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Solve this Maths question, or else, you aren't good enough for Oxbridge Watch

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    (Original post by LightBlueSoldier)
    Generally you start with a big matrix and then split it up case by case. I think it's useful to get into a mindset of rank=equations as when you get into some of the applications where it is unclear what the natural rank and dimension should be you can generate a rank in various ways (amongst other situations: this is the first that came to mind)


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    TBH I rarely do any linear algebra on systems of linear equations. But when I do I think of the rank as the degeneracy of the system, so higher rank means more degeneracy. This intuition applies nicely to geometry as well, where the rank of a matrix determines how nicely a vector field behaves near a point (e.g. maximal rank implies locally invertible).
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    I click on maths threads far too often... I don't have a clue what is going on
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    (Original post by Gwilym101)
    Taken from the Legendre's three square theorem page on wikipedia.

    In mathematics, Legendre's three-square theorem states that a natural number can be represented as the sum of three squares of integers

    n = x^2 + y^2 + z^2

    if and only if n is not of the form n = 4^a(8b + 7) for integers a and b.

    The first numbers that cannot be expressed as the sum of three squares (i.e. numbers that can be expressed as n = 4^a(8b + 7)) are

    7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71

    As trolls go I'd say this was a fairly sophisticated one. However it has the failing that anyone that is willing to attempt this question is more likely to know about the Legendre's three square theorem.

    So I give points for intelligence, points for sophisticatedness of the trap but take away points for mis-judging your targets.

    All in all I give this a 8/10. Wear it with pride.
    I have no idea about legendres square theorem but id have said this was a fairly easy question only difficult part is theres a fair amount of manipulation to do. Ive seen similar questions...actually i did the exact same question in year 12 i think with different numbers of course.
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    (Original post by rayquaza17)
    So you get to use maths in an actual job?? :zomg:
    Yeah. If you look at any of the more technical hedge fund or investment banks they tend to use lots of maths (and very highly value quantitative skills...there are various hedge funds that will only hire hire maths or physics PhDs)


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    (Original post by newblood)
    I have no idea about legendres square theorem but id have said this was a fairly easy question only difficult part is theres a fair amount of manipulation to do. Ive seen similar questions...actually i did the exact same question in year 12 i think with different numbers of course.
    It isn't an easy question, it is an impossible one.
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    (Original post by newblood)
    I have no idea about legendres square theorem but id have said this was a fairly easy question only difficult part is theres a fair amount of manipulation to do. Ive seen similar questions...actually i did the exact same question in year 12 i think with different numbers of course.
    It's not an easy question though, it's impossible (impossible to get an integer solution, that is). A couple of different people have posted about how no integer solutions exist.
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    (Original post by newblood)
    I have no idea about legendres square theorem but id have said this was a fairly easy question only difficult part is theres a fair amount of manipulation to do. Ive seen similar questions...actually i did the exact same question in year 12 i think with different numbers of course.
    The whole point of the theorem is that 39 can not be a result of x^2+y^2+z^2 if x, y and z are whole numbers. With decimals its possible but whole numbers won't ever give 39 as a result.
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    (Original post by james22)
    X
    You're definitely right in everything you've said.

    For example go to http://en.wikipedia.org/wiki/System_of_linear_equations

    The "number of equations" in a system is clearly as you defined it and not the rank of the system, however important the rank might be in dictating what the solution space looks like.

    The x+y=1 and ax+y=1 system you gave consists of two equations for all values of a. The rank is 2 except when a=1 in which case the rank is 1.
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    (Original post by james22)
    It isn't an easy question, it is an impossible one.
    (Original post by rayquaza17)
    It's not an easy question though, it's impossible (impossible to get an integer solution, that is). A couple of different people have posted about how no integer solutions exist.
    ''I have no idea about legendres square theorem but id have said this was a fairly easy question only difficult part is theres a fair amount of manipulation to do. Ive seen similar questions...actually i did the exact same question in year 12 i think with different numbers of course''
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    (Original post by james22)
    TBH I rarely do any linear algebra on systems of linear equations. But when I do I think of the rank as the degeneracy of the system, so higher rank means more degeneracy. This intuition applies nicely to geometry as well, where the rank of a matrix determines how nicely a vector field behaves near a point (e.g. maximal rank implies locally invertible).
    Do you really mean "higher rank means more degneracy"? I would have said the opposite, and this also tallies with your final comment about maximal rank implying locally invertible.
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    (Original post by RichE)
    You're definitely right in everything you've said.

    For example go to http://en.wikipedia.org/wiki/System_of_linear_equations

    The "number of equations" in a system is clearly as you defined it and not the rank of the system, however important the rank might be in dictating what the solution space looks like.

    The x+y=1 and ax+y=1 system you gave consists of two equations for all values of a. The rank is 2 except when a=1 in which case the rank is 1.
    Well obviously I disagree with you. His thought process is a flawed way of viewing algebra in my opinion.

    For the record, you also believe that

    x+y=1

    2(x+y)=2

    Is a two equation system?

    And that second system is deterministic in all cases except a=1 when it is a one equation system.

    I'll give you an example of what I mean. When you perform Gaussian elimination you are operating on a system of equations and as you shed equations you gain free variables. If you view systems as being of a fixed numbers of equations then you are losing the essence of what Gaussian elimination does. The number of equations in a system is dependent on the values that parameters take and to me it is crucial that is viewed that way.


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    (Original post by DFranklin)
    Do you really mean "higher rank means more degneracy"? I would have said the opposite, and this also tallies with your final comment about maximal rank implying locally invertible.
    Yes, that is what I meant.
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    (Original post by LightBlueSoldier)
    Well obviously I disagree with you. His thought process is a flawed way of viewing algebra in my opinion.

