Algebraic and Transcendental numbers proof

1. Sep 27, 2012

SMA_01

1. The problem statement, all variables and given/known data

Let α be a transcendental number and β an algebraic number. Prove that α+β is transcendental.

3. The attempt at a solution

It's known that adding two algebraic numbers results in an algebraic number.

Since β is algebraic, it is a root of a polynomial with integer coefficients.

That is p(x)=Ʃbix^i i=0 to n, p(β)=0.

So,

p(x)= b+b1x1+b2x2+b3x3+...+b(n-1)xn-1+bnxn

I'm stuck, I'm not sure where to go from here...

Any help is appreciated.

Thanks.

2. Sep 27, 2012

voko

By contradiction. Assume the sum is algebraic.

3. Sep 27, 2012

SMA_01

Okay, assuming the sum is algebraic, that means the root of p(x) added to α will give me a number that can be expressed as a root of another polynomial. How would I proceed?

4. Sep 27, 2012

voko

You have polynomial R(x) with integer coefficients whose root is α+β. Now consider R(x - β).

5. Sep 27, 2012

SMA_01

Would x represent any real number? I'm sorry, it's not really clicking with me...

6. Sep 27, 2012

SMA_01

Are you implying that I should take x=α+β? So that R(x-β) would be R(α+β-β)=R(α)?

7. Sep 27, 2012

voko

x is the argument of the polynomial R(x - β). What do its coefficients look like?

8. Sep 27, 2012

SMA_01

R(x-β)=b0+b1(x-β)+b2(x-β)2+....+bn(x-β)n

But since β is a transcendental number, we can further distribute the bn coefficients right? So for b1(x-β)=b1x-b1β

9. Sep 27, 2012

voko

β is algebraic according to your original post.

10. Sep 27, 2012

SMA_01

Oh yeah, sorry I made a mistake. It is algebraic, but I am still trying to figure out what this implies.

11. Sep 27, 2012

voko

Are the coefficients of R(x - β) integer, rational, algebraic, transcendental? What does that mean with regard to its roots?

12. Sep 27, 2012

SMA_01

The coefficients are algebraic, and the roots must also be algebraic.

13. Sep 27, 2012

voko

To prove the statement, you need to consider R(x + β) in the same way. Can you now connect all the pieces?

14. Sep 27, 2012

SMA_01

So, it R(x-β) and R(x+β) have algebraic coefficients then R(x) mus also have algebraic coefficients right?

15. Sep 27, 2012

voko

R(x) is defined in #4. Its definition follows from the assumption that α+β is algebraic.

16. Sep 27, 2012

HallsofIvy

Staff Emeritus
You started by assuming that $\alpha+ \beta$ was algebraic so that it was a root of a polynomial equation $R(x)= 0$ where R is a polynomial with integer coefficients. There is no point is now saying that they must be algebraic!

17. Sep 27, 2012

SMA_01

I know that, I think I worded what I was trying to say wrong. I meant that algebraic coefficients yield algebraic roots. So R(x) would be a contradiction to this, or am I completely off?

18. Sep 27, 2012

voko

R(x) does not contradict anything. You assume that $\alpha + \beta$ is algebraic, so there must be a polynomial with integer coefficients it is a root of. This is R(x). Then you consider another polynomial, which is $R(x + \beta)$. $\alpha$ obviously is its root. What does that mean, if $\beta$ is algebraic?

19. Sep 27, 2012

SMA_01

Do you mean α is a root of R(x+β)? Sorry, I know the solution must be obvious and I'm totally missing it. I'm not really familiar with algebraic and transcendental numbers.

20. Sep 27, 2012

voko

$P(x) = R(x + \beta)$. What coefficients does P have? What are its roots (algebraic, transcendental)? Is $\alpha$ its root? Why?