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Math Help - 2+2= 8 ; is this proof correct?

  1. #16
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    Re: 2+2= 8 ; is this proof correct?

    I get that, but it doesn't explain why there is difference between "two square roots" and "the square root" appart from convention - is there a mathamatical difference.
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  2. #17
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    Re: 2+2= 8 ; is this proof correct?

    Quote Originally Posted by Mytutoing View Post
    I get that, but it doesn't explain why there is difference between "two square roots" and "the square root" appart from convention - is there a mathamatical difference.
    There are two square roots of 4, 2~\&~-2.
    The square root of 4 is \sqrt{4}=2..
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  3. #18
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    Re: 2+2= 8 ; is this proof correct?

    OK by why is this just symantics and convention or is there a mathamatical difference
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    Re: 2+2= 8 ; is this proof correct?

    Quote Originally Posted by Mytutoing View Post
    OK by why is this just symantics and convention or is there a mathamatical difference
    There a mathematical difference: \sqrt{4}\ne -2.
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  5. #20
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    Re: 2+2= 8 ; is this proof correct?

    so the opperator/ symbol for taking the square root of a number is differnet to the square root of the number?!

    Sorry still dont get it - the mathamatical notation seems to just be rewriting the previous MOD version in another way.
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  6. #21
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    Re: 2+2= 8 ; is this proof correct?

    besides if I solve the problem to find the intersect of the two lines y=5 and y=x^2 I expect two answers although I would represent the solution as [root symbol]5=x. Your argument suggests that the answer is only the positive values.
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  7. #22
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    Re: 2+2= 8 ; is this proof correct?

    Quote Originally Posted by Mytutoing View Post
    so the opperator/ symbol for taking the square root of a number is differnet to the square root of the number?! Sorry still dont get it - the mathamatical seems to just be rewriting the previous MOD version in another way.
    If x^2=4 then x=2\text{ or }x=-2. There are two roots of that equation.

    \sqrt{4}=2 and -\sqrt{4}=-2.

    If |x|=2 then x=2\text{ or }x=-2.
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