    For the record, you also believe that

    x+y=1

    2(x+y)=2

    Is a two equation system?

    And that second system is deterministic in all cases except a=1 when it is a one equation system.

    I'll give you an example of what I mean. When you perform Gaussian elimination you are operating on a system of equations and as you shed equations you gain free variables. If you view systems as being of a fixed numbers of equations then you are losing the essence of what Gaussian elimination does. The number of equations in a system is dependent on the values that parameters take and to me it is crucial that is viewed that way.


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    I am not disputing your understanding of the mathematics behind the linear algebra - I am disputing your language. You are using language in a way that runs contrary to every linear algebra book I have ever read. A linear system of m equations in n variables is precisely that; it doesn't matter whether the system is consistent or not, uniquely solvable or not. These are standard definitions and as far as I am concerned you are ignoring the standard language, irrespective of how well you understand the maths. (You obviously think Wikipedia is also wrong on this point as well then.)

    (For the record it is a two equation system for all values of a - its the rank that is 1 when a = 1.)
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    (Original post by LightBlueSoldier)
    Well obviously I disagree with you. His thought process is a flawed way of viewing algebra in my opinion.

    For the record, you also believe that

    x+y=1

    2(x+y)=2

    Is a two equation system?

    And that second system is deterministic in all cases except a=1 when it is a one equation system.

    I'll give you an example of what I mean. When you perform Gaussian elimination you are operating on a system of equations and as you shed equations you gain free variables. If you view systems as being of a fixed numbers of equations then you are losing the essence of what Gaussian elimination does. The number of equations in a system is dependent on the values that parameters take and to me it is crucial that is viewed that way.


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    Ignoring everything else, why do you think that my way of viewing it is a flaw in my understanding of algebra? The only difference between you and me is the terminology used to describe systems of equations.
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    (Original post by RichE)
    I am not disputing your understanding of the mathematics behind the linear algebra - I am disputing your language. You are using language in a way that runs contrary to every linear algebra book I have ever read. A linear system of m equations in n variables is precisely that; it doesn't matter whether the system is consistent or not, uniquely solvable or not. These are standard definitions and as far as I am concerned you are ignoring the standard language, irrespective of how well you understand the maths. (You obviously think Wikipedia is also wrong on this point as well then.)

    (For the record it is a two equation system for all values of a - its the rank that is 1 when a = 1.)
    I think you are misapplying the definitions. But anyway this argument is going nowhere. Neither of us is going to budge.

    (Original post by james22)
    Ignoring everything else, why do you think that my way of viewing it is a flaw in my understanding of algebra? The only difference between you and me is the terminology used to describe systems of equations.
    My use of the word flawed was harsh and not really what I meant. I just find your way of thinking quite odd. But if it works for you then great.


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    (Original post by LightBlueSoldier)
    I think you are misapplying the definitions. But anyway this argument is going nowhere. Neither of us is going to budge.


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    But some of us are going to provide references to show that our language is the one that is commonly employed.

    PS I hope I am not misapplying the definitions as I teach and lecture this material.
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    (Original post by RichE)
    But some of us are going to provide references to show that our language is the one that is commonly employed.

    PS I hope I am not misapplying the definitions as I teach and lecture this material.
    Can you provide a specific example in a linear algebra textbook, or even a piece of serious mathematics online, of someone carrying a redundant equation throughout the evaluation of a system?


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    (Original post by LightBlueSoldier)
    Can you provide a specific example in a linear algebra textbook, or even a piece of serious mathematics online, of someone carrying a redundant equation throughout the evaluation of a system?


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    Why ask that question? That is not what started this debate. The debated point was how many equations are there in the system

    x + y = 1
    2x + 2y = 2

    The answer is 2, I would suggest indisputably so. Wikipedia agrees with me; the multiple linear algebra texts I have agree with me.

    We all agree how we might go about parametrising that system's solution space. That doesn't justify you calling James wrong for saying entirely right things about that system.
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    (Original post by RichE)
    Why ask that question? That is not what started this debate. The debated point was how many equations are there in the system

    x + y = 1
    2x + 2y = 2

    The answer is 2, I would suggest indisputably so.
    I agree this is beyond dispute. I'd say it's beyond dispute even if the two equations are

    x + y = 1
    x + y = 1

    or even

    x + y = 1
    0x + 0y = 0

    Apart from the obvious "What is an equation? And how many of them have I just written down?" argument, there's the natural (and important) equivalence between sets of n equations in k unknowns and n x k matrix equations. For sure we can't start saying "that matrix doesn't have n rows because two of them are the same", so we really don't want to start saying that about sets of equations either.
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    (Original post by RichE)
    Why ask that question? That is not what started this debate. The debated point was how many equations are there in the system

    x + y = 1
    2x + 2y = 2

    The answer is 2, I would suggest indisputably so. Wikipedia agrees with me; the multiple linear algebra texts I have agree with me.

    We all agree how we might go about parametrising that system's solution space. That doesn't justify you calling James wrong for saying entirely right things about that system.
    I think you just completely ignored my point, which is that systems have variable numbers of equations. I don't know what level of linear algebra you ended up a studying but this is the only way of thinking about it. To argue that there is some kind of intrinsic measure of the number of equations in a system is totally flawed.

    I completely agree that when first look at a system with n equations, then it is an n equation system, but as you learn more about it it might become an m equation system, and r equation system. There are even stochastic processes iirc that force such a system to become a more than n equation system as it evolves.

    This is my last post on the matter. Clearly we do not agree, although I think that is rather more because you have misinterpreted what I am saying than because either of us are technically incorrect.


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