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    (Original post by Euclidean)
    But can't figure out how to factorise that, at least not without seeing the answer already.
    Well, you certainly took the painful way through the problem, row reduction works better. But anywho, it's not hard, looking at the polynomial, to guess that b=c is a solution, and indeed:

    c^3 - c^3 - ac^2 + ca^2 + ac^2 - ca^2 = 0, so by symmetry c=a and b=a are also solutions, given that the polynomial has a highest power of a^2, b^2, c^2 then (b-a)(b-c)(c-a) = 0 is the only possible factorisation (since the coefficients are all 1).

    In fact, a rather harder version of this sort of technique comes up in STEP II 2016 Q2.
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    (Original post by Zacken)
    Well, you certainly took the painful way through the problem, row reduction works better. But anywho, it's not hard, looking at the polynomial, to guess that b=c is a solution, and indeed:

    c^3 - c^3 - ac^2 + ca^2 + ac^2 - ca^2 = 0, so by symmetry c=a and b=a are also solutions, given that the polynomial has a highest power of a^2, b^2, c^2 then (b-a)(b-c)(c-a) = 0 is the only possible factorisation (since the coefficients are all 1).

    In fact, a rather harder version of this sort of technique comes up in STEP II 2016 Q2.
    **** that question


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    (Original post by Euclidean)
    det(D) = \left| \begin{array}{ccc}1 & 1 & 1 \\a & b & c & a^2 & b^2 & c^2 \end{array} \right| = 0 \\ \\

 

\therefore (b - c)(c - a)(a - b) = 0

    Is there any nice way to deal with the factorisation of the determinant of D? The textbook method uses EROs to change the 1s to 0s in the rightmost two columns:

    det(D) = \left| \begin{array}{ccc} 1 & 0 & 0 \\a & b - a & c - a & a^2 & b^2 - a^2 & c^2 - a^2 \end{array} \right| = (b - a)(c - a)\left| \begin{array}{ccc} 1 & 0 & 0 \\a & 1 & 1 & a^2 & b + a & c + a \end{array} \right|

    But this method isn't really obvious to me, I tried just working with the determinant as it comes so I end up with:

    \left| \begin{array}{cc} b & c & b^2 & c^2 \end{array} \right| - \left| \begin{array}{cc} a & c & a^2 & c^2 \end{array} \right| + \left| \begin{array}{cc} a & b & a^2 & b^2 \end{array} \right| = 0 \\ \\



\therefore bc^2 - cb^2 - ac^2 + ca^2 + ab^2 - ba^2 = 0

    But can't figure out how to factorise that, at least not without seeing the answer already.

    Is there a good technique for spotting factorisations like that? physicsmaths
    Maybe think of it in a factor theorem kind of way. With expressions like these which are kind of repetitive/have a bit of symmetry going on you can check what happens if you make two of the letters equal. e.g. if you set b = c you get b^3 - b^3 -ab^2 + ba^2 + ab^2 - ba^2 = 0 Then you can divide by b - c. Besides unless the question asks for factorised form there's no need to factorise. You could also note that you have bc(c-b) + a^2(c-b) + a(b^2 - c^2) | Note to self: Don't leave maths pages like these open and then not check that someone has replied lol
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    Does anyone have any tips on how to write proofs to something that seems obvious? For example q5 page 21 of Beardon:

    Let the order of X be N
    I just started off with supposing (a) is true so  f(x' )=f(x) \Rightarrow x'=x so there are atleast N different elements in our image set (as at most 1 x value maps to each element in image set). But our image set is once again X with N elements, so it seem pretty clear each element maps to only 1 different image element so the function is one-to-one. it is therefore both injective and surjective so bijective.

    But to me this seems clunky and too wordy, how should I write it properly?
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    (Original post by EnglishMuon)
    Does anyone have any tips on how to write proofs to something that seems obvious? For example q5 page 21 of Beardon:

    Let the order of X be N
    I just started off with supposing (a) is true so  f(x' )=f(x) \Rightarrow x'=x so there are atleast N different elements in our image set (as at most 1 x value maps to each element in image set). But our image set is once again X with N elements, so it seem pretty clear each element maps to only 1 different image element so the function is one-to-one. it is therefore both injective and surjective so bijective.

