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Thread: Algebra math tricks/puzzle - Equivalences in Algebra

  1. #1
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    Exclamation Algebra math tricks/puzzle - Equivalences in Algebra

    I have an equivalent algebra math problem and I still don't know how you solve it.
    It starts like:
    1. Take a number
    2. Multiply the two numbers on either side of your number
    3. Add 1 to your number
    4. Take the square root of this answer
    5. What do you notice?
    6. Prove this is true for any number

    Now I know that the prove means it has to have some type of algebra in it. But what I get stuck on is 5. What do you notice
    Can somebody help please because it has really been annoying me
    Thank you
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    Re: Algebra math tricks/puzzle - Equivalences in Algebra

    1. Take a number
    $\displaystyle n$

    2. Multiply the two numbers on either side of your number
    $\displaystyle (n - 1)(n + 1) = n^2 - 1$

    3. Add 1 to your number
    $\displaystyle n^2$

    4. Take the square root of this answer
    $\displaystyle n$

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    Re: Algebra math tricks/puzzle - Equivalences in Algebra

    Quote Originally Posted by topsquark View Post
    1. Take a number
    $\displaystyle n$

    2. Multiply the two numbers on either side of your number
    $\displaystyle (n - 1)(n + 1) = n^2 - 1$

    3. Add 1 to your number
    $\displaystyle n^2$

    4. Take the square root of this answer
    $\displaystyle n$
    Example where the number chosen doesn't equal the final number:

    The number I take is -9.

    (-9 - 1)(-9 + 1) = (-10)(-8) = 80

    80 + 1 = 81

    $\displaystyle 81 = 9^2$

    $\displaystyle \sqrt{9^2} = |9| = 9 $ **



    ** $\displaystyle \sqrt{x^2} = |x|$
    Last edited by greg1313; Jan 17th 2017 at 10:32 AM.
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    Re: Algebra math tricks/puzzle - Equivalences in Algebra

    Thank you so much this has really helped me!!!
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    Re: Algebra math tricks/puzzle - Equivalences in Algebra

    Quote Originally Posted by greg1313 View Post
    Example where the number chosen doesn't equal the final number
    1. Take a number: $\displaystyle \pi$
    2. Multiply the two numbers on either side of your number: "the two numbers on either side of your number" is not well defined so the multiplication cannot be done.
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    Re: Algebra math tricks/puzzle - Equivalences in Algebra

    1 3 5 7 ...
    1 4 9 16...
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    Re: Algebra math tricks/puzzle - Equivalences in Algebra

    Quote Originally Posted by Archie View Post
    1. Take a number: $\displaystyle \pi$
    2. Multiply the two numbers on either side of your number: "the two numbers on either side of your number" is not well defined so the multiplication cannot be done.
    True but I would have interpreted "the two numbers on either side of your number" as indicated that the "numbers" referred to here are integers.
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