deducing that e is irrational (1 Viewer)

[duy.le]

Hey guys there just this question that ive been having trouble getting my head around. I can do part i), but part ii) i cant seem to do without fudging it and its probably wrong.

Q8.C)ii)

The number e is given by the value of the limiting sum

e=1+1+1/2!+1/3!+......

i) If n is a positive integer, and a=n!{e-(1+1+1/2!+1/3!+...+1/n!)}, show that
0 < a < 1/n (you can assume the answer if u want, its not that hard)

[3 marks]

ii) Hence deduce that e is irrational

[2 marks]

from the 2007 independent trial paper

I think that its relevant because it appears every 3 or 4 years and the last one was pi i think and in 2005 so i think we are due for another one this year. oh nos.

thanks.

 

[shaon0]

Hey guys there just this question that ive been having trouble getting my head around. I can do part i), but part ii) i cant seem to do without fudging it and its probably wrong.

Q8.C)ii)

The number e is given by the value of the limiting sum

e=1+1+1/2!+1/3!+......

i) If n is a positive integer, and a=n!{e-(1+1+1/2!+1/3!+...+1/n!)}, show that
0 < a < 1/n (you can assume the answer if u want, its not that hard)

[3 marks]

ii) Hence deduce that e is irrational

[2 marks]

from the 2007 independent trial paper

I think that its relevant because it appears every 3 or 4 years and the last one was pi i think and in 2005 so i think we are due for another one this year. oh nos.

thanks.

You can use Joseph Fourier's proof.
This might help a little.
If you want both proofs.
http://pirate.shu.edu/~wachsmut/ira/infinity/irrat_nm.html (this is an advanced proof though)

 

[conics2008]

e is a slut..

go prove that pi cant be written in the form of p/q ..

its alot better than e.

reference 2003 hsc paper.

 

[duy.le]

thanks for the answer, where did u get it?

 

[shaon0]

thanks for the answer, where did u get it?

somewhere, i downloaded it ages ago for some weird reason.

 

[Affinity]

Five bucks that this will not come up in the HSC this year

 

[vds700]

Five bucks that this will not come up in the HSC this year

you're on

 

[kony]

yeah they've got rid of all the random stuff in the exams. this e irrationality is not going to happen.

 

[shaon0]

yeah they've got rid of all the random stuff in the exams. this e irrationality is not going to happen.

I'll laugh if pi comes up.

 

[tommykins]

回复: Re: deducing that e is irrational

Lol. I can imagine in the exam and its like prove sqrt2 is irrational.

Be like 'fuck.'

 

[shaon0]

Re: 回复: Re: deducing that e is irrational

Lol. I can imagine in the exam and its like prove sqrt2 is irrational.

Be like 'fuck.'

that's actually very easy

 

[tommykins]

回复: Re: 回复: Re: deducing that e is irrational

be my guest. post up a proof without wikiing it.

 

[Iruka]

There are many rational numbers that do not have terminating decimal expansions.

 

[tommykins]

回复: Re: deducing that e is irrational

1/9 ? lol.

 

[Templar]

Re: 回复: Re: 回复: Re: deducing that e is irrational

Standard proof by contradiction/infinite descent, very simple, hardly looking up on Wiki.

 

[lolokay]

There are many rational numbers that do not have terminating decimal expansions.

whoops..I'll just get rid of that now

 

[tommykins]

回复: Re: deducing that e is irrational

Still waiting on a non-wiki proof by shaon0

 

[lolokay]

Re: 回复: Re: deducing that e is irrational

so sqrt2 = p/q
2 = (p/q)² and p and q are integers - and have no common factors
= p²/q²
if p doesn't have 2 as a factor, then as p and q are integers, and the square root of 2 isn't, the number can't be divisible by 2.
so p must be divisble by 2
so p² must be divisble by 4
so (p²/2)/q² must be divisble by 2
but (p²/2)/q² = 2/2 = 1
and 1 isn't divisible by 2, so (p²/2)/q² also isn't
therefore sqrt2 is not rational (by contradiction)

which would also prove that all square roots of integers that don't have integer square roots are irrational (if it's correct..)

 

[Affinity]

Re: 回复: Re: deducing that e is irrational

Still waiting on a non-wiki proof by shaon0

If you don't like the vanilla flavoured proof, it's also an easy consequence of Eisenstein's criterion.

 

[tommykins]

回复: Re: 回复: Re: deducing that e is irrational

If you don't like the vanilla flavoured proof, it's also an easy consequence of Eisenstein's criterion.

nah it's not that i do'nt accept wiki's proof or antyhing as i don't know how to prove it myself.

but shaon0 is implying the proof is 'easy' when he hasn't been able to show his own proof which is devised by his own thought, and not wiki's.