    But to me this seems clunky and too wordy, how should I write it properly?
    Notation makes everything look far more elegant


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    (Original post by EnglishMuon)
    Does anyone have any tips on how to write proofs to something that seems obvious? For example q5 page 21 of Beardon:

    Let the order of X be N
    I just started off with supposing (a) is true so  f(x' )=f(x) \Rightarrow x'=x so there are atleast N different elements in our image set (as at most 1 x value maps to each element in image set). But our image set is once again X with N elements, so it seem pretty clear each element maps to only 1 different image element so the function is one-to-one. it is therefore both injective and surjective so bijective.

    But to me this seems clunky and too wordy, how should I write it properly?
    To me, at least, your proof seems to be too much "obviously this and clearly this, so..." which doesn't really constitute a proof. Instead, whenever a problem includes "finite" or naturals only, you should always have induction in the back of your mind. Rigour, eh? :lol:

    So, induct on |X|, the base case is |X| = 1, so any injection is a surjection.

    Now, suppose that |X| \leq n means that X \to X is a surjection, then let |X| = n+1. Try and fill in the rest of the details yourself.

    Check this if you're still stuck.
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    (Original post by Zacken)
    To me, at least, your proof seems to be too much "obviously this and clearly this, so..." which doesn't really constitute a proof. Instead, whenever a problem includes "finite" or naturals only, you should always have induction in the back of your mind. Rigour, eh? :lol:

    So, induct on |X|, the base case is |X| = 1, so any injection is a surjection.

    Now, suppose that |X| \leq n means that X \to X is a surjection, then let |X| = n+1. Try and fill in the rest of the details yourself.

    Check this if you're still stuck.
    ughhhh ok. yea Im terrible at knowing how to write this proof stuff
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    (Original post by Euclidean)

    \therefore bc^2 - cb^2 - ac^2 + ca^2 + ab^2 - ba^2 = 0[/latex]

    But can't figure out how to factorise that, at least not without seeing the answer already.
    The hint here is to write it as a polynomial in c:

    (b-a) c^2 + (a^2 - b^2) c + ab(b - a) = (b - a)(c^2 - (a + b)c + ab)

    Can you take it from here?

    Incidentally, these (and their generalizations) are called Van der Monde determinants). You will re-visit these in the future
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    (Original post by EnglishMuon)
    Does anyone have any tips on how to write proofs to something that seems obvious? For example q5 page 21 of Beardon:

    Let the order of X be N
    I just started off with supposing (a) is true so  f(x' )=f(x) \Rightarrow x'=x so there are atleast N different elements in our image set (as at most 1 x value maps to each element in image set). But our image set is once again X with N elements, so it seem pretty clear each element maps to only 1 different image element so the function is one-to-one. it is therefore both injective and surjective so bijective.

    But to me this seems clunky and too wordy, how should I write it properly?
    In my opinion this is a bit too trivial to require a formal proof (unless you really, really desired a written check of your intuition). If I really needed to, though, I would write:

    Take injective f : X -> X where |X| = n. Im(f) has exactly n elements due to injectivity. The codomain X has exactly n elements. Since Im(f) is a subset of X, Im(f) = X, so surjectivity.


    You certainly don't need induction. 'Rigor' does not mean 'write the most convoluted proof you can muster'. It just means to justify each step of your argument clearly and precisely.

