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

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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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.

So you get to use maths in an actual job??

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.

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.

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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It isn't an easy question, it is an impossible one.

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.

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).

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.

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.

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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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.)

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.

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.

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.