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 is a solution, and indeed:(Original post by Euclidean)
But can't figure out how to factorise that, at least not without seeing the answer already.
, so by symmetry and are also solutions, given that the polynomial has a highest power of then 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.

Zacken
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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 is a solution, and indeed:
, so by symmetry and are also solutions, given that the polynomial has a highest power of then 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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 04072016 14:29
(Original post by Euclidean)
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:
But this method isn't really obvious to me, I tried just working with the determinant as it comes so I end up with:
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? physicsmathsLast edited by math42; 04072016 at 14:31. 
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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 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 onetoone. 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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 04072016 14:52
(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 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 onetoone. 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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 04072016 15:05
(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 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 onetoone. it is therefore both injective and surjective so bijective.
But to me this seems clunky and too wordy, how should I write it properly?
So, induct on , the base case is , so any injection is a surjection.
Now, suppose that means that is a surjection, then let . Try and fill in the rest of the details yourself.
Check this if you're still stuck. 
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 04072016 15:12
(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?
So, induct on , the base case is , so any injection is a surjection.
Now, suppose that means that is a surjection, then let . Try and fill in the rest of the details yourself.
Check this if you're still stuck. 
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 04072016 15:24
(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.
Can you take it from here?
Incidentally, these (and their generalizations) are called Van der Monde determinants). You will revisit these in the future 
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 04072016 15:27
(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 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 onetoone. it is therefore both injective and surjective so bijective.
But to me this seems clunky and too wordy, how should I write it properly?
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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 04072016 15:28
(Original post by EnglishMuon)
ughhhh ok. yea Im terrible at knowing how to write this proof stuff 
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 04072016 15:32
(Original post by Ecasx)
...
(Original post by Zacken)
I'm worse, trust me. I've just got the benefit of having done these problems some time ago. 
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 04072016 15:38
(Original post by Ecasx)
Since Im(f) is a subset of X, Im(f) = X, so surjectivity. 
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 04072016 15:40
(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. 
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 04072016 15:45
(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 is a solution, and indeed:
, so by symmetry and are also solutions, given that the polynomial has a highest power of then 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.
(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(cb) + a^2(cb) + 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
Note to self: Don't leave maths pages like these open and then not check that someone has replied lol
(Original post by Gregorius)
The hint here is to write it as a polynomial in c:
Can you take it from here?
Incidentally, these (and their generalizations) are called Van der Monde determinants). You will revisit these in the future
I'll be sure to remember that 
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 04072016 15:48
(Original post by Ecasx)
A function f : A > B is surjective if Im(f) = B. That is the definition of surjectivity. 
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 04072016 15:55
(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.
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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 04072016 16:26
G unit.
Fuk zacken up. Zqckens retarded
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 04072016 16:56
(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. 
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 04072016 17:16
I'm a subset, and I'm looking for cosets.

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 05072016 13:21
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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