Discrete Math: Proof of Rational Numbers

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rd_wingman

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Ok I have two questions from my discrete math class,

1.) Suppose a, b, and c are integers and x, y and z are nonzero real numbers that satisfy the following equations:
(xy)/(x+y) = a, (xz)/(x+z)=b, (yz)/y+z)-c
Is x rational? If, so express it as a ratio of two numbers

2.) Prove that if one solution for a quadratic equation of the form x^2+bx+c =0 is rational (where b and c are rational) then the other solution is also rational. (Use the fact that if the solutions of the equations are r and s, then x^2+bx+c =(x-r)(x-s).)

any help would be appreciated. thanks.
 
The second one is easy:

Factor x^2+bx+c as (x-r)(x-s) where r is rational such that x-r=0.

Then x^2+bx+c = x^2-(r+s)x+rs.

Thus -(r+s)=b, and s=-b-r. The sum/difference of two rational numbers is rational.
 
For (2)

Assume r is the one known to be rational:

(x - r)(x - s) = x^2 -(r+s)x + rs

Then c = rs and c is rational. Will that do it for you?
 
rd_wingman said:
Thanks, they both will work. I found the answer to the 1st one in my notes.
Click to expand...

Great! Send it along. I'd like to see it.
 
Suppose a, b, and c are integers and x, y and z are nonzero real numbers that satisfy the following equations:
(xy)/(x+y) = a, (xz)/(x+z)=b, (yz)/y+z)-c

Answer from class:
1/x + 1/y = 1/a, 1/x + 1/z = 1/b, and 1/y + 1/z = 1/c

1/y - 1/z = 1/a - 1/b

z/y = 1/a - 1/b + 1/c 1/y = 1/2[1/a - 1/b + 1/c ]

1/x + 1/2[1/a + 1/b + 1/c ] = 1/a

1/x = 1/a - 1/(2a) + 1/(2b) - 1(2c) (all are rational numbers)

1/x is rational therefor x is rational
 
rd_wingman said:
Suppose a, b, and c are integers and x, y and z are nonzero real numbers that satisfy the following equations:
(xy)/(x+y) = a, (xz)/(x+z)=b, (yz)/y+z)-c

Answer from class:
1/x + 1/y = 1/a, 1/x + 1/z = 1/b, and 1/y + 1/z = 1/c

1/y - 1/z = 1/a - 1/b

z/y = 1/a - 1/b + 1/c 1/y = 1/2[1/a - 1/b + 1/c ]

1/x + 1/2[1/a + 1/b + 1/c ] = 1/a

1/x = 1/a - 1/(2a) + 1/(2b) - 1(2c) (all are rational numbers)

1/x is rational therefor x is rational
Click to expand...

Thanks. I like it. I think maybe you got one or two things mixed up from the notes, but I see the clue now. The proof goes like this:

Suppose a, b, and c are integers and x, y and z are nonzero real numbers that satisfy the following equations:
(xy)/(x+y) = a
(xz)/(x+z) = b
(yz)/(y+z) = c

Invert the first equation and get:

(x+y)/xy = 1/a
x/xy + y/xy = 1/a
1/y + 1/x = 1/a

Likewise the second and third:

1/x + 1/y = 1/a, (I) << repeated
1/x + 1/z = 1/b, (II)
1/y + 1/z = 1/c (III)

Subtract to eliminate 1/z:

1/x - 1/y = 1/b - 1/c << II - III
1/x + 1/y = 1/a, << I, repeated
---------------------- << add them.
2/x = 1/a + 1/b - 1/c << we're practically there.

1/x = 1/2a + 1/2b - 1/2c

x = 1/(1/2a + 1/2b - 1/2c) and is rational.
 
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