Another "proof"

[ultrafilter]

I ran into this on another message board:

The solution method shown for these simultaneous equations
yields an incorrect answer. Where is the logic flaw ?

x^2 + xy + y^2 = 3
x + xy + y = -1

Put both equations in terms of xy:

xy = -x^2 - y^2 + 3
xy = -x - y - 1

Since xy = xy, set these two equations equal to each other:

-x - y - 1 = -x^2 - y^2 + 3

Add x^2 and y^2 to both sides:

x^2-x + y^2-y - 1 = 3

Add 1 to both sides:

x^2-x + y^2-y = 4

Factor:

x(x-1) + y(y-1) = 4

Now, one solution should be (2,2):

2(2-1) + 2(2-1) = 4
2(1) + 2(1) = 4
2 + 2 = 4
4 = 4

So, we seem to have a solution, but then we try it in the
original equations:

x^2 + xy + y^2 = 3
2^2 + (2)2 + 2^2 = 3
4 + 4 + 4 = 3
12 = 3

x + xy + y = -1
2 + 2(2) + 2 = -1
2 + 4 + 2 = -1
8 = -1

So, it doesn’t check out in either equation. What went wrong ?
Where is the logic error ?

Note that the 3 correct solutions (derived by other means) are
(2,-1), (-1,-1), and (-1,2). These solutions fit the method shown
above as well. So this method finds the correct solutions, and
some junk solutions, two. Are these “imaginary solutions” or
something comparable ?

I think I kinda understand what’s going on here. By eliminating the xy term from the second equation, you’re reducing its degree from two to one. That’s an irreversible step, so it’s not guaranteed to preserve the solution set of the original two equations.

Does that sound right, or is there something else I’m missing here?

[tim314]

The problem is definitely in going from
xy = -x^2 - y^2 + 3
xy = -x - y - 1
to
-x - y - 1 = -x^2 - y^2 + 3

since that’s the point at which (x,y)=(2,2) suddenly becomes a solution.

The way I see it, we can only eliminate an equation if we completely incorporate its content into the remaining equation. That’s kind of sloppy language, but what I mean is, we can view the equation:
xy = -x - y - 1
as a constraint on the value of y. For x not equal to zero, we can just take -1 -y/x -1/x and plug it in for y everywhere we see it in the first equation. If we do that and eliminate y, then we can throw away our constraint on y. But what’s done in this proof is we’ve basically done the substitution on the left side of the equation but not on the right side. So we haven’t completely incorporated equation 2 into equation 1, but we just threw it away anyway (thus introducing new solutions).

Is this the same as what you’re saying? I’m not totally happy with the way I’m phrasing it, but I think it boils down to a problem of throwing away a constraint without eliminating the variable it constrains. Someone else can probably state it better.

[bitwise]

@ultrafilter

I ran into this on another message board:

I think I kinda understand what’s going on here. By eliminating the xy term from the second equation, you’re reducing its degree from two to one. That’s an irreversible step, so it’s not guaranteed to preserve the solution set of the original two equations.

Does that sound right, or is there something else I’m missing here?

What the argument shows is that

x(x-1) + y(y-1) = 4

is a necessary condition for

x^2 + xy + y^2 = 3
x + xy + y = -1

That does not mean it is sufficient.

Whether a “reduction in degree” is responsible for the irreversibility, I am not sure. I am not really sure what you mean by degree here. All of following are polynomials of degree 2 in two in x and y:

x(x-1) + y(y-1) - 4
x^2 + xy + y^2 - 3
x + xy + y + 1

[bitwise]

@bitwise

What the argument shows is that

x(x-1) + y(y-1) = 4

is a necessary condition for

x^2 + xy + y^2 = 3
x + xy + y = -1

That does not mean it is sufficient.

Whether a “reduction in degree” is responsible for the irreversibility, I am not sure. I am not really sure what you mean by degree here. All of following are polynomials of degree 2 in two in x and y:

x(x-1) + y(y-1) - 4
x^2 + xy + y^2 - 3
x + xy + y + 1

Ignore the bolded part; that was a typo.

[bitwise]

A simpler example:

The solutions to

x^2 = xy
y^2 = xy

must satisfy x^2 = y^2, which is true whenever x and y have the same absolute value. However, for x and y, to satisfy the original two equations, they must also have the same sign and thus must be equal.

[Chronos]

Here’s an even simpler example: Consider the equations x + y = 3 and x - y = 1. I can combine these equations to give the equation x - y = 1 (this is, of course, a very trivial combination, but it’s perfectly valid). One solution to this equation is x = 11, y = 10, but this does not satisfy our original set of equations. This is not surprising, and neither should the OP’s example be.

[ftg]

Some good other examples given, but I’d like to highlight the root (!) of the problem. Take:

If c is a solution to x=1 then c is also a solution to x[sup]2[/sup]=1.

I highlighted the “if” and “then” to make it clear that an implication is being made. One extremely common failing, even for those trained in logic, is to mentally try to reverse an implication. (Infer the converse instead of the contrapositive.) What makes the OP’s example seemingly a puzzler is that the implication part is not so clearly stated and that we are used to performing many algebraic manipulations that are of the “if and only if” variety.

