Trancendental number

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Homework Statement



Prove that [tex]e^{\frac{n}{m}}[/tex] is trancendental, where m>0 and n are integers.

Homework Equations



e is trancendental

The Attempt at a Solution



A hint I got said this:
"Let ln(m\n)=m\n"
so I did this:
then[tex] \frac{m}{n} = e^{\frac{m}{n}}[/tex]
so then [tex](\frac{m}{n})^n=(e^{\frac{m}{n}})^n[/tex]
and then [tex](\frac{m}{n})^n-e^m=0[/tex]
this contradicts the trancendence of e, therefore e^n\m is trancendental.

I'm confused about the hint, and I'm not convinced that this is actually valid. This hint came from one of my classmates and aside from pulling out if the thin air, I don't see how he arrived at this.
Any input will be appreciated.
CC
 
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Answers and Replies

  • #2
StatusX
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ln(m/n)=m/n is never true. You probably mean ln(m/n)=p/q, ie, assume ln(m/n) is rational and reach a contradiction.
 
  • #3
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Hi,
So if I change the m\n to p\q and say ln(m\n) is rational, does the argument above hold, or do I need to completely start over?
 
  • #4
StatusX
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Why don't you try it?
 
  • #5
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Let ln(m\n)=p\q, where m>0,n,p and q are integers. So then ln(m\n) is rational.

then[tex] \frac{m}{n} = e^{\frac{p}{q}}[/tex]

so then [tex](\frac{m}{n})^q=(e^{\frac{p}{q}})^q[/tex]

and then [tex](\frac{m}{n})^q-e^p=0[/tex]

this contradicts the trancendence of e, so e^(n\m) must be trancendental.
Am I closer?
CC
 
  • #6
StatusX
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You've reached a contradiction, so you have to reject you assumption, which was that ln(m/n) was rational. This is different from e^n/m. But it shouldn't be hard to modify your proof to get what you want (hint: switch m/n and p/q)

EDIT: Sorry, I just realized I suggested doing it the wrong way around. Anyway, like I said, it shouldn't be hard to fix.
 
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  • #7
Dick
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Never mind. I see.
 
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  • #8
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I thought that if I got a polynomial with e as a root that I would get a contradiction. That was my idea. I don't know if the little invented equation up there is of any use. This is one of those problems that I have stared at so long I can't think of any other method. Please help me go the right way of you can.
Thanks,
CC
 
  • #9
StatusX
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You're right, I didn't notice that.

Let's start over. Note that if a satisfies a polynomial f(x), then any nth root of a satisfies the polynomial f(x^n). Thus if a number is algebraic, so are all its nth roots, and so, rearranging things a little, if a number is transcendental, so are all its nth powers (do you see how to get this?). If you can similarly show that the mth power of an algebraic number is algebraic, you'll be done.
 
  • #10
Dick
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Let ln(m\n)=p\q, where m>0,n,p and q are integers. So then ln(m\n) is rational.

then[tex] \frac{m}{n} = e^{\frac{p}{q}}[/tex]

so then [tex](\frac{m}{n})^q=(e^{\frac{p}{q}})^q[/tex]

and then [tex](\frac{m}{n})^q-e^p=0[/tex]

this contradicts the trancendence of e, so e^(n\m) must be trancendental.
Am I closer?
CC

This is a proof that ln(m/n) must be irrational, not that e^(n/m) is transcendental. I think your hint may apply to some other problem. Follow StatusX's suggestion!
 
  • #11
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Ok,
Assume that [tex]e^{\frac{n}{m}}[/tex] is algebraic.
Then it satisfies a polynomial of the form
[tex]x^m-e^n=0[/tex] for every value of n and m.
Now let m=n=1

then [tex]x-e=0[/tex]

contradiction
Is that valid?
CC
 
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  • #12
Dick
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No it's not valid. In fact, the more I read it the less I understand it, sorry. StatusX suggests that you prove the result by first proving that if c^(n/m) is algebraic then c is algebraic. That puts you one step away from a proof by contradiction.
 
  • #13
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I'm not sure how to prove that if c^(n\m) is algrbraic then c is algebraic.

I was considering something like 2^(5/6). It satisfies x^6-2^5=0, and any other algebraic number of that form c^(n\m) satisfies a polynomial of that looks like x^m-c^n=0.

Is it an inductive type proof that if c^(n\m) is algebraic then c is algebraic? I'm not really sure where to start. I can see why that's true, but I don't know where to start with a proof. Any input will be appreciated.
CC
 
  • #14
Dick
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Ok. The easy case is c^n algebraic implies c is algebraic. Can you do that one? The 'harder' case is c^(1/m) algebraic implies c is algebraic. Don't try to construct an explicit polynomial in this case. Have you proved that algebraic numbers are closed under multiplication?
 
  • #15
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Yes, we have that proof in the book.Ok. I'm going to work on the c^n thing and try out the c^1\m. Then I just invoke algebraic times algebraic is algebraic?
 
  • #16
Dick
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If you already know closure, then the c^(1/m) is the easy case. c=c^(1/m)*c^(1/m)*...*c^(1/m) (m times). So if c^(1/m) is algebraic, c is ???. Now try the other case. Then put them together to prove c^(n/m) algebraic implies c algebraic.
 
  • #17
dextercioby
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One can give a direct proof

Hypothesis

e is transcendental, i.e.

