Exponential function

outermeasure · **Posted:** Sun, 27 May 2012 08:13:13 UTC

Shadow wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

Because that's equivalent to knowing if $\log_2(e)$ is rational...

Why is that equivalent???

BTW, $\log_2(e)$ is irrational because

is transcendental.

outermeasure · **Posted:** Sun, 27 May 2012 08:19:16 UTC

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

In general, deciding whether a specific number is irrational or transcendental or ... is a very difficult thing, and indeed for most numbers it is probably impossible (heuristically, proofs are finite strings from a countable collection of letters, so only countably many numbers can be proven to be irrational/transcendental/...).

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Shadow wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

Because that's equivalent to knowing if $\log_2(e)$ is rational...a very difficult thing to decide in general. Understanding which numbers we can easily express are or are not algebraic is a VERY difficult prospect in general.

I do not understand. e is irrational, so ${2^e}$ cannot be rational, because like you said ${2^e} = r$ if $e = {\log _2}(r)$ .

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

outermeasure wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

In general, deciding whether a specific number is irrational or transcendental or ... is a very difficult thing, and indeed for most numbers it is probably impossible (heuristically, proofs are finite strings from a countable collection of letters, so only countably many numbers can be proven to be irrational/transcendental/...).

So it all goes back to Godel's theorem?

outermeasure · **Posted:** Sun, 27 May 2012 09:11:29 UTC

glebovg wrote:

outermeasure wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

In general, deciding whether a specific number is irrational or transcendental or ... is a very difficult thing, and indeed for most numbers it is probably impossible (heuristically, proofs are finite strings from a countable collection of letters, so only countably many numbers can be proven to be irrational/transcendental/...).

So it all goes back to Godel's theorem?

No. It doesn't depend on Godel's theorem. Just basic counting.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

outermeasure wrote:

glebovg wrote:

outermeasure wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

In general, deciding whether a specific number is irrational or transcendental or ... is a very difficult thing, and indeed for most numbers it is probably impossible (heuristically, proofs are finite strings from a countable collection of letters, so only countably many numbers can be proven to be irrational/transcendental/...).

So it all goes back to Godel's theorem?

No. It doesn't depend on Godel's theorem. Just basic counting.

I presume when you say countable you mean the cardinality of the set of numbers which can be proven to be irrational or transcendental is ${\aleph _0}$ , but the Lindemann–Weierstrass theorem, for instance, states that ${e^\alpha }$ is transcendental if $\alpha$ is a non-zero algebraic number. This set forms an infinite subset of the reals, and therefore is not countable.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

outermeasure wrote:

Shadow wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

Because that's equivalent to knowing if $\log_2(e)$ is rational...

Why is that equivalent???

BTW, $\log_2(e)$ is irrational because

is transcendental.

If ${\log _2}(e) = \frac{a}{b}$ , then e would be a root of ${2^a} - {x^b}$ , but e is transcendental. That is exactly what I thought.

outermeasure · **Posted:** Sun, 27 May 2012 16:55:43 UTC

glebovg wrote:

outermeasure wrote:

glebovg wrote:

outermeasure wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

In general, deciding whether a specific number is irrational or transcendental or ... is a very difficult thing, and indeed for most numbers it is probably impossible (heuristically, proofs are finite strings from a countable collection of letters, so only countably many numbers can be proven to be irrational/transcendental/...).

So it all goes back to Godel's theorem?

No. It doesn't depend on Godel's theorem. Just basic counting.

I presume when you say countable you mean the cardinality of the set of numbers which can be proven to be irrational or transcendental is ${\aleph _0}$ , but the Lindemann–Weierstrass theorem, for instance, states that ${e^\alpha }$ is transcendental if $\alpha$ is a non-zero algebraic number. This set forms an infinite subset of the reals, and therefore is not countable.

