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Math Help - Rearranging Surds / Proof

  1. #1
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    Rearranging Surds / Proof

    Hey..

    Sorry for the horrible appearence of this question but here it goes..


    Show that:
    17(1-(1/17^2))^0.5 = n(sqrt(2))

    Where n is an integer, whose value is to be found


    the closest ive gotten so far is
    17(sqrt(1-17^-2))
    or
    17(sqrt((17^2-1)/(17^2))


    ... but I'm not sure where to start since 17 is prime and I don't think the brackets can be expanded because the sqrt() surrounding the right-most part of the expression doesn't allow this as far as I'm aware


    I'd prefer to have the steps to complete the problem since the value for n is given at the back of the textbook (n=12).

    Thanks in advance,
    Kwah =]
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    Quote Originally Posted by kwah View Post
    Hey..

    Sorry for the horrible appearence of this question but here it goes..


    Show that:
    17(1-(1/17^2))^0.5 = n(sqrt(2))

    Where n is an integer, whose value is to be found


    the closest ive gotten so far is
    17(sqrt(1-17^-2))
    or
    17(sqrt((17^2-1)/(17^2))


    ... but I'm not sure where to start since 17 is prime and I don't think the brackets can be expanded because the sqrt() surrounding the right-most part of the expression doesn't allow this as far as I'm aware


    I'd prefer to have the steps to complete the problem since the value for n is given at the back of the textbook (n=12).

    Thanks in advance,
    Kwah =]
    one way to do this:

    17 \sqrt{1 - \frac 1{17^2}} = \sqrt{17^2 - 1}

    = \sqrt{288}

    = \sqrt{2(144)}

    = \sqrt{2} \sqrt{144}

    = 12 \sqrt{2}

    got it?
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  3. #3
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    You missed ENTIRELY where I was having trouble with - getting

    17 \sqrt{1 - \frac 1{17^2}} = \sqrt{17^2 - 1}

    Everything after that I would have been entirely happy with ...

    After sitting down at it for another 5 minutes i remembered that it becomes

    \sqrt{289} * \sqrt{1 - \frac {1}{289}}

    which is then

    \sqrt{289 - 289(\frac 1{289})}

    \sqrt{289 - 1}

    \sqrt{288}

    .. then the steps you outlined =]
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    Quote Originally Posted by kwah View Post
    You missed ENTIRELY where I was having trouble with - getting

    17 \sqrt{1 - \frac 1{17^2}} = \sqrt{17^2 - 1}

    Everything after that I would have been entirely happy with ...

    After sitting down at it for another 5 minutes i remembered that it becomes

    \sqrt{289} * \sqrt{1 - \frac {1}{289}}

    which is then

    \sqrt{289 - 289(\frac 1{289})}

    \sqrt{289 - 1}

    \sqrt{288}

    .. then the steps you outlined =]
    the trick is this: if x is positive, then \sqrt{x^2} = x. so if we square a positive number and then square root it, we simply undo the square root, so that makes sense. furthermore, it is a law of surds that: \sqrt{x} \sqrt{y} = \sqrt{xy}

    so, 17 \sqrt{1 - \frac 1{17^2}} = \sqrt{17^2} \sqrt{1 - \frac 1{17^2}} ............by the first thing i said

    = \sqrt{17^2 \left( 1 - \frac 1{17^2}\right)} .........by the law of surds that i mentioned

    the rest follows
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    the trick is this: if x is positive, then \sqrt{x^2} = x. so if we square a positive number and then square root it, we simply undo the square root, so that makes sense. furthermore, it is a law of surds that: \sqrt{x} \sqrt{y} = \sqrt{xy}

    so, 17 \sqrt{1 - \frac 1{17^2}} = \sqrt{17^2} \sqrt{1 - \frac 1{17^2}} ............by the first thing i said

    = \sqrt{17^2 \left( 1 - \frac 1{17^2}\right)} .........by the law of surds that i mentioned

    the rest follows
    uhh... that is simply a less-clear explanation of the steps i did and naming it as the law of surds lol (dull)



    btw, forgive me if ive misunderstood, but none of this quote from your post is accurate...
    the trick is this: if x is positive, then \sqrt{x^2} = x.
    For \sqrt{x^2} = x, it does not matter if x is positive since it gets squared first...
    Obviously, if it gets rooted first then squared, \sqrt{x} ^2 = x x has to be positive


    so if we square a positive number and then square root it, we simply undo the square root, so that makes sense.
    As for this second sentence, it does not make sense that squaring a number THEN rooting it = undoing the root ((in fact, applying a root is undoing the square - which came first))


    regards,
    Kwah =]
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    Quote Originally Posted by kwah View Post
    uhh... that is simply a less-clear explanation of the steps i did and naming it as the law of surds lol (dull)
    it is a law of surds. and my explanation is more general than yours was. you used 17 as the specific example, my x could be anything.

    btw, forgive me if ive misunderstood, but none of this quote from your post is accurate...


