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Math Help - Please help with exam prep! - Proofs

  1. #1
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    Please help with exam prep! - Proofs

    1. Some natural numbers can be expressed as a difference of two squares, but others cannot. For example, 12=4^2 - 2^2, but it is impossible to write 10 as a difference of two squares. Find a way to determine whether or not a given natural number can be expressed as the difference of two perfect squares.


    2. In rectangle ABCD, M is the midpoint of side AD and N is the midpoint of side DC. Segments AN and CM intersect at E
    a) Prove that <AEM=<MBN
    b)Determine how the angles in part a are related to the length: width ratio of the rectangle.


    3. A sequence is defined by the equation below:
    a) Prove that every term of the sequence is less than 3.
    b) Prove that every term of the sequence is greater than the preceding term.


    Any help you can give me on any/all of these questions would be great, thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by omgmath View Post

    3. A sequence is defined by the equation below:
    a) Prove that every term of the sequence is less than 3.
    b) Prove that every term of the sequence is greater than the preceding term.
    The sequence equation is missing
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  3. #3
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    Sorry I forgot the last equation, here it is:
    Attached Thumbnails Attached Thumbnails Please help with exam prep! - Proofs-receqn.bmp  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by omgmath View Post

    3. A sequence is defined by the equation below:

    t_1 = 1, t_{n + 1} = \sqrt {2 t_n + 1}

    a) Prove that every term of the sequence is less than 3.
    Now i'm one to always be off by some technicality, so use these answers as a general outline and fix it up to your liking.



    Let the sequence \left( t_n \right) be defined recursively as t_1 = 1, t_{n + 1} = \sqrt { 2 t_n + 1} for all integers n \geq 1

    We show that t_n < 3 for all integers n \geq 1

    We proceed by induction.

    Let P(n): " t_n < 3 for all integers n \geq 1"

    For P(1) we have t_1 = 1 < 3, so P(1) is true.

    Assume P(k) is true for some integer k > 1

    Since P(k) is true, we have

    t_k < 3

    \sqrt {2 t_{k - 1} + 1} < 3

    \Rightarrow 2 \sqrt {2 t_{k - 1} + 1} < 2 \cdot 3 = 6

    \Rightarrow 2 \sqrt {2 t_{k - 1} + 1} + 1 < 6 + 1 = 7

    \Rightarrow \sqrt { 2 \sqrt {2 t_{k - 1} + 1} + 1} < \sqrt {7}

    \Rightarrow \sqrt {2 t_k + 1} < \sqrt {7}

    \Rightarrow t_{k + 1} < \sqrt {7} < 3

    So we have t_{k + 1} < 3, which is P(k + 1)

    Thus the induction proof is complete, and we have t_n < 3 for all integers n \geq 1


    b) Prove that every term of the sequence is greater than the preceding term.
    Let \left( t_n \right) be the sequence defined above. We show that t_{n + 1} > t_n for all integers n \geq 1

    We proceed by induction.

    Let P(n): " t_{n + 1} > t_n for all integers n \geq 1"

    We have for P(1): t_{1 + 1} = t_2 = \sqrt {3} > 1 = t_1 . So P(1) is true.

    Assume P(k) is true for some integer k \geq 1

    Then we have:

    t_{k + 1} > t_k

    \Rightarrow 2 t_{k + 1} > 2 t_k

    \Rightarrow 2 t_{k + 1} + 1 > 2 t_k + 1

    \Rightarrow \sqrt {2 t_{k + 1} + 1} > \sqrt {2 t_k + 1}

    \Rightarrow t_{k + 2} > t_{k + 1}, which is P(k + 1)

    Thus the induction proof is complete and we have t_{n + 1} > t_n for all integers n \geq 1
    Last edited by Jhevon; June 17th 2007 at 06:33 PM.
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  5. #5
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    Hello, omgmath!

    3. A sequence is defined by: . t_1 = 1,\;t_{n+1} \:=\:\sqrt{2t_n + 1}

    a) Prove that every term of the sequence is less than 3.

    b) Prove that every term of the sequence is greater than the preceding term.
    How about an inductive proof?

    (a) S(n)\!: \;t_n < 3 for all n.

    Verify S(1)\!:\;t_1 \,=\,1 \:<\:3 . . . true


    Assume S(k) is true: . t_k \:< \:3

    Multiply both sides by 2: . 2\!\cdot\!t_k \:<\:6

    Add 1 to both sides: . 2\!\cdot\!t_k + 1 \:<\:7

    Take the square root of both sides: . \sqrt{2\!\cdot\!t_k + 1} \:< \:\sqrt{7}


    The left side is: t_{k+1}
    The right side is: \sqrt{7} \:=\:2.645... \:<\:3

    Therefore: . t_{k+1} \:<\:3
    . . The inductive proof is complete.

