prove ln2 is irrational.

Since we're on the topic, I thought I'd share this question:

Prove that sqrt(2) + e is irrational.

This was one of the harder bonus questions from one of my courses last term. I only managed to come up with a very handwavy argument, but I couldn't solve it. So naturally I'm very interested in seeing a solution. (I promise rep, too. :smile:

)

Reply 5

yusufu

16

dvs

Since we're on the topic, I thought I'd share this question:

Prove that sqrt(2) + e is irrational.

This was one of the harder bonus questions from one of my courses last term. I only managed to come up with a very handwavy argument, but I couldn't solve it. So naturally I'm very interested in seeing a solution. (I promise rep, too. :smile:

)

along a similar line: i was reading a book a few days back which said that it is not known whether pi + e is rational or irrational. will give rep for proof either way. :biggrin:

Reply 6

evariste_3086

dvs

Since we're on the topic, I thought I'd share this question:

Prove that sqrt(2) + e is irrational.

This was one of the harder bonus questions from one of my courses last term. I only managed to come up with a very handwavy argument, but I couldn't solve it. So naturally I'm very interested in seeing a solution. (I promise rep, too. :smile:

)

if rt(2)+e is rational then (rt(2)+e)^2 is also rational
we have
(rt(2)+e)^2
=2+e^2+2rt(2)e
=2+e(rt(2)+e) is rational

Reply 7

if rt(2)+e is rational then (rt(2)+e)^2 is also rational
we have
(rt(2)+e)^2
=2+e^2+2rt(2)e
=2+e(rt(2)+e) is rational

How is 2+e(rt2+e) rational?

Reply 8

davros

16

dvs

ln(2) = p/q (p/q in lowest form - in particular, q is nonzero)
2 = e^(p/q)
2^q = e^p

If this is true, then e satisfies the equation e^p - 2^q = 0 i.e. e is the root of a polynomial equation with integer coefficients. But by a famous theorem of Hermite, e is transcendental, i.e. it cannot be the root of such an equation.

Therefore the original assumption was false and ln 2 cannot be rational.

Reply 9

evariste_3086

Esquire

How is 2+e(rt2+e) rational?

because if a is rational then clearly a^2 is rational.
Ive assumed rt2+e is rational so [r(2)+e]^2=2+e(rt2+e) is also rational.

Reply 10

BCHL85

12

can we say e is transcendental number while sqrt(2) is algebraic number. So e + sqrt(2) is transcendental as well (prove by contradiction).
And transcendental numbers are irrational since any rational p/q is solution of qx = p

Reply 11

dvs

BCHL85

can we say e is transcendental number while sqrt(2) is algebraic number. So e + sqrt(2) is transcendental as well (prove by contradiction).
And transcendental numbers are irrational since any rational p/q is solution of qx = p

That was my instant proof, but we weren't allowed to use the fact the e is transcendental unless we could prove it ourselves.

I like evariste's proof though. It's so simple! :smile:

It strikes me as odd though, because I remember my professor saying that this statement was very difficult to prove and we had to rely on the power series expansion of e^x...

Reply 12

BCHL85

12

a small mistake in evariste's proof:
(rt(2) + e)^2 = 2 + e(2rt(2) + e), not 2 + e(rt(2) + e)

Ah.. :|

BCHL85

a small mistake in evariste's proof:
(rt(2) + e)^2 = 2 + e(2rt(2) + e), not 2 + e(rt(2) + e)

i hang my head in shame and blame the hang-over. :redface:

Reply 15

because if a is rational then clearly a^2 is rational.
Ive assumed rt2+e is rational so [r(2)+e]^2=2+e(2rt2+e) is also rational.

I don't see how this tells us anything. You are saying "If this is rational, then this is rational" but what if it isn't rational? Can you prove that 2+e(2rt2+e) is rational? "if God exists, then God exists" proved... :biggrin:

Reply 16

meef cheese

Esquire

I don't see how this tells us anything. You are saying "If this is rational, then this is rational" but what if it isn't rational? Can you prove that 2+e(2rt2+e) is rational? "if God exists, then God exists" proved... :biggrin:

It's proof by contradiction. He assumes it's rational then goes onto show that this leads to a contradiction (that (rt2+e)² is irrational) therefore his assumption was wrong and it is irrational.

Reply 17

e-unit

It's proof by contradiction. He assumes it's rational then goes onto show that this leads to a contradiction (that (rt2+e)2 is irrational) therefore his assumption was wrong and it is irrational.

I know that this is meant to be proof by contradiction. Where did he prove that (rt2+e)^2 is irrational? Where is the contradiction in this statement...

evariste

Ive assumed rt2+e is rational so [r(2)+e]^2=2+e(2rt2+e) is also rational.

Surely he can only prove by contradiction if there is a contradiction. He is trying to show that rt2+e is irrational, this means that:
(1)If rt2+e is rational then (rt2+e)^2 is also rational.
(2)If rt2+e is irrational then (rt2+e)^2 is also irrational.
Therefore he assumes that rt2+e is rational because if (rt2+e)^2 is irrational then he has proved rt2+e as being irrational. The problem is: how did he prove that (rt2+e)^e is irrational or rational?
Now, evariste himself has said that...
(1) he assumed that (rt2+e) is rational
(2) he found that (rt2+e)^2 is rational
... There is no contradiction here and he was trying to prove that (rt2+e) was irrational anyway... Do you see my problem?

Reply 18

dvs

(2)If rt2+e is irrational then (rt2+e)^2 is also irrational.

This is not true. Take for example sqrt(2) which is irrational while (sqrt(2))^2 = 2 is rational.

The point is if x is rational then so is x*x = x^2. However if x^2 is irrational, x cannot be rational (by the previous statement).

The contradiction in evariste's proof is that his assumption led him to say that "2+e(rt(2)+e)" is rational, when it isn't (because then e would be rational).

Anyway it doesn't matter because he had an arithmetic slip which invalidated his proof.

Reply 19