# Transcendental numbers

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#1
Prove that e^3 is transcendental. You may assume that e is transcendental

I think this easier than I'm making it, can anybody help me?
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6 years ago
#2
(Original post by 010c)
Prove that e^3 is transcendental. You may assume that e is transcendental

I think this easier than I'm making it, can anybody help me?
Suppose e were not transcendental. What follows?
ETA: typo, should be "Suppose e^3…"
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6 years ago
#3
(Original post by Smaug123)
Suppose e were not transcendental. What follows?
I'm not sure this is helpful.

On the other hand:

"Suppose e^3 is not transcendental. Then we can find ..."
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6 years ago
#4
(Original post by DFranklin)
I'm not sure this is helpful.

On the other hand:

"Suppose e^3 is not transcendental. Then we can find ..."
Sorry - typo. (Actually was a typo - I even wrote out a proof first.)
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6 years ago
#5
If isn't transcendental then it solves equal to zero some finite polynomial.

But then what does this imply about , which is already known to be transcendental?
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#6
(Original post by Zii)
If isn't transcendental then it solves equal to zero some finite polynomial.

But then what does this imply about , which is already known to be transcendental?

Does it imply that e is rational which is a contradiction?
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6 years ago
#7
(Original post by 010c)
Does it imply that e is rational which is a contradiction?
No.

But rational is not the opposite of transcendental...
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6 years ago
#8
(Original post by Zii)
If isn't transcendental then it solves equal to zero some finite polynomial.

But then what does this imply about , which is already known to be transcendental?
You need to make some statement about the coefficients of that polynomial, otherwise no numbers would be transcendental!

(Original post by 010c)
Does it imply that e is rational which is a contradiction?
See DFranklin's response - the opposite of transcendental is algebraic.
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6 years ago
#9
Lindemann-Weierstrass theorem?

(may be "overthinking")
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6 years ago
#10
(Original post by Hasufel)
Lindemann-Weierstrass theorem?

(may be "overthinking")
Unless I'm being totally stupid, this is massive overkill. It's literally a 1 - 2 line proof from the definitions.
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6 years ago
#11
(Original post by DFranklin)
Unless I'm being totally stupid, this is massive overkill. It's literally a 1 - 2 line proof from the definitions.
Yup! - thought so, thanks!

(in the zone on something else and just glanced)

will +rep when can
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6 years ago
#12
(Original post by davros)
You need to make some statement about the coefficients of that polynomial, otherwise no numbers would be transcendental!
Yes sorry I missed out the fact the polynomial is over but honestly I took that for granted given the definition of transcendental numbers.
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6 years ago
#13
(Original post by Zii)
Yes sorry I missed out the fact the polynomial is over but honestly I took that for granted given the definition of transcendental numbers.
Again, the issue is the coefficients of the polynomial.

For example, is a root of the polynomial , so if you allow the coefficients to be arbitrary real numbers, it would turn out that there were NO real transcendental numbers.
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6 years ago
#14
(Original post by DFranklin)
Again, the issue is the coefficients of the polynomial.

For example, is a root of the polynomial , so if you allow the coefficients to be arbitrary real numbers, it would turn out that there were NO real transcendental numbers.
I agree, when I say polynomial over , I am talking in the strictly algebraic sense, that is the coefficients are in .

Sorry for not making that clear
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6 years ago
#15
I have absolutely no worthwhile contribution to make to the OP here but coincidentally I literally just watched a video by Numberphile on transcendental numbers. Here is a link if anyone is interested:

Main Video

Extra Footage
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6 years ago
#16
(Original post by Zii)
I agree, when I say polynomial over , I am talking in the strictly algebraic sense, that is the coefficients are in .

Sorry for not making that clear
That's clear, but we are very doubtful that it's actually the right thing.

The point being that if you allow arbitrary real coefficients, then every number is algebraic, so no number is transcendental.
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6 years ago
#17
(Original post by Zii)
I agree, when I say polynomial over , I am talking in the strictly algebraic sense, that is the coefficients are in .

Sorry for not making that clear
The coefficiants need to be in the rationals (or equivalently the integers), not just any real number.
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