    Part (b) is proved in a similar manner.


    [Edit: Jhevon beat me to it . . .]
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    [Edit: Jhevon beat me to it . . .]
    you were a lot smoother than me though. i went back to defining t_k in terms of t_{k - 1} , which was totally unnecessary and unaesthetic.

    you can go after the other two if you want
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    Thanks guys, any ideas on the first 2?
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  8. #8
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    Quote Originally Posted by omgmath View Post
    1. Some natural numbers can be expressed as a difference of two squares, but others cannot. For example, 12=4^2 - 2^2, but it is impossible to write 10 as a difference of two squares. Find a way to determine whether or not a given natural number can be expressed as the difference of two perfect squares.
    This is a number theory question.

    Say,
    n = a^2 - b^2 = (a+b)(a-b)
    A possible factorization of n is n\cdot 1

    And hence it is reasonable to search for solutions to:
    a+b=n
    a-b=1

    The solution is given by (a,b) = \left( \frac{n+1}{2}, \frac{n-1}{2} \right).

    Now the thing that we need is for a,b to be integers. And that certainly happens if n is an odd positive integer.

    What happens if it is even? The possibilities is that n has one of the four forms: 4k,4k+1,4k+2,4k+3. Note we covered the cases 4k+1,4k+3. Let us do n=4k. Then since it is divisible by 4 we can write:
    n = \left( \frac{n}{4} + 1 \right)^2 - \left( \frac{n}{4} - 1 \right)^2

    And what happens if n=4k+2. Well, the is impossible because a square modulo 4 must be either 1 or 3. Hence a^2 - b^2 = 0,1,3 (\bmod 4). Not 2! Hence such a representation is impossible.
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    Is there any way you could explain that to me a bit better, I have no clue how you got the solution for (a,b), and I dont understand what you were getting at when you said they need to be integers if n i odd...Im basically completely lost
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    Quote Originally Posted by omgmath View Post
    Is there any way you could explain that to me a bit better, I have no clue how you got the solution for (a,b), and I dont understand what you were getting at when you said they need to be integers if n i odd...Im basically completely lost
    Add the two equations to eliminate b.

    Subtract the two equation to eliminate a.

    Do you know how to solve simulatenous equations?
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  11. #11
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    Quote Originally Posted by omgmath View Post
    2. In rectangle ABCD, M is the midpoint of side AD and N is the midpoint of side DC. Segments AN and CM intersect at E
    a) Prove that <AEM=<MBN
    b)Determine how the angles in part a are related to the length: width ratio of the rectangle.
    I have an ugly way of doing it, but I think it works.

    Look at picture below. We want to show the blue angles are equal to eachother.

    In coordinate geometry you perhaps learn the following theorem, let l_2,l_1 be two lines (non-vertical). And let l_2 has bigger angle measurement then l_1 with the x-axis. Then if \theta is the angle between them we have:
    \tan \theta = \frac{m_2 - m_1}{1+m_2m_1} where m_2 is the slope of l_2 and m_1 is the slope of l_1.

    Here is how we use it.

    Let \alpha be the angle <AEM. And \beta be the angle <MBN. We will show that \tan \alpha = \tan \beta and deduce \alpha = \beta.

    Now we find \tan \alpha and \tan \beta by using the theorem above.

    Let me do \alpha and leave \beta for you to do.

    In the picture \alpha is formed by lines AN and CM with side CM creating a bigger angle than AN with x-axis (in this case x-axis is AD).

    Thus,
    \tan \alpha = \frac{m_{CM} - m_{AN}}{1 + m_{CM}\cdot m_{AN}}.

    To compute the slope we just find the ratio between the the height and width.

    m_{CM} = \frac{CD}{DM} = \frac{2DN}{DM}

    m_{AN} = \frac{DN}{AD} = \frac{DN}{2DM}

    Thus,
    \tan \alpha = \frac{\frac{2DN}{DM} - \frac{DN}{2DM}}{1+ \frac{2DN}{DM}\cdot \frac{DN}{2DM}} = \frac{\frac{3DN}{2DM}}{\frac{DM^2+DN^2}{DM^2}} = \frac{3DN\cdot DM}{2MN^2}

    This answers the second part of the question on how to compute the angle in terms of the sides.

    Now do the same thing with \tan \beta.
    And show they give the same value.
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  12. #12
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    Quote Originally Posted by omgmath View Post
    1. Some natural numbers can be expressed as a difference of two squares, but others cannot. For example, 12=4^2 - 2^2, but it is impossible to write 10 as a difference of two squares. Find a way to determine whether or not a given natural number can be expressed as the difference of two perfect squares.

    Suppose integer X can be written as the difference of two squares, then:

    X = a^2-b^2 = (a-b)(a+b)

    then if we suppose a>b>0, then X must have two non-zero factors u, v,
    u!=v, where

    u=a+b

    v=a-b

    adding these we see that u+v=2a, and subtracting these we see that
    u-v=2b.

