expansion is true (equal to LHS) for all real x and y.
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    (Original post by 111davey111)
    Thanks, so is it that when using complex numbers and expanding the bracket this - rt(x)rt(x) is not equal to rt(x^2) so no absolute value here.
    Yes.
    In C, rt(x)rt(x) is not rt(x^2), and rt(x^2) is not abs(x).
    So there is no equivalence between rt(x)rt(x) and abs(x)

    Also, in R, rt(x)rt(x) is not rt(x^2) but rt(x^2) is indeed abs(x)

    In positive R, both relations hold and hence rt(x)rt(x)=abs(x)

    But if you don't use complex numbers then you can use the modulus but there is no need to as x is greater than or equal to 0? sorry if I'm not getting it
    You initially asked about the domain of x,y.

    If we restrict the range of the function down to R, then we must have x,y positive in R.
    If we allow the range to be C, then we can have x,y entirely in R.

    You can use modulus for the first case if you want but it’s obsolete.
    You cannot use modulus in the second case because rt(x)rt(x) is not rt(x^2) for x in R, as said in previous part of this post.
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    (Original post by RDKGames)
    Yes.
    In C, rt(x)rt(x) is not rt(x^2), and rt(x^2) is not abs(x).
    So there is no equivalence between rt(x)rt(x) and abs(x)

    Also, in R, rt(x)rt(x) is not rt(x^2) but rt(x^2) is indeed abs(x)

    In positive R, both relations hold and hence rt(x)rt(x)=abs(x)



    You initially asked about the domain of x,y.

    If we restrict the range of the function down to R, then we must have x,y positive in R.
    If we allow the range to be C, then we can have x,y entirely in R.

    You can use modulus for the first case if you want but it’s obsolete.
    You cannot use modulus in the second case because rt(x)rt(x) is not rt(x^2) for x in R, as said in previous part of this post.
    ah thanks i get it now
 
 
 
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