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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- January 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? - January 12th 2010, 12:46 AMtonio
- January 12th 2010, 01:34 AMcrossbone
Tonio,

what does "|" symbol mean? - January 12th 2010, 01:46 AMtonio
- January 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.

- January 12th 2010, 03:45 AMtonio
- January 12th 2010, 03:55 AMHallsofIvy
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. - January 12th 2010, 04:03 AMMoo
- January 12th 2010, 06:46 AMtonio

Not if the fractions are reduced:

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

, and now add these two eq's:

.Since n divides the right hand n divides the left hand as well, contradicting the fact that is reduced...unless ,and the same for q, of course.

Tonio - January 12th 2010, 07:01 AMJhevon
Another alternative: We wish to show that the roots of are integers. By the quadratic formula, it suffices to show that the expression is always an integer provided it is rational (which means is a perfect square, call it ) and .

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. - January 12th 2010, 01:21 PMcrossbone
- January 12th 2010, 01:31 PMJhevon
- January 12th 2010, 02:16 PMcrossbone
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.

- January 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 was important to**tonio**'s argument. he could show that by showing that

we have

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

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. - January 12th 2010, 04:10 PMcrossbone