1. Originally Posted by crossbone
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?
oh no, who told you this was precalc? and i very much doubt that this kind of math is routinely done in U.S. high schools, you might encounter them if you are in some sort of advanced placement or math competition. you may note though, that my and HallsOfIvy's solutions do not involve the | symbol and may be easier for a high school student to grasp, so go with what makes sense to you. i should move this thread to another forum... but which?

2. Originally Posted by jarman007
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?
This, in fact, is true of a polynomial of arbitrary degree; it was proven first by Gauss, in this much more general form :

Let $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ be a monic polynomial with integer coefficients.
Suppose $f(x)=p(x)q(x)$ with $p(x),q(x) \in \mathbb{Q}[x]$. Then in fact $p(x),q(x) \in \mathbb{Z}[x]$.

The proof is quite simple using algebra. Suppose $p(x),q(x) \notin \mathbb{Z}[x]$; then there is a least positive integer $m$ and a least positive integer $n$ such that $mp(x)$ and $np(x)$ are in $\mathbb{Z}[x]$. It's easy to see that there is no integer $>1$ which divides all the coefficients of $mp(x)$ (or else $m$ could be made smaller), and similarily for $np(x)$.

Take any prime $p$ which divides the product $mn$, and let $h(x) \mapsto \overline{h(x)}$ denote the canonical projection $\mathbb{Z}[x] \longrightarrow (\mathbb{Z}/p\mathbb{Z})[x]$. Then $0 = \overline{mnf(x)} = \overline{mnp(x)q(x)} = \overline{mp(x)}\ \overline{nq(x)}$. Since $(\mathbb{Z}/p\mathbb{Z})[x]$ is an integral domain, we must have $\overline{mp(x)} = 0$ or $\overline{nq(x)} = 0$. But $\overline{mp(x)} \neq 0$ and $\overline{nq(x)} \neq 0$ or else all the coefficients of $mp(x)$ or all the coefficients of $nq(x)$ would be divisible by $p$, which contradicts the choice of $m,n$, as described above. $\square$

3. Originally Posted by Jhevon
oh no, who told you this was precalc? and i very much doubt that this kind of math is routinely done in U.S. high schools, you might encounter them if you are in some sort of advanced placement or math competition. you may note though, that my and HallsOfIvy's solutions do not involve the | symbol and may be easier for a high school student to grasp, so go with what makes sense to you. i should move this thread to another forum... but which?
actually it's only Tonio's explanation that eluded me. the other answers were pretty clear. I'm glad to learn something which is why I'm here

4. Originally Posted by tonio
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
Yeah, but the way he said it, it seemed obvious that it had to be equal to $\pm 1$, while all this reasoning is necessary !

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