    For \sqrt{x^2} = x, it does not matter if x is positive since it gets squared first...
    Obviously, if it gets rooted first then squared, \sqrt{x} ^2 = x x has to be positive
    you are mistaken. in general, \sqrt{x^2} = |x|

    by the way, x cannot be rooted first if it was negative. you'd get a complex number in that case
    As for this second sentence, it does not make sense that squaring a number THEN rooting it = undoing the root ((in fact, applying a root is undoing the square - which came first))
    the square is under the square root, so we apply it first. \sqrt{x^2} = (x^2)^{1/2}, you do what is in the brackets first.
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    Quote Originally Posted by kwah View Post
    As for this second sentence, it does not make sense that squaring a number THEN rooting it = undoing the root ((in fact, applying a root is undoing the square - which came first))
    yes, i meant to type square, not square root. but that typo seems obvious, as you pointed out. my apologies
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    I'm really confused where to begin with replying to this ..

    first off, i don't agree that \sqrt{x^2} = |x|
    regardless of the value of x, positive or negative, it results in \sqrt{positive number}, yeah?

    roots can have a positive OR negative value, therefore \sqrt{x^2} does not necessarily equate to |x|


    If, as I stated in my last post, the ^2 is OUTSIDE the root, then x is forced to become positive for two reasons..
    \sqrt{x}^2 = (\sqrt{x})^2 = |x| is true (note: the ^2 outside the root)

    for the reasons we have both stated, you cannot root negative numbers therefore x is required to be positive..
    AND, squaring it afterwards would force it to become positive regardless..


    the only other confusions / point I have to make about this is
    \sqrt{x^2} = (x^2)^0.5 = x^1
    \sqrt{(-10)^2} = ((-10)^2)^0.5 = (-10)^1 (negative)


    There was more that I wanted to say but it has gotten lost in my mind lol..

    regards,
    Kwah =]
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    Quote Originally Posted by kwah View Post
    I'm really confused where to begin with replying to this ..

    first off, i don't agree that \sqrt{x^2} = |x|
    regardless of the value of x, positive or negative, it results in \sqrt{positive number}, yeah?

    roots can have a positive OR negative value, therefore \sqrt{x^2} does not necessarily equate to |x|


    If, as I stated in my last post, the ^2 is OUTSIDE the root, then x is forced to become positive for two reasons..
    \sqrt{x}^2 = (\sqrt{x})^2 = |x| is true (note: the ^2 outside the root)

    for the reasons we have both stated, you cannot root negative numbers therefore x is required to be positive..
    AND, squaring it afterwards would force it to become positive regardless..


    the only other confusions / point I have to make about this is
    \sqrt{x^2} = (x^2)^0.5 = x^1
    \sqrt{(-10)^2} = ((-10)^2)^0.5 = (-10)^1 (negative)
    \sqrt{x^2} = |x| by definition. it is not something i made up. there are two standard definitions for |x|, and that is one of them.

    (exercise: try graphing \sqrt{x^2} and see what you get. it will be the absolute value function for x NOT the line y = x as you are claiming)

    counter example to what you said.

    let x = -2

    compute \sqrt{x^2}

    we get: \sqrt{(-2)^2} = \sqrt{4} = 2 \ne x

    also, i never said anything about (\sqrt{x})^2, so i do not see why you are arguing with me about it. yes indeed, the domain of this function is x \ge 0. that is if we are considering real numbers. i have no problem with you saying that x has to be nonnegative here for real numbers.

    There was more that I wanted to say but it has gotten lost in my mind lol..
    don't worry about it

    try to remember and come back
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    counter example to what you said.

    let x = -2

    compute \sqrt{x^2}

    we get: \sqrt{(-2)^2} = \sqrt{4} = 2 \ne x
    jeez lol ..

    \sqrt{(-2)^2} = \sqrt{4} = 2 \ne x is NOT correct ...
    \sqrt{(-2)^2} = \sqrt{4} = 2 OR -2 (im not sure how to get the plus minus symbol)

    Think about it .. how did you get -2^2 = 4 ??

    (exercise: try graphing \sqrt{x^2} and see what you get. it will be the absolute value function for x NOT the line y = x as you are claiming)
    err.. graphing \sqrt{x^2} would give the two lines y = x and y = -x



    also, i never said anything about (\sqrt{x})^2, so i do not see why you are arguing with me about it.
    I'm merely offering an alternative where it DOES have to equal |x|
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    Quote Originally Posted by kwah View Post
    I'm really confused where to begin with replying to this ..

    first off, i don't agree that \sqrt{x^2} = |x|
    regardless of the value of x, positive or negative, it results in \sqrt{positive number}, yeah?

    roots can have a positive OR negative value, therefore \sqrt{x^2} does not necessarily equate to |x|
    i thought i might respond to this specifically.