    Now suppose we are given any non zero u and v, u>v, such that their sum is even (then their difference is even as well), we can write:

    a=(u+v)/2

    and:

    b=(u-v)/2

    then uv=(a+b)(a-b).

    So we see that for any X to be written as the difference of two squares
    it is necessary and sufficient that X have two distinct factors whoes sum
    is even.

    RonL
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  13. #13
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    Arrow

    Hello, omgmath!

    I have an even uglier solution for #2 . . .


    2. In rectangle ABCD, M is the midpoint of side AD and N is the midpoint of side DC.
    . . Segments AN and CM intersect at E
    a) Prove that <AEM=<MBN
    b) Determine how the angles in part a are related to the length: width ratio of the rectangle.
    Let L = AD,\;W = CD

    Let \angle CMD = \alpha\quad\Rightarrow\quad\angle MCD \,=\, 90 -\alpha

    Let \angle NAD = \beta\quad\Rightarrow\quad\angle AND = 90-\beta\quad\Rightarrow\quad\angle ANC = 90 + \beta

    In \Delta CEN:\;\angle CEN \;=\;180^o - (90 - \alpha) - (90 + \beta) \;=\;\alpha - \beta

    . . Hence: . \angle AEM \;=\;\alpha - \beta

    We have: . \tan(\angle AEM) \;=\;\tan(\alpha - \beta) \;=\;\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\cdot\tan\beta}

    \text{Since: }\tan\alpha \:=\:\frac{W}{\frac{L}{2}} \:=\:\frac{2W}{L}\;\text{ and }\;\tan\beta \:=\:\frac{\frac{W}{2}}{L} \:=\:\frac{W}{2L}

    . . we have: . \tan(\angle AEM) \;=\;\frac{\frac{2W}{L} - \frac{W}{2L}}{1 + \left(\frac{2W}{L}\right)\left(\frac{W}{2L}\right)  }

    . . which simplifies to: . \boxed{\tan(\angle AEM) \:=\:\frac{3LW}{2(L^2+W^2)}} . [1]


    Here's the uglier part . . .

    In right triangle BAM\!:\;BM^2 \:=\:\left(\frac{L}{2}\right)\!^2 + W^2\quad\Rightarrow\quad BM \:=\:\frac{\sqrt{L^2+4W^2}}{2}
    In right triangle BCN\!:\;BN^2 \:=\:L^2 + \left(\frac{W}{2}\right)\!^2\quad\Rightarrow\quad BN \:=\:\frac{\sqrt{4L^2+W^2}}{2}
    In right triangle NDM\!:\;MN^2 \:=\:\left(\frac{L}{2}\right)\!^2 + \left(\frac{W}{2}\right)\!^2\quad\Rightarrow\quad MN \:=\:\frac{\sqrt{L^2+W^2}}{2}

    \text{Then: }\;\cos(\angle MBN) \:=\:\frac{BM^2 + BN^2 - MN^2}{2(BM)(BN)} \:=\:\frac{\frac{L^2+4W^2}{4} +\frac{4l^2+W^2}{4} - \frac{L^2+W^2}{4}}{2\left(\frac{\sqrt{L^2+4W^2}}{2  }\right)\left(\frac{\sqrt{4L^2+W^2}}{2}\right)}

    . . which simplifies to: . \cos(\angle MBN) \;=\;\frac{2(L^2+W^2)}{\sqrt{L^2+4W^2}\sqrt{4L^2+W  ^2}} \;=\;\frac{adj}{hyp}

    Since: opp^2 \:=\:hyp^2 - adj^2\quad\Rightarrow\quad opp^2 \;=\;\left[\sqrt{L^2+4W^2}\sqrt{4L^2+W^2}\right]^2 - \left[2(L^2+W^2)\right]^2

    . . Then: . opp^2 \:=\:9L^2W^2\quad\Rightarrow\quad opp \:=\:3LW

    And we have: . \boxed{\tan(\angle MBN) \;=\;\frac{3LW}{2(L^2+W^2)}} . [2]


    Therefore, from [1] and [2]: . \angle AEM \:=\:\angle MBN

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Let r \,=\,\frac{L}{W}
    We have: . \tan\theta \;=\;\frac{3LW}{2(L^2+W^2)}
    \text{Divide top and bottom by }W^2\!:\;\;\tan\theta \;=\;\frac{3\left(\frac{L}{W}\right)}{2\left[\left(\frac{L}{W}\right)^2 + 1\right]} \;=\;\frac{3r}{2(r^2+1)}
    Therefore: . \boxed{\theta \;=\;\tan^{-1}\!\left(\frac{3r}{2(r^2+1)}\right)}

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