    With facts that are this obvious, it can be difficult at first to write something that resembles a proof.
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    (Original post by EnglishMuon)
    ughhhh ok. yea Im terrible at knowing how to write this proof stuff
    I'm worse, trust me. I've just got the benefit of having done these problems some time ago.
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    (Original post by Ecasx)
    ...
    Ah ok, thanks. I think Ill write as much as I can out formally for now though just for the practise

    (Original post by Zacken)
    I'm worse, trust me. I've just got the benefit of having done these problems some time ago.
    Ah ok, but thanks anyway I really struggle with these sorts of things though. I find it easy to imagine but I always miss the correct way of putting it down. Hopefully I'll be less crap by the end of the summer
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    (Original post by Ecasx)
    Since Im(f) is a subset of X, Im(f) = X, so surjectivity.
    I would argue that the whole point of this question (bear in mind that it comes up right after defining injections and surjections) is to get you to prove that if Im(X) = X then surjective and not just state it.
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    (Original post by Zacken)
    I would argue that the whole point of this question (bear in mind that it comes up right after defining injections and surjections) is to get you to prove that if Im(X) = X then surjective and not just state it.
    A function f : A -> B is surjective if Im(f) = B. That is the definition of surjectivity.
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    (Original post by Zacken)
    Well, you certainly took the painful way through the problem, row reduction works better. But anywho, it's not hard, looking at the polynomial, to guess that b=c is a solution, and indeed:

    c^3 - c^3 - ac^2 + ca^2 + ac^2 - ca^2 = 0, so by symmetry c=a and b=a are also solutions, given that the polynomial has a highest power of a^2, b^2, c^2 then (b-a)(b-c)(c-a) = 0 is the only possible factorisation (since the coefficients are all 1).

    In fact, a rather harder version of this sort of technique comes up in STEP II 2016 Q2.
    Ah thanks very much! I'll take a look PRSOM

    (Original post by 13 1 20 8 42)
    Maybe think of it in a factor theorem kind of way. With expressions like these which are kind of repetitive/have a bit of symmetry going on you can check what happens if you make two of the letters equal. e.g. if you set b = c you get b^3 - b^3 -ab^2 + ba^2 + ab^2 - ba^2 = 0 Then you can divide by b - c. Besides unless the question asks for factorised form there's no need to factorise. You could also note that you have bc(c-b) + a^2(c-b) + a(b^2 - c^2) | Note to self: Don't leave maths pages like these open and then not check that someone has replied lol
    Cheers 13! I can't rep you yet

    Note to self: Don't leave maths pages like these open and then not check that someone has replied lol
    Happens to all of us Not a bad thing though, means there are lots of helpful users

    (Original post by Gregorius)
    The hint here is to write it as a polynomial in c:

    (b-a) c^2 + (a^2 - b^2) c + ab(b - a) = (b - a)(c^2 - (a + b)c + ab)

    Can you take it from here?

    Incidentally, these (and their generalizations) are called Van der Monde determinants). You will re-visit these in the future
    Thank you very much!

    I'll be sure to remember that
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    (Original post by Ecasx)
    A function f : A -> B is surjective if Im(f) = B. That is the definition of surjectivity.
    Not as it's given in the book. A function f : A -> B is surjective if, for each b in B, f(a) = b for at least one a in A. This is obviously equivalent to the definition you've given, but you need to demonstrate the equivalence.
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    (Original post by Zacken)
    Not as it's given in the book. A function f : A -> B is surjective if, for each b in B, f(a) = b for at least one a in A. This is obviously equivalent to the definition you've given, but you need to demonstrate the equivalence.
    Which is so trivial a matter it does not need mentioning. It really isn't the point of the proof, since, for example, we could write

    Take injective f : X -> X where |X| = n. Im(f) has exactly n elements due to injectivity. The codomain X has exactly n elements. If there is x in X s.t. there is no y in X for f(y) = x, then Im(f) has < n elements. Contradiction.

    without explicit mention of Im(f) = X.
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    G unit.
    Fuk zacken up. Zqckens retarded


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    (Original post by Ecasx)
    Which is so trivial a matter it does not need mentioning. It really isn't the point of the proof, since, for example, we could write

    Take injective f : X -> X where |X| = n. Im(f) has exactly n elements due to injectivity. The codomain X has exactly n elements. If there is x in X s.t. there is no y in X for f(y) = x, then Im(f) has < n elements. Contradiction.

    without explicit mention of Im(f) = X.
    Yeah, alright. I agree.
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    I'm a subset, and I'm looking for cosets.
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    can anyone working through the vector and matrices help me with this example please using the equation of the line used in the attachments
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