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Math Help - Quadratic Equations

  1. #1
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    Quadratic Equations

    Standard form of quadratic equations is ax^2+bx+c=0

    if a=1 , b and c are integers and the roots are rational numbers

    then prove that roots must be integers?
    Last edited by jarman007; January 11th 2010 at 09:44 PM.
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    Quote Originally Posted by jarman007 View Post
    Standard form of quadratic equations is ax^2+bx+c=0

    if a=1 , b and c are integers and the roots are rational numbers

    then prove that roots must be integers?

    Suppose \frac{r}{s} is a root of x^2+bx+c=0 , and that (r,s)=1 (this means: the fraction \frac{r}{s} is reduced) , then

    \frac{r^2}{s^2}+b\frac{r}{s}+c=0\Longrightarrow r^2+brs+cs^2=0 (multiplication by common denominator). As s|(brs+cs^2) then also s|r^2\Longleftrightarrow s|r , which is possible (by the reduced fraction assumption) iff  s=\pm 1 ... end now the argument

    Tonio
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    Tonio,
    what does "|" symbol mean?
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    Quote Originally Posted by crossbone View Post
    Tonio,
    what does "|" symbol mean?

    "|" = divide : a|b\Longrightarrow \exists c\in\mathbb{Z}\,\,\,s.t.\,\,\,b=ac

    Tonio
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    how you get from the equation? i'm just trying to follow your thought process and argument.
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    Quote Originally Posted by crossbone View Post
    how you get from the equation? i'm just trying to follow your thought process and argument.

    brs+cs^2=s(br+cs) ,and everything's integer here, of course.

    Tonio
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    A slightly different proof. Let \frac{m}{n} and \frac{p}{q} be the rational roots (m, n, p, and q integer, of course), reduced to lowest terms. Then we must have \left(x- \frac{m}{n}\right)\left(x- \frac{p}{q}\right)= 0. Multiplying that out, x^2- \left(\frac{m}{n}- \frac{p}{q}\right)x+ \frac{mp}{nq} = x^2- \frac{mq- np}{nq}x+ \frac{mp}{nq}= x^2+ bx+ c= 0.

    Since b and c are integers, we must have nq= 1. Since n and q are integers, we must have n= \pm 1 and q= \pm 1. Then the roots \frac{m}{n} and \frac{p}{q} are integer.
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    Hi,
    Quote Originally Posted by HallsofIvy View Post
    A slightly different proof. Let \frac{m}{n} and \frac{p}{q} be the rational roots (m, n, p, and q integer, of course), reduced to lowest terms. Then we must have \left(x- \frac{m}{n}\right)\left(x- \frac{p}{q}\right)= 0. Multiplying that out, x^2- \left(\frac{m}{n}- \frac{p}{q}\right)x+ \frac{mp}{nq} = x^2- \frac{mq- np}{nq}x+ \frac{mp}{nq}= x^2+ bx+ c= 0.

    Since b and c are integers, we must have nq= 1. Since n and q are integers, we must have n= \pm 1 and q= \pm 1. Then the roots \frac{m}{n} and \frac{p}{q} are integer.
    Why the sentence in red ?
    We could have nq|(mq-np) and nq|(mp) without nq being 1 (or -1)
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    Quote Originally Posted by Moo View Post
    Hi,

    Why the sentence in red ?
    We could have nq|(mq-np) and nq|(mp) without nq being 1 (or -1)

    Not if the fractions are reduced:

    nq\mid(mq-np)\Longrightarrow mq-np=xnq

    nq\mid mp\Longrightarrow mp =ynq

    multiply the first eq. above by m and the second one by n:

    m^2q-mnp=xmnq

    mnp=yn^2q , and now add these two eq's:

    m^2q=(xm+yn)nq\Longrightarrow m^2=(xm+yn)n .Since n divides the right hand n divides the left hand as well, contradicting the fact that \frac{m}{n} is reduced...unless n=\pm 1 ,and the same for q, of course.

