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    (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?
    yup i am. looking forward to graph theory. did abit of it at A levels alrdy.
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    (Original post by miml)
    FOUNDATIONS

    January 2006

    1) Every \displaystyle n>1 \in \mathbb{N} can be written as a product of prime numbers. This factorisation is unique up to order.

    2) \displaystyle 7\mathbb{Z} \cap 3\mathbb{Z}=21\mathbb{Z}

    3) \displaystyle (A \cap B)^c = A^c \cup B^c and \displaystyle (A \cup B)^c = A^c \cap B^c

    4)  \displaystyle (\not P) \land Q

    5) (x-3) and (x-1)

    6) \displaystyle \binom{n}{k-1} + \binom{n}{k} = \frac{n!}{(k-1)!(n-k+1)!}+\frac{n!}{(k)!(n-k)!}=\dfrac{n!k + n!(n-k+1)}{k!(n-k+1)!} =\displaystyle \frac{n!(k+n-k+1)}{k!(n+1-k)!} = \frac{(n+1)!}{k!(n+1-k)!} = \binom{n+1}{k}

    7) \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where \displaystyle f(x)=(x+1)x(x-1)
    \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where \displaystyle f(x)=?
    \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where  \displaystyle f(x)=x^2
    \displaystyle f:\mathbb{Z}\rightarrow\mathbb{Z  } where \displaystyle f(x)=2x

    8) There is a bijection \displaystyle f:A \rightarrow B

    9) (134)(25)

    10) Closure: \displaystyle \forall a,b \in G , ab=ba\in G
    Associativity: \displaystyle  \forall a,b,c \in G, (ab)c=a(bc)
    Identity Element: \displaystyle  \exists e \in G : \forall a \in G ae = ea = a
    Inverse Element: \displaystyle \forall a \in G \exists a^{-1} \in G: aa^{-1} = a^{-1}a = e

    11) \displaystyle 17^{18} = 17^{(19-1)} = 1 (mod 19) by FLT.


    January 2007

    1) a|b means 'a divides b' ie \displaystyle \exists k \in \mathbb{N} : b=ak . If a|b and b|c then \displaystyle k_1, k_2 \in \mathbb{N} : b=ak_1 and \displaystyle c=bk_2 \implies c = ak_1k_2 \implies a|c

    2) Every \displaystyle n>1 \in \mathbb{N} can be written as a product of prime numbers. This factorisation is unique up to order.

    3) \displaystyle  5\mathbb{Z} \cap 3\mathbb{Z} = 15\mathbb{Z}
    \displaystyle  5\mathbb{Z} + 3\mathbb{Z} = {Z}
    \displaystyle  5\mathbb{Z} \cup 3\mathbb{Z} is not a subgroup.

    4) Venn diagrams

    5)  \diplaystyle\begin{matrix}

(P & \wedge  & Q) & \vee   & (\setminus   & P  & \wedge  & Q)  & \Leftrightarrow   & Q \\ 

 1& 1 &1  &1  &0  &1  &0  &1  &1  &1 \\ 

 1& 0& 0 & 0 & 0 & 1 & 0 &  0&1  &0 \\ 

 0& 0& 1 & 1&  1&  0&  1&  1&  1& 0\\ 

 0& 0& 0 & 0&  1&  0&  1& 0 &  1&0 

\end{matrix}

    6) If P(x) = (x-a)Q(x) + R(x) then R is the constant polynomial P(a)

    7) (x+4) and (x-1)

    8) Coefficient of x^k is \displaystyle \binom{n}{k}a^{n-k}b^k
    Let a=1, bx = -1 in \displaystyle (a+bx)^n = \sum_{k=0}^{\infty} \binom{n}{k}a^{n-k}(bx)^k \implies (-1+1)^n = 0 = \sum_{k=0}^{\infty} (-1)^k \binom{n}{k}

    9) \displaystyle f:\mathbb{N}\rightarrow\mathbb{N  } where  \displaystyle \left\{\begin{matrix}

f(1)=1

\\ f(x)=x-1 \forall x>1



\end{matrix}\right.
    \displaystyle f:\mathbb{N}\rightarrow\mathbb{N  } where  \displaystyle f(x)=x^2
    \displaystyle f:\mathbb{N}\rightarrow\mathbb{N  } where  \displaystyle f(x)=5

    10) There is a bijection \displaystyle f:\mathbb{N} \rightarrow \mathbb{Q} but no bijection \displaystyle f:\mathbb{N} \rightarrow \mathbb{R}


    January 2008

    1) \displaystyle 8\mathbb{Z} + 12\mathbb{Z}=4\mathbb{Z}

    2) \displaystyle \varnothing , {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}

    3) A\(B u C)

    4) ~P V ~Q

    5) (x+3) and (x-2)

    6) Let a=b=1 in \displaystyle (a+b)^n = \sum_{k=0}^{n} a^kb^{n-k} \binom{n}{k} \implies (1+1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k}

    7) ?

    8) There is an injection (but no bijection)  \displaystyle f: A \rightarrow B

    9) (1324)

    10)  \displaystyle 15^{22} = (15^2)^{11} = 15^2 (mod 11) by FLT  \displaystyle 15^2 = 4^2 (mod 11) = 16 (mod 11) = 5 (mod 11)


    January 2009

    1)  \displaystyle 9\mathbb{Z} + 15\mathbb{Z} = 3\mathbb{Z}

    2) A\(B u C)

    3) (~P) ^ (~Q)

    4) (2x+1) and (x+4)

    5)  f:\mathbb{N}^3 \rightarrow \mathbb{N} where f(a,b,c) = a(b+1)(c+2) [/latex]

    6) Yes, a surjection A->B has a right inverse (which is also an injection) B->A.

