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Another challenge question: Irrationality of e (2 Viewers)
- Thread starter seanieg89
- Start date
k02033
Member
so
clearly
factorizing gives
therefore
since
Now
where
therefore
as required
k02033
Member
Do we use proof by contradiction for the 2nd part?
Yeah, probs let e=p/q where p,q E Z+
k02033
Member
dam stuck, will try again tomorrow
k02033
Member
1st thing i tried, didnt get anywhere >.>
A bit late to be doing maths now? lol
k02033
Member
more like forgot my discrete maths already, but yea i will run with the lame late excuse.
e-Sn=1/(n+1)!+...
So, 0<1/(n+2)!+...<1/(n(n+1)!)
Does the above help?
The proof is long-winded. It could've been proved a lot quicker if there was an "otherwise."
seanieg89
Well-Known Member
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- Aug 8, 2006
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- HSC
- 2007
Nice work with the inequality, and yes contradiction is pretty much the only way to go for the second part. Its not very long, just a little tricky.
Hint: Consider the proven inequality for certain special partial sums.
Hint: Consider the proven inequality for certain special partial sums.
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k02033
Member
underestimated the question, i tried to look at specific sums and using the good old gcd contradiction, doesnt seem to work...
gurmies
Drover
That would be likely if it were |e-S_n| < ...
k02033
Member
ha ha the answer is funny, its so obvious
Theorem: e is irrational.
Proof by contradiction:
Let
where p,q, are positive integers.
From part 1 we know
!}< \frac{p}{q}-[1+1+\frac{1}{2!}+\frac{1}{3!}....\frac{1}{n!}]< \frac{1}{n\times n!})
Now let n=q and multiply throughout by q! gives
!-[2q!+1+\frac{q!}{2!}+\frac{q!}{3!}+...]<\frac{1}{q})
Now the middle term is an integer! And you cant have an integer satisfying that inequality therefore e is irrational, as required. I was way too stubborn with the gcd approach, 20 pages of trial and error ...
Theorem: e is irrational.
Proof by contradiction:
Let
From part 1 we know
Now let n=q and multiply throughout by q! gives
Now the middle term is an integer! And you cant have an integer satisfying that inequality therefore e is irrational, as required. I was way too stubborn with the gcd approach, 20 pages of trial and error ...
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Fuck! I essentially had the last line but didn't deduce anything from it, so i chucked the paper in the binha ha the answer is funny, its so obvious
Theorem: e is irrational.
Proof by contradiction:
Let
From part 1 we know
Now let n=q and multiply throughout by q! gives
Now the middle term is an integer! And you cant have an integer satisfying that inequality therefore e is irrational, as required. I was way too stubborn with the gcd approach, 20 pages of trial and error ...
Kudos, k02033. I would rep you again but need to spread the rep.k02033
Member
yay! i had it ages ago too, how embarrassing hahhaha ( i threw my pen couple of times)Fuck! I literally, had the last line but didn't deduce anything from it, so i chucked the paper in the bin
k02033
Member
Oh by the way, how do we put a space while using the latex equation editor?
Idk, i never use LaTeX. There's a LaTeX guide stickied somewhere in the Math forums.
k02033
Member
Yeah, nw.