Results 1 to 8 of 8

Math Help - Beginning Discrete math, help plz.

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    175

    Beginning Discrete math, help plz.

    I have a couple of questions 1 of which I more or less know what should be the proof, I just dont really know how to prove it.

    1. My first question is how would I go about proving that m^2 = n^2
    if and only if m = n or m = -n.

    I know I have to prove it both ways, but am not quite sure how to construct it.

    2. My next problem is proving that either 2x10^500 + 15 or 2x10^500 + 16
    is not a perfect square.

    Now I am fairly certain that perfect squares do not appear sequentially, thus proving that one of them is not square, but I am not sure how to exactly go about it.

    The first question should probably be a proof that is fairly simple, these questions are both coming from chapters 1.6 and 1.7 of my book, so they shouldnt be too difficult.

    help appreciated. Extra points to those who can help me understand a little bit more.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, p00ndawg!

    Here is half of #2 . . .


    \text{2. Prove that neither }\,2\cdot10^{500} + 15\,\text{ nor }\,2\cdot10^{500} + 16\,\text{ is a perfect square.}
    We observe the following . . .

    \text{If }n\text{ is an even number, its square is a multiple of 4.}
    . . m \,= \,2k\quad\Rightarrow\quad m^2 \,=\,4k^2

    \text{If }n\text{ is odd, its square is one more than a mutliple of 4.}
    . . n^2 \:= \:(2k+1)^2\:=\:4k^2 + 4k + 1 \:=\:4k(k+1) + 1


    We have: . M \;=\;2\cdot10^{500} + 15 \;=\;2\cdot10\!\cdot\!10^{499} + 12 + 3 \;=\;2\!\cdot\!2\cdot5\!\cdot\!10^{499} + 4\cdot 3 + 3

    . . Hence: . M \;=\;4\left(5\!\cdot\!10^{499} +3\right) + 3


    Therefore, M is three more than a mutiple of 4 . . . It cannot be a square.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by p00ndawg View Post
    I have a couple of questions 1 of which I more or less know what should be the proof, I just dont really know how to prove it.

    1. My first question is how would I go about proving that m^2 = n^2
    if and only if m = n or m = -n.
    show that m^2 = n^2 \implies m = n \mbox{ or } m = -n. we can get this by square rooting both sides of the equation

    then show that m = n \mbox { or } m = -n \implies m^2 = n^2. we have to deal with each case separately. we get the desired result if we square both sides of the equation
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2008
    Posts
    175
    thank you guys for your help, what I am wondering although, is how in the HECK can you come up with some of these things?

    Because I just started this class, should It be this kind of awkward for me to do, or come up with some of these proofs?

    or rather, will it get easier as I practice?

    Because I must tell you, some of these things that are coming up in these proofs are kind of ridiculoid, and while they make PERFECT sense, I am just not quite sure how I could come up with them myself.

    And to tell you the truth, I am by no means bad at math in anyway, but this type of thinking really makes my head hurt.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,390
    Thanks
    1476
    Awards
    1
    Quote Originally Posted by p00ndawg View Post
    And to tell you the truth, I am by no means bad at math in anyway, but this type of thinking really makes my head hurt.
    You may be dismayed to learn that “this type of thinking” is really what mathematics is all about. What you may have thought was mathematics was just preliminary to the real heart of mathematics.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2008
    Posts
    175
    Quote Originally Posted by Plato View Post
    You may be dismayed to learn that “this type of thinking” is really what mathematics is all about. What you may have thought was mathematics was just preliminary to the real heart of mathematics.
    Yea I totally understand what you're saying, so If I clarify what I would mean to say is that I am rather good at standard mathematics.

    Of course, I only started learning this, so one would assume I wont be very good at this, Im pretty sure its safe to also assume that not many people were very good at the start of this type of mathematics as well.

    of course, I have been wrong before.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    (1)

    \begin{array}{lrcl}<br />
{} & m^2 &=& n^2\\<br />
\Leftrightarrow & m^2-n^2 &=& 0\\<br />
\Leftrightarrow & (m-n)(m+n) &=& 0\\<br />
\Leftrightarrow & m-n=0 &\mbox{or}& m+n=0\\<br />
\Leftrightarrow & m=n &\mbox{or}& m=-n<br />
\end{array}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jan 2008
    Posts
    175
    thank you for the reply.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Beginning Discrete Proof
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: March 10th 2009, 09:26 AM
  2. Help with discrete math
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: June 27th 2008, 09:25 AM
  3. Discrete Math PLEASE HELP!!
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 2nd 2008, 10:20 AM
  4. discrete math
    Posted in the Discrete Math Forum
    Replies: 13
    Last Post: June 19th 2007, 06:46 PM
  5. Discrete Math
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 26th 2007, 05:24 PM

Search Tags


/mathhelpforum @mathhelpforum