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Why does a / root(a) = root(a) ? Watch

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    we have

    3 / root(3)

    A.

    which is identical to

    (3^1) / (3^1/2)

    which is identical to

    (3^1) x (3^-1/2)

    as

    1 x -(1/2) = -(1/2)

    it follows that

    (3^1) x (3^-1/2) = 3^(-1/2)

    which is the same as

    1 / (3^1/2)

    B.

    3 / root(3) x root(3) / root(3) = 3 *root(3) / root(3)^2

    which is equal to

    3*root(3) / 3

    .

    but

    root(3) does not equal 1/root(3).

    Have I misunderstood something?
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     \frac{3\sqrt 3}{3} = \sqrt 3 .
    The 3 on top and bottom cancel.
    I'm not sure if I have read the question right.
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    index laws. a^m \times a^n = a^{m+n}, you have to add the powers not multiply.
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    (Original post by Athematica)
    (3^1) x (3^-1/2)

    it follows that

    (3^1) x (3^-1/2) = 3^(-1/2)
    I have marked in bold your first mistake.

    Think about the index laws : when you multiply two numbers with the same base, you add the powers, not multiply.
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    (Original post by Athematica)
    we have

    3 / root(3)

    A.

    which is identical to

    (3^1) / (3^1/2)

    which is identical to

    (3^1) x (3^-1/2)

    as

    1 x -(1/2) = -(1/2)

    it follows that

    (3^1) x (3^-1/2) = 3^(-1/2)

    which is the same as

    1 / (3^1/2)

    B.

    3 / root(3) x root(3) / root(3) = 3 *root(3) / root(3)^2

    which is equal to

    3*root(3) / 3

    .

    but

    root(3) does not equal 1/root(3).

    Have I misunderstood something?
    \sqrt{a} = a^{1/2}

    So \displaystyle \frac{a}{\sqrt{a}}=\frac{a^1}{a^  {1/2}}=a^1 \cdot a^{-1/2}=a^{(1-1/2)}=a^{1/2}=\sqrt{a}

    (Original post by Athematica)
    Have I misunderstood something?
    Yeah only one of the most important basic rules concerning indices.
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    It was a silly mistake to make. Thank you. It is coherent.

    (Original post by RDKGames)
    Yeah only one of the most important basic rules concerning indices.
    There is no need for snark. I made careless error.
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    (Original post by Athematica)
    There is no need for snark. I made careless error.
    Live and learn, at least you won't make the same mistake again.
    Spoiler:
    Show
    I hope.
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    I also made another error in typing when showing I understand that it works. This is not my evening.
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    (Original post by Athematica)
    I also made another error in typing when showing I understand that it works. This is not my evening.
    I was just about to respond to that. Days like this happen sometimes.
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    a / root(a) = a^1 x a^(-1/2) = a^(1/2) = root(a)

    That is the correct answer

    I also learned through play with these that the series

    s = 1/root(1)-root(2) + 1/root(2)-root(3) + 1/root(3)-root(4) +1/root(4)-root(5) + ... + 1/root(99)-root(100)

    is s = 9, which is pretty cool
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    (Original post by Athematica)
    a / root(a) = a^1 x a^(-1/2) = a^(1/2) = root(a)

    That is the correct answer

    I also learned through play with these that the series

    s = 1/root(1)-root(2) + 1/root(2)-root(3) + 1/root(3)-root(4) +1/root(4)-root(5) + ... + 1/root(99)-root(100)

    is s = 9, which is pretty cool
    It's an example of a telescoping series once you rationalise
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    (Original post by Zacken)
    It's an example of a telescoping series once you rationalise
    Oh, I didn't realise this is a formal thing that happens.

    I noticed something similar when solving the paradox where you have a space to travel and can move at half the distance the step you did before, the first being half the distance of the space. You turn an infinite series into finite numbers which is quite satisfying
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    (Original post by Athematica)
    Oh, I didn't realise this is a formal thing that happens.

    I noticed something similar when solving the paradox where you have a space to travel and can move at half the distance the step you did before, the first being half the distance of the space. You turn an infinite series into finite numbers which is quite satisfying
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    Haha. Great channel!

    Yeah, that's the one :yep: Skimming through, they solve it the same way I did in class. Hadn't realised that is the usual way to do it so hi-five me?



    On another note. I am really, really bad at the maths challenge-y stuff my college are asking me to do before (October or November?) this year's senior one. Any tips?
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    (Original post by Athematica)
    Haha. Great channel!

    Yeah, that's the one :yep: Skimming through, they solve it the same way I did in class. Hadn't realised that is the usual way to do it so hi-five me?
    If you're talking about doing something like S=... and then 2S=... then yeah it's a pretty standard method to do these sums. It's where \sum\limits_{r=1}^{n}r = \frac{1}{2}n(n+1) comes from. Gauss was the first person to use it, I think, at least for the sum of integers.
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    x^2/x=x
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    (Original post by Maker)
    x^2/x=x
    Omg this means we can use difference of two squares on surds!!!
 
 
 
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