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?
A slightly different proof. Let and be the rational roots (m, n, p, and q integer, of course), reduced to lowest terms. Then we must have . Multiplying that out, .
Since b and c are integers, we must have nq= 1. Since n and q are integers, we must have and . Then the roots and are integer.
It suffices to show that is even. We consider two cases: (1) is even and (2) is odd. (Notice that the parity of is immaterial, since would be even whether is even or odd).
case 1: is even.
Then, is even, so that is even (since it is the difference of two even integers), which implies is even. But then, is even, since it is the sum of two even integers.
case 2 is no worse and will be left to the OP.
that's the problem. i'm trying to understand his logic from going from to . 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.
showing that was important to tonio's argument. he could show that by showing that
by dividing through by , we have
since the right side is an integer (because ), the left side must be also, because we have equality here. but is an integer if and only if
they are actually quite different symbols. "|" means "divides" while "/" means "divided by".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.
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.