    \sqrt{\mbox{positive number}} = \mbox{a positive number} ALWAYS

    it is never true that \sqrt{\mbox{positive number}} = \mbox{negative number}

    i think i see where your misunderstanding comes from

    for instance, if you are solving x^2 = 1 the solution is x = \pm 1 (remark here: how did i get +/- 1? if you take sqrt(x^2) to be |x| it makes sense! if we use say sqrt(x^2) = x, then the only solution would be x = 1, which we know is false)

    however, that is not the same as saying \sqrt{1} = \pm 1 which is in fact what you are claiming by saying the root of a positive number can be negative or positive.

    do you understand what i am saying, or is there something else you're not clear on?
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    Quote Originally Posted by kwah View Post
    jeez lol ..

    \sqrt{(-2)^2} = \sqrt{4} = 2 \ne x is NOT correct ...
    \sqrt{(-2)^2} = \sqrt{4} = 2 OR -2 (im not sure how to get the plus minus symbol)

    Think about it .. how did you get -2^2 = 4 ??
    i never said -2^2 = 4, i said (-2)^2 = 4. that is correct

    -2*-2 = +4

    also, what you have is not correct. the square root function does not return negative values! look at the graph of square root x, the range is y greater than or equal to zero

    err.. graphing \sqrt{x^2} would give the two lines y = x and y = -x
    no it will not. it will give the graph for |x|

    I'm merely offering an alternative where it DOES have to equal |x|
    you are mistaken my friend
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  13. #13
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    Quote Originally Posted by Jhevon View Post
    i thought i might respond to this specifically.

    \sqrt{\mbox{positive number}} = \mbox{a positive number} ALWAYS

    it is never true that \sqrt{\mbox{positive number}} = \mbox{negative number}

    i think i see where your misunderstanding comes from

    for instance, if you are solving x^2 = 1 the solution is x = \pm 1 (remark here: how did i get +/- 1? if you take sqrt(x^2) to be |x| it makes sense! if we use say sqrt(x^2) = x, then the only solution would be x = 1, which we know is false)

    however, that is not the same as saying \sqrt{1} = \pm 1 which is in fact what you are claiming by saying the root of a positive number can be negative or positive.

    do you understand what i am saying, or is there something else you're not clear on?
    it is not me that is misunderstanding ..

    use this numerical example ..

    (x)^2 = (-x)^2 = xx
    (10)^2 = (-10)^2 = 100

    now root everything

    ((10)^2)^{0.5} OR ((-10)^2)^{0.5} = \sqrt{100}
    (10)^{2*0.5} OR (-10)^{2*0.5} = \sqrt{100}
    (10)^1 OR (-10)^1 = \sqrt{100}
    (10) OR (-10) = \sqrt{100}
    \sqrt{100} = (10) OR (-10) ///// EDIT: Just added this line in so that if can be directly compared to its algebraic equivalent contained in the "therefore"

    therefore

    \sqrt{(x)^2} = (x) or (-x)


    EDIT: included the brackets to make it absolutely clear what I am intending here ..
    and by "OR", i mean that either value is true - i do not know the plus OR minus symbol (a + sign with a - placed on top or below it)

    Is this step-by-step proof clear enough?

    regards
    kwah =]
    Last edited by kwah; February 12th 2008 at 04:29 PM.
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    Quote Originally Posted by kwah View Post
    it is not me that is misunderstanding ..

    use this numerical example ..

    x^2 = -x^2 = xx
    10^2 = -10^2 = 100
    now root everything
    (10^2)^0.5 OR (-10^2)^0.5 = \sqrt{100}
    10^(2*0.5) OR -10^(2*0.5) = \sqrt{100}
    10^1 OR -10^1 = \sqrt{100}
    10 OR -10 = \sqrt{100}
    therefore
    \sqrt{x^2} = x or -x


    Is this step-by-step proof clear enough?

    regards
    kwah =]
    first of all x^2 \ne -x^2 so nothing you said after that is correct (in the sense that the proof started with an incorrect premise). you started with an incorrect assumption. if we take f(x) = x^2, then -x^2 = - f(x) not f(x)

    it is true that x^2 = (-x)^2 which is completely different from what you are saying

    it is funny though, you last statement agrees with me. |x| returns x or minus x depending on the sign of x
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  15. #15
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    Quote Originally Posted by Jhevon View Post
    i never said -2^2 = 4, i said (-2)^2 = 4. that is correct

    -2*-2 = +4
    yeah .. that is what i meant .. for (-2) to be squared..


    also, what you have is not correct. the square root function does not return negative values! look at the graph of square root x, the range is y greater than or equal to zero
    HAVING THE RANGE GREATER THAN OR EQUAL TO ZERO IS LIMITING TO NOT SHOW ANY VALUES LESS THAN ZERO WHICH IS WHY YOU ARE ONLY SEEING ONE VALUE, DUH!!
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