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Another Australian question: Proving that e is irrational watch

1. Attached.
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2. Question 7.pdf (143.1 KB, 87 views)
3. (Original post by constellarknight)
Attached.
Nice question. Is this a question you want answering btw?
4. (Original post by constellarknight)
Attached.
(Original post by zetamcfc)
Nice question. Is this a question you want answering btw?
I have a formal proof by contradiction
5. (Original post by TeeEm)
I have a formal proof by contradiction
So do I
6. Well you can have a go. The examiner's report says that there were "virtually no totally successful solutions".
7. (Original post by constellarknight)
Well you can have a go. The examiner's report says that there were "virtually no totally successful solutions".
Well I don't do the way as probably intended, but hey-ho.

Idea: All transcendental numbers are irrational. Therefore if is transcendental then is irrational.

Claim: is transcendental.

Proof: Assume is not transcendental. Then

Without the loss of generality we can assume and

Define

Where p is an arbitrary prime.Then is a polynomial in . Put

and note that

Hence for any

Multiply by and sum over to get

from the equation supposedly satisfied by by

We claim that for each is an integer, and that this integer is divisible by unless and . To establish the claim we just apply Liebniz's rule; the only non-zero term arising when come from the factor being differentiated exactly p times. Since , all such terms are integers divisible by . In the exceptional case , the frist non-zero term occurs when , and then

The next non-zero terms are all multiples of . The value of equation is therefore

for some . If , then the integer is not divisible by . So fot sufficiently large primes the value of the equation is an integer not divisible by , hence not zero.

Now estimating the integral. If then

so

which tends to zero as , thus we have a contradiction . Therefore is transcendental which implies it is also irrational.

(In no way the method they would want for the answer but was fun LaTeXing it )
8. (Original post by zetamcfc)
Well I don't do the way as probably intended, but hey-ho.

Idea: All transcendental numbers are irrational. Therefore if is transcendental then is irrational.

Claim: is transcendental.

Proof: Assume is not transcendental. Then

Without the loss of generality we can assume and

Define

Where p is an arbitrary prime.Then is a polynomial in . Put

and note that

Hence for any

Multiply by and sum over to get

from the equation supposedly satisfied by by

We claim that for each is an integer, and that this integer is divisible by unless and . To establish the claim we just apply Liebniz's rule; the only non-zero term arising when come from the factor being differentiated exactly p times. Since , all such terms are integers divisible by . In the exceptional case , the frist non-zero term occurs when , and then

The next non-zero terms are all multiples of . The value of equation is therefore

for some . If , then the integer is not divisible by . So fot sufficiently large primes the value of the equation is an integer not divisible by , hence not zero.

Now estimating the integral. If then

so

which tends to zero as , thus we have a contradiction . Therefore is transcendental which implies it is also irrational.

(In no way the method they would want for the answer but was fun LaTeXing it )
The intended solution: page 11 of exampaper.web.fc2.com/test/hs/y12/maths/HSC4USolutions/2001.pdf, starting at (b)(i).
9. (Original post by constellarknight)
The intended solution: page 11 of exampaper.web.fc2.com/test/hs/y12/maths/HSC4USolutions/2001.pdf, starting at (b)(i).
Yup, knew they wanted a solution linking to the question, but why be so boring?
10. (Original post by zetamcfc)
Yup, knew they wanted a solution linking to the question, but why be so boring?
Especially when they didn't say "hence"
11. (Original post by Zacken)
Especially when they didn't say "hence"
The best is the 'hence or otherwise' question, and you actually take the otherwise option
12. (Original post by zetamcfc)
The best is the 'hence or otherwise' question, and you actually take the otherwise option
13. (Original post by Zacken)

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