[ultrafilter]

@ftg

Some good other examples given, but I’d like to highlight the root (!) of the problem. Take:

If c is a solution to x=1 then c is also a solution to x[sup]2[/sup]=1.

I highlighted the “if” and “then” to make it clear that an implication is being made. One extremely common failing, even for those trained in logic, is to mentally try to reverse an implication. (Infer the converse instead of the contrapositive.) What makes the OP’s example seemingly a puzzler is that the implication part is not so clearly stated and that we are used to performing many algebraic manipulations that are of the “if and only if” variety.

Yeah, that’s what I ended up going with as a response over there. The trick isn’t so much seeing the problem as it is stating the general principle, which I think you do need some formal logic to come up with.

[nivlac]

The basic lesson from all this is to be careful when you apply totally legitimate algebraic operations when you seek the solution to an equation or set of equations. In the process, spurious solutions can crop up. That’s why I always preach the need to always plug in the “solutions” into the original equation(s) to make sure that they really work.

[Mathochist]

@ultrafilter

I ran into this on another message board:

I think I kinda understand what’s going on here. By eliminating the xy term from the second equation, you’re reducing its degree from two to one. That’s an irreversible step, so it’s not guaranteed to preserve the solution set of the original two equations.

Does that sound right, or is there something else I’m missing here?

More or less right. The degree’s not so much the essential bit, though. I think you’d get the same effect if you homogenized it to consider projective varieties. Basically, it’s that they turned the two polynomials into one and then just solved that. Plugging one expression for xy from one polynomial into the other polynomial gives another polynomial which will also be vanish on the zero set of both of the original polynomials, but that doesn’t mean that its zero set is the zero set of the first two.

[Tyrrell_McAllister]

@Chronos

Here’s an even simpler example: Consider the equations x + y = 3 and x - y = 1. I can combine these equations to give the equation x - y = 1 (this is, of course, a very trivial combination, but it’s perfectly valid). One solution to this equation is x = 11, y = 10, but this does not satisfy our original set of equations. This is not surprising, and neither should the OP’s example be.

Even simpler: consider the system
x = 1,
x = 2.
Adding these equations yields 2x = 3. Dividing both sides by 2 yields x = 3/2.

But x=3/2 doesn’t solve the original system.

More or less right. The degree’s not so much the essential bit, though. I think you’d get the same effect if you homogenized it to consider projective varieties. Basically, it’s that they turned the two polynomials into one and then just solved that. Plugging one expression for xy from one polynomial into the other polynomial gives another polynomial which will also be vanish on the zero set of both of the original polynomials, but that doesn’t mean that its zero set is the zero set of the first two.

Mathocist, you never cease to amaze me with your readiness to dive into details more technical than the matter at hand requires :).

As others have stated, the problem is that you can’t deduce that “If A then B” by showing that “If B then A”. The error is purely at the logical level. The particulars of algebraic geometry are incidental.

[Mathochist]

@Tyrrell McAllister

Even simpler: consider the system
x = 1,
x = 2.
Adding these equations yields 2x = 3. Dividing both sides by 2 yields x = 3/2.

But x=3/2 doesn’t solve the original system.

Mathocist, you never cease to amaze me with your readiness to dive into details more technical than the matter at hand requires :).

As others have stated, the problem is that you can’t deduce that “If A then B” by showing that “If B then A”. The error is purely at the logical level. The particulars of algebraic geometry are incidental.

I beg to differ. Really, any logic is only as good as the model of predicate calculus at hand. In this case, that associated to algebraic sheaves of sets over projective space; that is, maybe in another case implication is equivalent to its converse, but in this context it’s not :D.

Seriously, though, there’s only one mathematics out there, so there’s no such thing as “too technical”. It’s really just a different way of thinking about the same problem. You prefer a thousand different techniques for a million different problems while I prefer one natural, powerful framework that sees them all as special cases of one solution for a single class of problems.

[Chronos]

Even simpler: consider the system

x = 1,
x = 2.

No, that’s not an example. That’s an inconsistent system, one for which there exists no solution (or rather, for which if there is a solution, then I’m the Pope and all of mathematics goes tumbling down to its foundations). If you changed it to x = 1, y = 2, though, then you’d have a valid example. You could combine those to get x + y = 3, for which one valid solution (which is not a solution of the original equation) is x = 4, y = -1.

[Tyrrell_McAllister]

@Chronos

No, that’s not an example. That’s an inconsistent system, one for which there exists no solution (or rather, for which if there is a solution, then I’m the Pope and all of mathematics goes tumbling down to its foundations). If you changed it to x = 1, y = 2, though, then you’d have a valid example. You could combine those to get x + y = 3, for which one valid solution (which is not a solution of the original equation) is x = 4, y = -1.

Eh, I maintain that my example captured the most significant error of the original problem. The point is that we began with stronger conditions, deduced weaker conditions, and then found that something that satisfied the weaker conditions failed to satisfy the stronger conditions. I don’t think it matters that I began with conditions so strong that nothing could satisfy them.