1. ## transcendental vs. irrational.

an irrational number is a decimal fraction that doesn't repeat nor end.
pi falls into this category but i always heard that it is a transcendental number.

what is the difference between transcendental and irrational?

2. An irrational number may be the root of an algebraic equation. For example $x^2-2 =0$ has roots $\pm \sqrt{2}$ which are both irrational. A transcendental number can not be the root of an algebraic equation. Examples of transcendental numbers are e and $\pi$ although there are lots of others. Check out this wiki page for details.

3. Why is $x^2-\pi^2=0$ not algebraic?

4. Because in the definition of algebraic number only rational numbers are allowed as coefficients in the polynomial.

The rationals are the natural field that you get from the integers.
The algebraic numbers (including complex numbers) are what you get when you don't allow numbers to escape your grasp just because the root of a polynomial might not be a known number. So all the roots of $x^5 + i x^4 + \sqrt{2} x^3 - 3 x^2 + 4 x + \frac{1+\sqrt{5}}{2} = 0$ are algebraic numbers.

But $\pi$ is not a number you can get to by a finite number of additions, subtractions, multiplications, divisions and taking the roots of polynomials.

5. The coefficients have to be rational numbers.

EDIT: rpenner beat me to it.

6. Originally Posted by rpenner
Because in the definition of algebraic number only rational numbers are allowed as coefficients in the polynomial.

The rationals are the natural field that you get from the integers.
The algebraic numbers (including complex numbers) are what you get when you don't allow numbers to escape your grasp just because the root of a polynomial might not be a known number. So all the roots of $x^5 + i x^4 + \sqrt{2} x^3 - 3 x^2 + 4 x + \frac{1+\sqrt{5}}{2}$ are algebraic numbers.

But $\pi$ is not a number you can get to by a finite number of additions, subtractions, multiplications, divisions and taking the roots of polynomials.
100% correct.

7. Originally Posted by camilus
100% correct.
You'll find that this is typically the case when rpenner comments

8. Originally Posted by prometheus
The coefficients have to be rational numbers.
I might go so far as to say they must all be integers, since rationals would only be required if you restricted yourself to a monomial (ie the coefficient of the highest power of x is 1), without this restriction you just multiply through by the lower common multiple of all the denominators of the rational coefficients.

This means you can then 'define' your rationals as zeros of linear polynomials (ie $ax+b=0$ where $a \not= 0$), rather than requiring an apriori definition of rationals to do with equivalence classes which escapes me right now.

9. Absolutely. And multiplying the LCM of the denominators of the coefficients leaves the roots of the polynomial unchanged even though the polynomial is now expressed with integer coefficients. But I like the general concept of root taking as something that we don't have to worry about the exact nature of the coefficients since the algebraic numbers are closed under that operation.

But if you are interested in these fine distinctions, and many people are, then you can rank your algebraic numbers by the smallest polynomial with integer coefficients which gives them as a root. Thus rational numbers are algebraic numbers of degree 1, quadratic surds $\frac{a + \sqrt{b}}{c}$ are algebraic numbers of degree 2, and so on.

http://mathworld.wolfram.com/AlgebraicNumber.html
http://mathworld.wolfram.com/Algebra...olynomial.html

Perhaps, some kind person could be bothered to demonstrate that the minimal polynomials of the above quintic are of degree greater than 5 since I did not limit myself to rationals.

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