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Transcendental numbers

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?

Reply 1

Smaug123

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…"

(edited 9 years ago)

Reply 2

DFranklin

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 ..."

Reply 3

Smaug123

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.)

Reply 4

Zii

e^3

isn't transcendental then it solves equal to zero some finite polynomial.

But then what does this imply about

e

, which is already known to be transcendental?

Reply 5

010c

Original post by Zii

e^3

isn't transcendental then it solves equal to zero some finite polynomial.

But then what does this imply about

e

, which is already known to be transcendental?

Does it imply that e is rational which is a contradiction?

Reply 6

DFranklin

Original post by 010c

Does it imply that e is rational which is a contradiction?

No.

But rational is not the opposite of transcendental...

Reply 7

davros

Original post by Zii

e^3

isn't transcendental then it solves equal to zero some finite polynomial.

But then what does this imply about

e

, 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.

Reply 8

Hasufel

Lindemann-Weierstrass theorem?

(may be "overthinking")

Reply 9

DFranklin

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.

Reply 10

Hasufel

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

Reply 11

Zii

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

\mathbb{R}

but honestly I took that for granted given the definition of transcendental numbers.

Reply 12

DFranklin

Original post by Zii

Yes sorry I missed out the fact the polynomial is over

\mathbb{R}

but honestly I took that for granted given the definition of transcendental numbers.

Again, the issue is the coefficients of the polynomial.

For example,

\pi

is a root of the polynomial

p(x) = x - \pi

, so if you allow the coefficients to be arbitrary real numbers, it would turn out that there were NO real transcendental numbers.

Reply 13

Zii

Original post by DFranklin

Again, the issue is the coefficients of the polynomial.

For example,

\pi

is a root of the polynomial

p(x) = x - \pi

, 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

\mathbb{R}

, I am talking in the strictly algebraic sense, that is the coefficients are in

\mathbb{R}

.

Sorry for not making that clear

Reply 14

Gilo98

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

Reply 15

DFranklin

Original post by Zii

I agree, when I say polynomial over

\mathbb{R}

, I am talking in the strictly algebraic sense, that is the coefficients are in

\mathbb{R}

.

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.