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?

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- Jan 11th 2010, 08:50 PMjarman007Quadratic 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? - Jan 12th 2010, 12:46 AMtonio

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

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

Tonio - Jan 12th 2010, 01:34 AMcrossbone
Tonio,

what does "|" symbol mean? - Jan 12th 2010, 01:46 AMtonio
- Jan 12th 2010, 02:12 AMcrossbone
how you get http://www.mathhelpforum.com/math-he...85bf90ab-1.gif from the equation? i'm just trying to follow your thought process and argument.

- Jan 12th 2010, 03:45 AMtonio
- Jan 12th 2010, 03:55 AMHallsofIvy
A slightly different proof. Let $\displaystyle \frac{m}{n}$ and $\displaystyle \frac{p}{q}$ be the rational roots (m, n, p, and q integer, of course), reduced to lowest terms. Then we must have $\displaystyle \left(x- \frac{m}{n}\right)\left(x- \frac{p}{q}\right)= 0$. Multiplying that out, $\displaystyle x^2- \left(\frac{m}{n}- \frac{p}{q}\right)x+ \frac{mp}{nq}$$\displaystyle = 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 $\displaystyle n= \pm 1$ and $\displaystyle q= \pm 1$. Then the roots $\displaystyle \frac{m}{n}$ and $\displaystyle \frac{p}{q}$ are integer. - Jan 12th 2010, 04:03 AMMoo
- Jan 12th 2010, 06:46 AMtonio

Not if the fractions are reduced:

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

$\displaystyle nq\mid mp\Longrightarrow mp =ynq$

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

$\displaystyle m^2q-mnp=xmnq$

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

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

Tonio - Jan 12th 2010, 07:01 AMJhevon
Another alternative: We wish to show that the roots of $\displaystyle x^2 + bx + c = 0$ are integers. By the quadratic formula, it suffices to show that the expression $\displaystyle \frac {-b \pm \sqrt{b^2 - 4c}}2$ is always an integer provided it is rational (which means $\displaystyle b^2 - 4c$ is a perfect square, call it $\displaystyle k^2$) and $\displaystyle b,c \in \mathbb{Z}$.

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

**case 1:**$\displaystyle b$ is even.

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

**case 2**is no worse and will be left to the OP. - Jan 12th 2010, 01:21 PMcrossbone
- Jan 12th 2010, 01:31 PMJhevon
- Jan 12th 2010, 02:16 PMcrossbone
that's the problem. i'm trying to understand his logic from going from $\displaystyle r^2+brs+cs^2=0$ to $\displaystyle s/(brs+cs^2)$. 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.

- Jan 12th 2010, 02:28 PMJhevon
actually, he was applying basic properties of integers that you can learn about in a number theory or abstract algebra class.

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

we have

$\displaystyle r^2 + brs + cs^2 = 0$

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

by dividing through by $\displaystyle s$, we have

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

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

Quote:

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.

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. - Jan 12th 2010, 04:10 PMcrossbone