Prove the following is irrational

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In summary, the conversation discusses a method for proving that a given number is irrational. The method involves finding the polynomial with integer coefficients satisfied by the number and then checking for rational roots using the factors of the leading coefficient and the constant term. If none of the possible rational roots work, then the number is proven to be irrational. The conversation also mentions considering the factors of the leading coefficient and constant term when checking for rational roots.
  • #1
Pinkk
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Prove that http://img705.imageshack.us/img705/2408/aaa12.png is irrational. A user on another forum suggested the following:

http://img130.imageshack.us/img130/1352/abcde.png

I follow that up to the last sentence. Can anyone clarify for me how to show this proof?
 
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  • #2
if x^6 is not element Q x is not element Q too, you can see it...So it is a contradiction and x is irrational.
 
  • #3
Yes, but how do I prove that x^6 is irrational. I cannot simply say so since it is possible for the sum of two irrationals to be rational.
 
  • #4
But the sum isn't.
 
  • #5
The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.

More detail. Your number satisfies
[tex]
{X}^{12}-{X}^{11}-{X}^{10}+{X}^{9}-3\,{X}^{7}+2\,{X}^{6}+2\,{X}^{5}+{X
}^{4}-{X}^{3}+{X}^{2}+2\,X-5 = 0
[/tex]
To find rational roots, consider the factors of [itex]-5[/itex] (and the factors of the leading coefficient, 1) so you just have to try [itex]5, -5, 1, -1[/itex]. Since none of these is a solution of that polynomial, all solutions are irrational.
 
  • #6
g_edgar said:
The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.

More detail. Your number satisfies
[tex]
{X}^{12}-{X}^{11}-{X}^{10}+{X}^{9}-3\,{X}^{7}+2\,{X}^{6}+2\,{X}^{5}+{X
}^{4}-{X}^{3}+{X}^{2}+2\,X-5 = 0
[/tex]
To find rational roots, consider the factors of [itex]-5[/itex] (and the factors of the leading coefficient, 1) so you just have to try [itex]5, -5, 1, -1[/itex]. Since none of these is a solution of that polynomial, all solutions are irrational.

If I may ask a stupid question, what rational roots would you check for to see if

15X^{12} + ... + 14 had rational roots.
 
  • #7
g_edgar said:
The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.

Dang that's so frustrating. I saw this yesterday and was thinking of the rational root theorem, but was too tired to work it out :)
 
  • #8
ramsey2879 said:
If I may ask a stupid question, what rational roots would you check for to see if

15X^{12} + ... + 14 had rational roots.

The factors of 14 are 1,2,7,14 and the factors of 15 are 1,3,5,15, so you have to try just 32 possibilities: plus or minus a fraction, the numerator is one of 1,2,4,7 and the denominator is one of 1,3,5,15.
 

Related to Prove the following is irrational

What does it mean to prove that a number is irrational?

Proving that a number is irrational means showing that it cannot be written as a ratio of two integers, or whole numbers. In other words, it is a number that cannot be expressed as a fraction.

Why do we need to prove that a number is irrational?

In mathematics, proving a number is irrational is important because it helps us understand the nature of numbers and their relationships. It also helps us to differentiate between rational and irrational numbers and to identify patterns and properties of these numbers.

What are some common examples of irrational numbers?

Some common examples of irrational numbers include pi (3.1415926...), the square root of 2 (1.4142135...), and the golden ratio (1.6180339...). These numbers are all non-repeating and non-terminating decimals, meaning they go on forever without repeating a pattern.

How do you prove that a number is irrational?

To prove that a number is irrational, you can use the proof by contradiction method. This involves assuming that the number is rational and then showing that this assumption leads to a contradiction or impossibility. Therefore, the number must be irrational.

Are all irrational numbers also transcendental?

No, not all irrational numbers are transcendental. While all transcendental numbers are irrational, not all irrational numbers are transcendental. For example, pi is both irrational and transcendental, but the square root of 2 is irrational but not transcendental.

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