[tex] a_{1} e^{k} +a_{2}e^{k-1}+...+a_{k}e+a_{k+1}=0 \rightarrow \left\{a_{i}\right\}_{i=1}^{i=k+1}=0 \ , \ \forall a_{i}\in\mathbb{Z} \ , \forall k\in\mathbb{N}} [/tex]

Conclusion

[tex] a_{1} e^{\left(\frac{m}{n}\right)k} +a_{2}e^{\left(\frac{m}{n}\right)(k-1)}+...+a_{k}e^{\frac{m}{n}}+a_{k+1}=0 \rightarrow \left\{a_{i}\right\}_{i=1}^{i=k+1}=0 \ , \ \forall a_{i}\in\mathbb{Z} \ , \forall k\in\mathbb{N} [/tex]

Proof: Pick k=n p, p arbitrary in [itex] \mathbb{N} [/itex] Rename mp=k' still in [itex] \mathbb{N} [/itex] Endproof.
 
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  • #18
Dick
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Sorry. I don't really buy the proof of your conclusion. To show e^(n/m) transcendental, you have to show the implication holds for ALL k, not selected ones.
 
  • #19
dextercioby
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"k" is arbitrary, it's n/m times an arbitrary natural number. When k' generates N, k generates N, since m,n are natural numbers (coprime if you prefer).
 
  • #20
Dick
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Then what is the k=n*p restriction? Furthermore, this still doesn't reduce all of the exponents to integers. Isn't that what you are trying to do?
 
  • #21
dextercioby
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It needn't reduce all exponents, you might choose a_{2}, a_{3},...,a_{k} =0 from the very beginning. I've written the most general polynomial of "k-th order"(apparently superfluously) but it's enough to prove for the a_{1}x^{k}+a_{k+1} polynomial...
 
  • #22
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Can I just say that if a^n is algebraic over a finite field F then a is algebraic because [tex] a\in F[/tex] and every element of a finite field is algebraic?
 
  • #23
Dick
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You can't restrict the choice of polynomials "from the beginning". It has to hold for ALL polynomials.
 
  • #24
dextercioby
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Yes, but what good would a finite field do to you ?
 
  • #25
Dick
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Can I just say that if a^n is algebraic over a finite field F then a is algebraic because [tex] a\in F[/tex] and every element of a finite field is algebraic?

It's true, but it's a bit circular. If a^n solves a polynomial, then 'a' solves a VERY similar polynomial.
 
  • #26
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If that's valid, then I can say that c^n being algebraic implies that c is algebraic and as Dick pointed out, c^1\m algebriac implies c algebraic, then
c^n*c^(1\m)=c^(n\m) is algebraic. So then I can say that e^(n\m) can't be algebraic because that implies that e is algebraic, which is a LIE!
 
  • #27
Dick
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Can I just say that if a^n is algebraic over a finite field F then a is algebraic because [tex] a\in F[/tex] and every element of a finite field is algebraic?

Actually, it's a bit not true as well depending on what finite dimensional field extension F over Q (NOT finite field) you are talking about. F might be defined to contain a^n, but not a. Stick to the direct path.
 
  • #28
Dick
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If that's valid, then I can say that c^n being algebraic implies that c is algebraic and as Dick pointed out, c^1\m algebriac implies c algebraic, then
c^n*c^(1\m)=c^(n\m) is algebraic. So then I can say that e^(n\m) can't be algebraic because that implies that e is algebraic, which is a LIE!

Not quite. Your exponent arithmetic is off. You want c^(n/m)=(c^n)^(1/m) algebraic implies c^n is algebraic (1/m rule) implies c is algebraic (n rule). The LIE part is now correct.
 
  • #29
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But every finite extension of a field is algebraic.
 
  • #30
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I'm not sure now if that applies. I can see if a is in F then a finit extension would contain a^n, but maybe not the other way.
I see my exponent error.
 
  • #31
Dick
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I'm not sure now if that applies. I can see if a is in F then a finit extension would contain a^n, but maybe not the other way.
I see my exponent error.

Yes, it doesn't apply. If I extend Q by sqrt(2), I have an extension field of dimension 2. But it does not include 2^(1/4).
 
  • #32
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Ok, so I can go from a being algebraic to a^n being algebraic. I get that, but does it HAVE to go the other way? I've confused myself here.
I'm taking a break and getting a Dr. Pepper.
CC
 
  • #33
Dick
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Ok, so I can go from a being algebraic to a^n being algebraic. I get that, but does it HAVE to go the other way? I've confused myself here.
I'm taking a break and getting a Dr. Pepper.
CC

Yeah, take a break. And forget about finite field extensions. Just consider that a^n algebraic means there is a polynomial satisfied by a^n e.g.:
a_0+a_1*a^n+a_2*(a^n)^2+...a_m*(a^n)^m=0

Can you think of a polynomial satisfied by 'a'?
 
  • #34
HallsofIvy
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Do you mean "if an is algebraic, then a is algegraic"? Certainly that's true. If an is algebraic, of order m, then there exist a polynomial equation of degree m, with integer coefficients, such that
[tex]k_m(a^n)^m+ k_{m-1}(a^n)^{m-1}+ ...+ k_0= 0[/tex]
but then
[tex]k_m a^{n+m}+ k_{m-1}a^{n+m-1}+ ...+ k_0= 0[/tex]
a is algebraic of order (less than or equal to) n+m.
 
  • #35
Dick
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Do you mean "if an is algebraic, then a is algegraic"? Certainly that's true. If an is algebraic, of order m, then there exist a polynomial equation of degree m, with integer coefficients, such that
[tex]k_m(a^n)^m+ k_{m-1}(a^n)^{m-1}+ ...+ k_0= 0[/tex]
but then
[tex]k_m a^{n+m}+ k_{m-1}a^{n+m-1}+ ...+ k_0= 0[/tex]
a is algebraic of order (less than or equal to) n+m.

You gave it away! But your exponent arithmetic could use some work too. :wink:
 

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