yup i am. looking forward to graph theory. did abit of it at A levels alrdy.(Original post by placenta medicae talpae)
*looks this up*
Appaz 9.9 and 9.10 are not being tested
Which is a shame, because I really like 9.9.
And how 9.10 could be tested anyway beats me!
Btw, you doing discrete this term?

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 08012011 00:23

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 08012011 11:57
(Original post by miml)
FOUNDATIONSJanuary 2006January 2007
1) ab means 'a divides b' ie . If ab and bc then and
2) Every can be written as a product of prime numbers. This factorisation is unique up to order.
3)
is not a subgroup.
4) Venn diagrams
5)
6) If P(x) = (xa)Q(x) + R(x) then R is the constant polynomial P(a)
7) (x+4) and (x1)
8) Coefficient of is
Let a=1, bx = 1 in
9) where
where
where
10) There is a bijection but no bijectionJanuary 2008

placenta medicae talpae
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 08012011 20:02
Anyone any ideas how to show 2.24 of MORSE analysis booklet?
And also, I keep telling myself I must be missing something obvious, but how is it true that every real has two decimal expansions?
I completely understand the argument for rationals, but what about the irrationals?Last edited by placenta medicae talpae; 08012011 at 20:11. 
matt2k8
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 09012011 12:35
(Original post by placenta medicae talpae)
Anyone any ideas how to show 2.24 of MORSE analysis booklet?
And also, I keep telling myself I must be missing something obvious, but how is it true that every real has two decimal expansions?
I completely understand the argument for rationals, but what about the irrationals? 
electriic_ink
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 09012011 14:38
Few questions I wouldn't mind help with:
1) From Analysis 1d Jan 2010:
If
If anyone wants my answers to the rest of Q1, just say and I'll post them.
2) Find a injection
Thanks.(Original post by placenta medicae talpae)
And also, I keep telling myself I must be missing something obvious, but how is it true that every real has two decimal expansions?
I completely understand the argument for rationals, but what about the irrationals?Last edited by electriic_ink; 09012011 at 15:21. 
tomthecool
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 09012011 15:00
(Original post by electriic_ink)
Few questions I wouldn't mind help with:
1) From Analysis 1d Jan 2010:
If
What if a_n = 1 + 1/n ?  Then a_n is always > 1, but a=1
Pick any two prime numbers, e.g. 2 and 3. You know that 2^a*3^b is unique, for each pair of values (a,b).
So f(a/b) = 2^a*3^b is injective. 
electriic_ink
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 09012011 15:25
(Original post by tomthecool)
I don't think that's even true...??
What if a_n = 1 + 1/n ?  Then a_n is always > 1, but a=1Pick any two prime numbers, e.g. 2 and 3. You know that 2^a*3^b is unique, for each pair of values (a,b).
So f(a/b) = 2^a*3^b is injective. 
placenta medicae talpae
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 09012011 19:29
(Original post by matt2k8)
It's false
I can't think of a counterexample, for example.
Also, sorry to be a pest, but does anyone know how to go about doing Jan 2005 question 4 part iii?(Original post by January 2005 Paper)
It has been known since the 17^{th} century that can be represented as:
In 1995, the following formula was discovered:
Estimate the number of terms required to approximate to within 10^{10} for both formulae.
Hint: use the fact (which you do not need to prove) that the term in brackets in the second formula lies between 0 and 4 for every value of k.Last edited by placenta medicae talpae; 09012011 at 19:32. 
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 09012011 21:50
(Original post by placenta medicae talpae)
Anyone any ideas how to show 2.24 of MORSE analysis booklet?
And also, I keep telling myself I must be missing something obvious, but how is it true that every real has two decimal expansions?
I completely understand the argument for rationals, but what about the irrationals?
and for your second question. it is false because 0 has only 1 decimal expansion. 
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 09012011 21:54
(Original post by placenta medicae talpae)
Yeah, but how do you show this?
I can't think of a counterexample, for example.
Also, sorry to be a pest, but does anyone know how to go about doing Jan 2005 question 4 part iii? 
tomthecool
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 10012011 01:06
(Original post by electriic_ink)
Should be more careful when asking for help. It says a>=1.
Assume a<1
Let e = min{a1, a2, a3, ...}  a
[So e > 0, since a1, a2, ... are all > 1 and a < 1]
Then for all n, a_n  a = a_n  a (since a_n > 1 > a) >= min{a1, a2, a3, ...}  a = e
Therefore a_n does not converge to a (CONTRADICTION)
Sorry for the lack of proper symbols due to me being too lazy to write in latex
By the way, I also realised a slight problem with my answer for the second question: if a is negative, then 2^a*3^b is not a natural number!
So to fix this, you could just say something like:
If a>=0, then f(a/b)=2^a*3^b
If a<0, then f(a/b)=5^(a)*3^b
Note: The reason I had to use a new arbitrary prime number (5) in this second case is that the function must be injective, so f(1/2) cannot = f(1/2), for example.Last edited by tomthecool; 10012011 at 01:37. 
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 10012011 03:46
Are you living in Tocil 3 by any chance?

