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Algebraic and Transcendental numbers proof

  • Thread starter SMA_01
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  • #1
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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.
 

Answers and Replies

  • #2
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By contradiction. Assume the sum is algebraic.
 
  • #3
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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
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You have polynomial R(x) with integer coefficients whose root is α+β. Now consider R(x - β).
 
  • #5
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Would x represent any real number? I'm sorry, it's not really clicking with me...
 
  • #6
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Are you implying that I should take x=α+β? So that R(x-β) would be R(α+β-β)=R(α)?
 
  • #7
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x is the argument of the polynomial R(x - β). What do its coefficients look like?
 
  • #8
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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
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β is algebraic according to your original post.
 
  • #10
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Oh yeah, sorry I made a mistake. It is algebraic, but I am still trying to figure out what this implies.
 
  • #11
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Are the coefficients of R(x - β) integer, rational, algebraic, transcendental? What does that mean with regard to its roots?
 
  • #12
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The coefficients are algebraic, and the roots must also be algebraic.
 
  • #13
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To prove the statement, you need to consider R(x + β) in the same way. Can you now connect all the pieces?
 
  • #14
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So, it R(x-β) and R(x+β) have algebraic coefficients then R(x) mus also have algebraic coefficients right?
 
  • #15
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R(x) is defined in #4. Its definition follows from the assumption that α+β is algebraic.
 
  • #16
HallsofIvy
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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 [itex]\alpha+ \beta[/itex] was algebraic so that it was a root of a polynomial equation [itex]R(x)= 0[/itex] where R is a polynomial with integer coefficients. There is no point is now saying that they must be algebraic!
 
  • #17
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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
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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
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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
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## P(x) = R(x + \beta) ##. What coefficients does P have? What are its roots (algebraic, transcendental)? Is ## \alpha ## its root? Why?
 
  • #21
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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...
 
  • #22
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What does this contradiction mean?
 
  • #23
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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
 
  • #24
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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.
 

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