Huh? $\mathbb{Q}$ is countable, so $\mathbb{Q}^{alg}$ is countable, so $\{e^\alpha\mid\alpha\in\mathbb{Q}^{alg},\alpha\neq 0\}$ is a countable subset of the set of transcendentals which (is computable and) can be proven to be transcendental.

Shadow · **Posted:** Sun, 27 May 2012 17:18:58 UTC

outermeasure wrote:

Shadow wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

Because that's equivalent to knowing if $\log_2(e)$ is rational...

Why is that equivalent???

BTW, $\log_2(e)$ is irrational because

is transcendental.

oops, Sorry, I kept unwrapping the same definition over and over in my memory and that time the wrong quantity came out. Thanks for the catch.

Shadow · **Posted:** Sun, 27 May 2012 17:20:01 UTC

glebovg wrote:

outermeasure wrote:

glebovg wrote:

outermeasure wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

In general, deciding whether a specific number is irrational or transcendental or ... is a very difficult thing, and indeed for most numbers it is probably impossible (heuristically, proofs are finite strings from a countable collection of letters, so only countably many numbers can be proven to be irrational/transcendental/...).

So it all goes back to Godel's theorem?

No. It doesn't depend on Godel's theorem. Just basic counting.

I presume when you say countable you mean the cardinality of the set of numbers which can be proven to be irrational or transcendental is ${\aleph _0}$ , but the Lindemann–Weierstrass theorem, for instance, states that ${e^\alpha }$ is transcendental if $\alpha$ is a non-zero algebraic number. This set forms an infinite subset of the reals, and therefore is not countable.

What are you talking about? $\mathbb{N}$ is an infinite subset of the reals.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Quote:

What are you talking about? $\mathbb{N}$ is an infinite subset of the reals.

I meant an uncountable infinite subset. Sorry.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Quote:

Huh? $\mathbb{Q}$ is countable, so $\mathbb{Q}^{alg}$ is countable, so $\{e^\alpha\mid\alpha\in\mathbb{Q}^{alg},\alpha\neq 0\}$ is a countable subset of the set of transcendentals which (is computable and) can be proven to be transcendental.

I do not understand why use chose ${\Bbb Q}$ , but I get the point. For some reason I thought algebraic numbers were uncountable.

Shadow · **Posted:** Sun, 27 May 2012 23:44:34 UTC

glebovg wrote:

Quote:

Huh? $\mathbb{Q}$ is countable, so $\mathbb{Q}^{alg}$ is countable, so $\{e^\alpha\mid\alpha\in\mathbb{Q}^{alg},\alpha\neq 0\}$ is a countable subset of the set of transcendentals which (is computable and) can be proven to be transcendental.

I do not understand why use chose ${\Bbb Q}$ , but I get the point. For some reason I thought algebraic numbers were uncountable.

Northcott's Theorem establishes their countability.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

glebovg wrote:

Shadow wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

Because that's equivalent to knowing if $\log_2(e)$ is rational...a very difficult thing to decide in general. Understanding which numbers we can easily express are or are not algebraic is a VERY difficult prospect in general.

I do not understand. e is irrational, so ${2^e}$ cannot be rational, because like you said ${2^e} = r$ if $e = {\log _2}(r)$ .

Shadow, could you reply to my previous post. I still do not understand. Thanks.

Shadow · **Posted:** Mon, 28 May 2012 00:03:03 UTC

glebovg wrote:

Shadow wrote:

glebovg wrote:

But why is it unknown whether ${2^e}$ is irrational?

Because that's equivalent to knowing if $\log_2(e)$ is rational...a very difficult thing to decide in general. Understanding which numbers we can easily express are or are not algebraic is a VERY difficult prospect in general.

I do not understand. e is irrational, so ${2^e}$ cannot be rational, because like you said ${2^e} = r$ if $e = {\log _2}(r)$ .

Shadow, could you reply to my previous post. I still do not understand. Thanks.

You should ignore that post, telling that is no easier than the original question.

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Exponential function

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