    7)  \displaystyle\alpha\beta = (1234)
     \displaystyle \beta \alpha = (1243)

    8) Closure: \displaystyle \forall a,b \in G , ab=ba\in G
    Associativity: \displaystyle  \forall a,b,c \in G, (ab)c=a(bc)
    Identity Element: \displaystyle  \exists e \in G : \forall a \in G ae = ea = a
    Inverse Element: \displaystyle \forall a \in G \exists a^{-1} \in G: aa^{-1} = a^{-1}a = e

    9) \displaystyle 15^{17} = 15 (mod 17) by FLT
    Can I make a guess for 7 of Jan2008 paper? f(r)=(2^m)*(3^n), where r=m/n and hcf(m,n)=1.
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    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?
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    (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?
    It's false
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    Few questions I wouldn't mind help with:

    1) From Analysis 1d Jan 2010:

    If a_n \rightarrow a \ \ \text{and} \ \ a_n > 1 \ \forall \ n>0, \ \ \text{show that} \ a\ge1 \ \ \ \text{(2 marks)}

    If anyone wants my answers to the rest of Q1, just say and I'll post them.

    2) Find a injection  f:\mathbb{Q} \rightarrow \mathbb{N}

    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?
    Yeah I agree with matt2k8, they don't.
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    (Original post by electriic_ink)
    Few questions I wouldn't mind help with:

    1) From Analysis 1d Jan 2010:

    If a_n \rightarrow a \ \ \text{and} \ \ a_n > 1 \ \forall \ n>0, \ \ \text{show that} \ a>1 \ \ \ \text{(2 marks)}
    I don't think that's even true...??
    What if a_n = 1 + 1/n ? -- Then a_n is always > 1, but a=1

    2) Find a injection  f:\mathbb{Q} \rightarrow \mathbb{N}
    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.
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    (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=1
    Should be more careful when asking for help. It says 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.
    OK. Thanks. I'll rep you tomorrow.
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    (Original post by matt2k8)
    It's false
    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?

    (Original post by January 2005 Paper)
    It has been known since the 17th century that \pi can be represented as:

    \pi = 4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}

    In 1995, the following formula was discovered:

    \pi = \sum_{k=0}^\infty \frac{1}{16^k} \Big( \frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6} \Big)

    Estimate the number of terms required to approximate \pi 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.
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    (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?
    Q2.24 is actually a strictly contracting sequence. l here is 1/2 which is less than 1. and we know that all strictly contracting sequences are Cauchy and hence converge.

    and for your second question. it is false because 0 has only 1 decimal expansion.
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    (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?
    Where did you get hold of this question? I haven't seen in anywhere!
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    (Original post by electriic_ink)
    Should be more careful when asking for help. It says a>=1.
    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)

    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.
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    Are you living in Tocil 3 by any chance?
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    (Original post by yeohaikal)
    Where did you get hold of this question? I haven't seen in anywhere!
    It's in the MORSE booklet!
    There's also one very similar to it a few years later.
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    (Original post by Narev)
    Are you living in Tocil 3 by any chance?
    Who, me?? I'm a 2nd year, in Leam
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    (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 + ... + (2k-1) = k^2
    then 1 + 3 + 5 + ... + (2k-1) + (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 P \in \mathbb{R}[x], then the remainder on division of P by

    (x-a) is equal to P(a)
    e. (x+1) only
    f. Suppose x is even so x = 2w for some integer w. then y^2 = 6 + 8w^3 = 2



(3+4w^3).
    So then 2|y^2, so 2|y as 2 is prime. So y = 2u for

    some integer u.
    so then 4u^2 = 6 + 8w^3, so 2u^2 = 3 + 4w^3
    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 Schroeder-Bernstein Theorem, there exists a bijection

    h:A->B.
    i. if p,m \in \mathbb{N}, p is prime and p does not divide m,
    m^{p-1} mod p = 1
    j. 10^{12} mod 11 = 10^{10}.10^2 mod 11
     = 10^2 mod 11 by FLT
     = 100 mod 11 = 1 as 100 = 11*9 + 1
    Hmm.. Got all the same answers even for h. But I'm not quite sure what they are looking out for given that it's 5 marks. I did show how to get the injection through a particular mapping. Under time-constraint I would quote the schroeder-bernstein theorem, but I have a feeling they might want to see a proof of it? or perhaps a sketch proof? not sure. anybody else tried and got anything to share on this question?
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    (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)
    a_n=(1+1/n). Then what exactly is e?
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    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)
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    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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    (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.
    I also messed up both of these parts of question 2.
    My "proof" of x^{\frac{1}{n}} was pretty much saying it gets close to one ... OH LOOK IT JUST TENDS TO ONE DOESN'T IT. :rolleyes:
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    (Original post by placenta medicae talpae)
    I also messed up both of these parts of question 2.
    My "proof" of x^{\frac{1}{n}} was pretty much saying it gets close to one ... OH LOOK IT JUST TENDS TO ONE DOESN'T IT. :rolleyes:
    Something to do with the quotient rule, right?

    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) :rolleyes:

    Also, have fun in G&M and analysis 2!
 
 
 
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