placenta medicae talpae
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 10012011 08:04
(Original post by yeohaikal)
Where did you get hold of this question? I haven't seen in anywhere!
There's also one very similar to it a few years later. 
tomthecool
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 10012011 13:57
(Original post by Narev)
Are you living in Tocil 3 by any chance? 
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 12012011 02:31
(Original post by matt2k8)
I'm not 100% sure about h, but the rest seems OK to me?
Foundations, 2010, Q1:Spoiler:Show
a. When n = 1, LHS = 1, RHS = 1^2 = 1 therefore true for n =1.
Now assume true for n = k that 1 + 3 + 5 + ... + (2k1) = k^2
then 1 + 3 + 5 + ... + (2k1) + (2(k+1) 1) = k^2 + 2k + 1 = (k+1)^2
Therefore, if true for n =k , true for n = k+1. true for n = 1 so true for all
natural numbers n by induction.
b. Every natural number greater than one can be expressed as a product of prime
numbers, unique up to the order of the terms.
c. h = 3, a = 4, b = 13
d. If , then the remainder on division of P by
(xa) is equal to P(a)
e. (x+1) only
f. Suppose x is even so x = 2w for some integer w. then .
So then , so as 2 is prime. So y = 2u for
some integer u.
so then , so
But this is a contradiction as the LHS is even but the RHS is odd. So x must be odd.
g. Only true if a is true, b is false, so equivalent to i. (dunno how to put a truth
table on here)
h. True. as g is a surjection A>B, it has a right inverse g^1 which is an
injection B>A. So by the SchroederBernstein Theorem, there exists a bijection
h:A>B.
i. if , p is prime and p does not divide m,
j.
by FLT
as 100 = 11*9 + 1 
IrrationalNumber
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 12012011 05:13
(Original post by tomthecool)
Ah, well that makes more sense then! In that case, there are many variations on how you could prove this, but here is an example: (Note  I'm using e instead of epsilon, since epsilon doesn't work without latex)
Assume a<1
Let e = min{a1, a2, a3, ...}  a
[So e > 0, since a1, a2, ... are all > 1 and a < 1]
Then for all n, a_n  a = a_n  a (since a_n > 1 > a) >= min{a1, a2, a3, ...}  a = e
Therefore a_n does not converge to a (CONTRADICTION) 
placenta medicae talpae
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 12012011 11:08
Anyone know whether we can use pencil in this exam?
(I normally go sick at any institution which judge candidates on what type of writing implement they use)Last edited by placenta medicae talpae; 12012011 at 11:10. 
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 12012011 19:51
I didn't even know this thread existed
I forgot the proof of 2a, the one about P1 = QP2 + R, and couldn't do the one about x^1/n > 1 (i vaguely remember the proof but i couldn't make any clever moves that seemed to help, bleh). Most other things seemed cool. I loved the groups/isomorphism question about the matrices I answered 4 questions on both so hopefully the 2 parts I did badly on won't count against me. 
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 12012011 21:01
(Original post by ziedj)
I didn't even know this thread existed
I forgot the proof of 2a, the one about P1 = QP2 + R, and couldn't do the one about x^1/n > 1 (i vaguely remember the proof but i couldn't make any clever moves that seemed to help, bleh). Most other things seemed cool. I loved the groups/isomorphism question about the matrices I answered 4 questions on both so hopefully the 2 parts I did badly on won't count against me.
My "proof" of was pretty much saying it gets close to one ... OH LOOK IT JUST TENDS TO ONE DOESN'T IT. 
tomthecool
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 13012011 01:39
(Original post by placenta medicae talpae)
I also messed up both of these parts of question 2.
My "proof" of was pretty much saying it gets close to one ... OH LOOK IT JUST TENDS TO ONE DOESN'T IT.
Oh, it's funny how you'll look back at this exam next year and think how easy it is! (Compared to what's coming up later)
Also, have fun in G&M and analysis 2!