    Tonio
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    Quote Originally Posted by jarman007 View Post
    Standard form of quadratic equations is ax^2+bx+c=0

    if a=1 , b and c are integers and the roots are rational numbers

    then prove that roots must be integers?
    Another alternative: We wish to show that the roots of x^2 + bx + c = 0 are integers. By the quadratic formula, it suffices to show that the expression \frac {-b \pm \sqrt{b^2 - 4c}}2 is always an integer provided it is rational (which means b^2 - 4c is a perfect square, call it k^2) and b,c \in \mathbb{Z}.

    It suffices to show that -b \pm \sqrt{b^2 - 4c} is even. We consider two cases: (1) b is even and (2) b is odd. (Notice that the parity of c is immaterial, since 4c would be even whether c is even or odd).

    case 1: b is even.

    Then, b^2 is even, so that k^2 = b^2 - 4c is even (since it is the difference of two even integers), which implies k is even. But then, -b \pm \sqrt{b^2 - 4c} = -b \pm k is even, since it is the sum of two even integers.

    case 2 is no worse and will be left to the OP.
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    Quote Originally Posted by tonio View Post
    brs+cs^2=s(br+cs) ,and everything's integer here, of course.

    Tonio
    the original equation is r^2+brs+cs^2=0
    why did you ignore r^2?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by crossbone View Post
    the original equation is r^2+brs+cs^2=0
    why did you ignore r^2?
    he didn't. you asked about one part though, so he answered that part. he did state that s|r^2...
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    that's the problem. i'm trying to understand his logic from going from r^2+brs+cs^2=0 to s/(brs+cs^2). it's like some leap of logic unless he's using some mathematical properties of polynomial coefficients that I'm missing. btw, I'm new here so quite curious why is everybody using "|" instead of "/" to signify division? there's no using the shift key to choose the latter.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by crossbone View Post
    that's the problem. i'm trying to understand his logic from going from r^2+brs+cs^2=0 to s/(brs+cs^2). it's like some leap of logic unless he's using some mathematical properties of polynomial coefficients that I'm missing.
    actually, he was applying basic properties of integers that you can learn about in a number theory or abstract algebra class.

    showing that s|r was important to tonio's argument. he could show that s|r by showing that s|r^2

    we have
    r^2 + brs + cs^2 = 0

    \Rightarrow r^2 = -(brs + cs^2)

    by dividing through by s, we have

    \frac {r^2}s = -(br + cs)

    since the right side is an integer (because s|(brs + cs^2)), the left side must be also, because we have equality here. but \frac {r^2}s is an integer if and only if s|r^2

    btw, I'm new here so quite curious why is everybody using "|" instead of "/" to signify division? there's no using the shift key to choose the latter.
    they are actually quite different symbols. "|" means "divides" while "/" means "divided by".

    the difference?

    2|4 means that 4 = 2k for some integer k. it means that 2 divides into 4 without leaving a remainder. while,

    2/4 = 1/2 = 0.5

    see the difference? one is making a statment about the relationship, namely a divisibility relationship, between two integers, while the other signifies performing and arithmetical operation, namely division, on two integers.
    Last edited by Jhevon; January 13th 2010 at 08:48 AM.
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    Quote Originally Posted by Jhevon View Post
    actually, he was applying basic properties of integers that you can learn about in a number theory or abstract algebra class.

    showing that s|r was important to tonio's argument. he could show that s|r by showing that s|r^2

    we have
    r^2 + brs + cs^2 = 0

    \Rightarrow r^2 = -(brs + cs^2)

    by dividing through by s, we have

    \frac {r^2}s = -(br + cs)

    since the right side is an integer (because s|(brs + cs^2)), the left side must be also, because we have equality here. but \frac {r^2}s is an integer if and only if s|r^2

    they are actually quite different symbols. "|" means "divides" while "/" means "divided by".

    the difference?

    2|4 means that 4 = 2k for some integer k. it means that 2 divides into 4 without leaving a remainder. while,

    2/4 = 1/2 = 0.5

    see the difference? one is making a statment about the relationship, namely a divisibility relationship, between two integers, while the other signifies performing and arithmetical operation, namely division, on two integers.
    i've never encountered the use of | in my high school math. i know the stuff here are all pre-calc but isn't the use of these esoteric explanation beyond the average high-school student?
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