# Algebraic and Transcendental numbers proof

## Homework Statement

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

## 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.

By contradiction. Assume the sum is algebraic.

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?

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

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

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

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

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

β is algebraic according to your original post.

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

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

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

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

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

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

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

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?

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?

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.

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

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

The coefficients are algebraic, and the roots are supposed to be algebraic. α is a root, but it's not algebraic...

Since α is transcendental, then by definition, it cannot be a root. And then α+β can't be a root either. So it's a contradiction. Is that correct?

As a side question, α is not a root of R(x-β). Why did I need to mention this polynomial then? Is it insufficient to just use R(x+β)?

Thank you

Yes, that is correct.

I mentioned initially R(x-β) because I hoped that would give you an idea and you would then come up with R(x+β), so you would have more done on your own. R(x-β) is not required for the proof.