# Irrational Equations

• Oct 4th 2008, 04:18 PM
BlakeRobertsonMD
Irrational Equations
Hi

I'm supposed to prove that all the solutions to X^2 = X + 1 are irrational.

I'm not sure exactly where to begin. I know by contradiction I would make it rational and replace x with A/B. Then work the problem through. After I worked it through I ended up with

0 = b/a + (b^2)/(a^2) and I'm not even sure if i'm on the right track even.

Do you any of you know?

Thanks for the help.
• Oct 4th 2008, 04:41 PM
Jhevon
Quote:

Originally Posted by BlakeRobertsonMD
Hi

I'm supposed to prove that all the solutions to X^2 = X + 1 are irrational.

I'm not sure exactly where to begin. I know by contradiction I would make it rational and replace x with A/B. Then work the problem through. After I worked it through I ended up with

0 = b/a + (b^2)/(a^2) and I'm not even sure if i'm on the right track even.

Do you any of you know?

Thanks for the help.

use the rational roots theorem. you know it, right?
• Oct 4th 2008, 04:44 PM
BlakeRobertsonMD
Nope. *goes to look it up* Can you explain a little bit?
• Oct 4th 2008, 05:13 PM
BlakeRobertsonMD
Well heres what I've done. Hoepfully it sounds about right. I actually get this roots test better then some of the area material. If this sounds wrong please tell me.

Proof that all the solutions to the equation are irrational. (hint – rational/irrational numbers)
Theorem: All solutions to the equation are irrational.
Proof: To solve this equation, let the equation = 0, so that the new equation is –X^2 + x + 1.

Then use the Rational Roots test to find any of the rational numbers that will make the equation = 0.

By using the Rational Roots test we find the factors of the first and last part of the equation. By doing this we get -1, 1.

By using the two rational numbers found in the equation we find that none equal to 0. Meaning that there are no rational solutions to the equation, thus there is an irrational solution to the equation and the conclusion is true.
• Oct 4th 2008, 05:17 PM
Jhevon
Quote:

Originally Posted by BlakeRobertsonMD
Well heres what I've done. Hoepfully it sounds about right. I actually get this roots test better then some of the area material. If this sounds wrong please tell me.

Proof that all the solutions to the equation are irrational. (hint – rational/irrational numbers)
Theorem: All solutions to the equation are irrational.
Proof: To solve this equation, let the equation = 0, so that the new equation is –X^2 + x + 1.

Then use the Rational Roots test to find any of the rational numbers that will make the equation = 0.

By using the Rational Roots test we find the factors of the first and last part of the equation. By doing this we get -1, 1.

By using the two rational numbers found in the equation we find that none equal to 0. Meaning that there are no rational solutions to the equation, thus there is an irrational solution to the equation and the conclusion is true.

it seems you have the basic idea. you just need to say it better. try to sound like a mathematician. but yes, that's the gist of it. by the rational roots theorem, the only possible rational solutions to the equation and +1 and -1. since neither of these are solutions, there are no rational solutions. thus, any real solution has to be irrational. now all you have to do, is prove that a real solution exists. you can do this using the intermediate value theorem, or by Descartes rule of signs (look those up now :))
• Oct 5th 2008, 03:41 AM
Plato
The equation $\displaystyle x^2 - x -1=0$ is equivalent.
The discriminate $\displaystyle \sqrt {b^2 - 4ac} = \sqrt 5$ is irrational.
The square root of any non-square natural number is irrational.
Therefore, the roots are irrational.