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Thread: Would a/i=ai?

  1. #1 Would a/i=ai? 
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    Would a/i=ai because of the following?
    a/i=( (a/i)2 )1/2=(a2/i2)1/2=(-a2)1/2=ai


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  3. #2  
    Forum Junior anticorncob28's Avatar
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    I assume that i is the imaginary unit.
    If a/i = ai, then multiplying both sides by i gives a = -a, so there is clearly something wrong with your proof.
    The flaw is in the second step: the property a^(bc) = (a^b)^c applies if a is a positive real number, but not necessarily if otherwise.


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  4. #3  
    Forum Junior anticorncob28's Avatar
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    The same flaw is also shown in the second to last step.
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  5. #4  
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    Quote Originally Posted by endercreeper01 View Post
    Would a/i=ai because of the following?
    a/i=( (a/i)2 )1/2=(a2/i2)1/2=(-a2)1/2=ai
    Bzzt.

    Look, using your "logic"......



    WOW

    PS If you use the TEX facility provided here, your posts would be easier to read
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  6. #5  
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    Just wondering.
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  7. #6  
    Forum Junior anticorncob28's Avatar
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    Just to give you some insight and not to leave you lost, I'll tell you the correct variation of your formula. If you want to figure it out on your own then sorry just don't read ahead.
    Since i*i = -1, then -(i*i) = 1 and (-i)*i = 1 and thus 1/i = -i. Multiplying both sides by a gives a/i = -ai. So your formula would've been correct if you would have just a minus sign in there.
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  8. #7  
    Bassaricyon neblina Olinguito's Avatar
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    if and only if .
    Strange likes this.
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  9. #8  
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  10. #9  
    KJW
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    Quote Originally Posted by endercreeper01 View Post
    Would a/i=ai because of the following?
    a/i=( (a/i)2 )1/2=(a2/i2)1/2=(-a2)1/2=ai
    This represents the danger of many-to-one operations: they are not invertible. The operation of squaring is a two-to-one operation, ie and square to the same value and therefore the sign information is lost. When the square-root is taken, this is strictly a one-to-two operation and the wrong sign was chosen in that it wasn't the original sign that was squared, thus leading to the incorrect answer.
    anticorncob28 and Brett like this.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  11. #10  
    KJW
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    Quote Originally Posted by KJW View Post
    This represents the danger of many-to-one operations: they are not invertible.
    In fact, one can "prove" any sort of nonsense by neglecting this point. This can be done for example by arranging the "proof" to contain the expression:



    where the is somehow obscured by being an expression that evaluates to zero. Then by dividing both sides by this zero expression, one goes from the above true expression to the nonsense:



    Note that multiplying by zero is a many-to-one operation and is ultimately why dividing by zero is an invalid operation.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  12. #11  
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    as we know i^4 = 1
    so a * 1 / i = a * i^4 / i = a * i^3

    because i^3 = i^2 * i = -i
    then
    a * i^3 = a * (-i) = -a * i
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