x Turn on thread page Beta
 You are Here: Home >< Maths

# rt(2)+e is irrational. watch

1. Made a big error in my post to this thread about the irrationality of rt(2)+e.[missed a 2 from one line to the next which gave a false proof]
Just to redeem myself,and earn the rep i received for it, here is a solution :

result 1) e^2 is irrational.
pf

assume not so we have

e^2=a/b

choose m large and set
p=2^(m)
q=p-m
we have
2^q divides p!
proof
highest power of 2 dividing p is given by
[p/2]+[p/2^2]+...[p/2^(2m)]
=2^(2m)-1 + 2^(2m-2)+.....+1
=(2^(2^2m)-1)/1
=2^p-1
but
2^q=2^(p-m)
and p-m is less than 2^(p)-1 so result follows.

we have

p! 2^n
------ is an integer for n<p
n! 2^q

proof
we need what power of 2 divides n!
let r be the highest power of 2 such that [n/2^r] is non-zero
the required power of 2 is
[n/2]+[n/2^2]+..[n/2^r]<n{1/2+1/4+....1/2^r}
=n(1-2(-r))
<n
so highest power of 2 dividing n! is bounded by 2^n
cancelling these out with the 2^n leaves 2^(k) with k positive.
The p! has enough powers of 2 to rid us of 2^q and the remaining
numbers of n! since n<p when we have factored out the powers of 2.
This leaves us integers on the numerator so the expression is an
integer.

now consider the
p! 2^n
the sum of the terms ------ for n=p+1 to oo
n! 2^q
=
2^(p+1-q) ... 2^(p+2-q)
--------- + --------- +.....
( p+1) ..... (p+1)(p+2)

but q=p-m so the sum becomes

2^(m+1) ... 2^(m+2)...
--------- + --------- +...
p+1 ......... (p+1)(p+2)

<
2^m[ 2 ...... 2^2
..... [ ---- + -----
...... [ p...... p^2
=
2^(m+1)
-------
( p-2)
which is less than 1 since p=2^2m

Since the sum of terms with n > p, is not even close to 1, we can multiply by b*p!/2^q (instead of just p!/2^q) and then we just have to choose m to be large enough that the tail is *still* smaller than 1 and result(1) follows.

result 2) 1,e,e^2 are linearly independent over Q.

pf.

assume not so there exist p,q,r in Q not all 0 s.t
p+qe+re^2=0
multiply to rid us of the denominators to get
q'e+r'e^2+p'=0 for p',q',r' integers
ie q'e+r'e^2 =p'' for some integer d
but then
q'+r'e=p''e^(-1)
so there exists integers a,b st
ae+be^(-1) is an integer
but

ae+be^(-1)=
2a+sum (a+b) + .. (a-b)
........... ----- ... -----
............ (2k)! .... (2k+1)!

let p>a+b+1

for k<=p

(2p+1)!(a+b) + ... (a-b)
.......... ----- .... ------ is an integer.
......... (2k)! .... (2k+1)!

for k>=p+1 we are left with the sum of terms
(a+b) +.. (a-b) ............ (a+b)
----- ... ---------- +..... --------------------- +......
(2p+2) .. (2p+2)(2p+3) .... (2p+2)(2p+3)(2p+4)

< a+b . a+b
...--- + --- + ....
...p.......p^2

=
a+b [ 1 + 1 + 1
---..[......-- . ---
..p...[......p.. p^2

=
a+b
----
p-1

which is less than 1 by our choice of p.

hence ae+be^(-1) is not integral.
Therefore 1,e,e^2 are linearly independent over Q.

Finally we have

result (3) rt(2)+e is irrational

pf
assume not so
rt(2)+e=a/b
brt(2)=a-be
2b^2=a^2+b^2e^-2abe
b^2e^2-2abe+(a^2-2b^2).1=0
this gives a linear dependence between 1,e,e^2 contradicting resul(2) hence
rt(2)+e is irrational.

Note also have rt(p)+e is irrational for all rt(p) irrational
2. now edited so it all lines up.
3. Wow... i guess that is redemption.
4. Will read it all soon. Excellent work!
5. (Original post by evariste)
rt(2)+e=a/b
brt(2)=a+be
2b^2=a^2+b^2e^+2abe
Surely thats brt(2)=a-be ??
6. (Original post by Matt_2K)
Surely thats brt(2)=a-be ??
Yes, but fortunately that doesn't affect the proof. We still have a dependece relation. Plus everything else in the proof is fine, as far as I can tell.
7. (Original post by Matt_2K)
Surely thats brt(2)=a-be ??
Yes you are correct, its now corrected.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 10, 2006
Today on TSR

### Loughborough better than Cambridge

Loughborough at number one

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams