Tuesday, October 23, 2007

And vs. Or: Developing Deeper Understanding in Algebra

While some are awaiting the remaining chapters in Alec Klein's interview, I know the rest of my readers are wondering why I've gone into early hibernation. I'm actually doing some independent mathematical research apart from anything on this blog but I will not bore you with the details at this time. That may change (i.e., I may bore you later!).

In the meantime, since the central theme of this blog has always been developing student conceptual understanding, here are a couple of problems for you to consider giving to your Algebra 2 students (or beyond). Conjunction vs. Disjunction is often misunderstood by students and these ideas appear in so many contexts in mathematics, from absolute values to inequalities and beyond. Consider giving these as warm-ups, for review, practice for SAT's, etc. I'm not suggesting these are difficult or challenging problems. Their purpose is to promote deeper reflection on the part of the student. Students who have strong background and understanding will simply solve these quickly and not see why anyone would make a big deal over them. However, you may find other students who don't grasp the ideas as readily or have forgotten. Comparing/contrasting is a powerful heuristic when trying to develop a more profound understanding of mathematics...

1. If (a -4)2 + (b+4)2 = 0, what is the least possible value of a2 + b2?

2. If (a-4)(b+4) = 0 , what is the least possible value of a2 + b2?
(A) 0 (B) 8 (C) 16 (D) 32 (E) 64

Ask students to explain to each other, why the word 'least' is irrelevant in Question 1 but not in Question 2. Also, how does question 1 relate to circles?


Totally_clueless said...

Nice problems. The insight is quite interesting.

The obvious generalizations are:
(1) How does question 2 relate to hyperbolas?

(2) Instead of the RHS being 0 in the two equations, change it to some value k and repeat the problem. Find a geometric interpretation in each case (which might help solve the problem in an easier fashion also).

(3) For a given curve f(x,y)=0, describe a method how to find the point on the curve closest to the origin.


Eric Jablow said...

You can extend this puzzle to Diophantine problems, though things can get tricky. The interesting point is that it isn't obvious to get 'not'. You instead have to rework the equations.

For example, n is composite if n = ab, and neither a nor b equals 1. But a≠1 and b≠1 aren't easily expressible as a Diophantine equation. So, you have to resort to techniques like

n is composite if n = (a^2+b^2+c^2+d^2+2)(e^2+f^2+g^2+h^2+2) for integers a–h. Every nonnegative integer is the sum of 4 squares, you know.

And then, you can mention without proof the result of Hilbert's tenth problem. That should confuse any high school student nicely, I guess.

Dave Marain said...

Nice generalizations, tc! Let's pursue this a bit further...

Start with xy = 0. The graph of this eqn in 2 variables is of course the coordinates axes whose equations are x=0, y=0.
Note that these lines can be obtained mechanically by starting with the rectangular hyperbola xy = 1 (or, as you suggested, xy = k) and replacing the right-hand side by zero. (This technique will appear more than once in this comment.) This 'method' produces the asymptotes x=0 and y=0. Why does it work? Solving for y, we obtain
y=1/x (or y=k/x). The horizontal asymptote is determined by allowing x to increase without bound, which leads to y=0. Similarly, for the other asymptote. Thus xy=0 can be thought of as a pair of lines, the asymptotes of the rectangular hyperbola xy=1 or a degenerate hyperbola!

Now consider the given problem, #2:
(a-4)(b+4) = 0 or
(x-4)(y+4) = 0.
One may think of this as a translation of xy = 0 to the 'origin' (4,-4). Thus again our graph is a pair of lines (or asymptotes or a degenerate huperbola). Thanks, tc, for suggesting this analytic geometry interpretation.

The original issue of disjunction sets up a nice interpretation of horizontal and vertical lines:
(x-4)(y+4)=0 leads to the disjunction:
(x=4, y can be any real) OR
(y=-4, x can be any real).
The first possibility is an interesting way of reinforcing the concept of a vertical line. Similarly for the 2nd possibility. Of course, the disjunction allows both factors to be zero, giving us the 'and' solution (4,-4). Remember the good old days when we taught union vs. intersection...

By the way, precalc students are often surprised by the 'cheating' method for finding asymptotes of the hyperbola (x^2)/(a^2) - (y^2)/(b^2) = 1, namely, replacing the right side by zero. As a challenge, ask them why this works. The traditional method is to solve for y and manipulate the radical, however, there's an easier way:

Since the horizontal asymptote is obtained by having x go to infinity, it makes sense to divide both sides by x^2:
As x goes to infinity, we obtain:
(b^2)(a^2) - (y^2)/(x^2) = 0 or
(x^2)/(a^2) - (y^2)/(b^2) = 0, voila!

Dave Marain said...

Nice, Eric! I was posting my comment but you beat me to it. I really feel we should rename that to Eric's Tenth Problem. And I was concerned that someone might now see an application for these two problems today -- how foolish of me!

cecil kirksey said...

To TC:
Re your number 3. Are you suggesting that this is simple? It generally involves taking the derivative of the distance and setting to zero and solving. Since the resultant equation can be anything but simple I would appreciate hearing what your solution would be.

To DM:
I guess I am missing the point your are trying make with your questions. If the answers are not 32 and 32 then I am really in the dark!!!

Dave Marain said...

I agree with you on #1 since the condition implies that both quantities must equal zero. That is, a = 4 AND b = -4.

However, the 2nd one is more subtle since either a-4 OR b+4 must equal zero, but it is not necessary for both to be zero at the same time. Thus, for example, we can choose a = 4 and b = 0. This will minimize a^2+b^2, so the least value is 16.

I believe tc's part 3 can be handled by ordinary differential calculus. If s^2 = x^2+y^2, then (2s)ds/dx = 2x + 2y(dy/dx). From f(x,y)=0, we can obtain an expression for dy/dx by implicit differentiation. The other option is partial differentiation but I'll leave it to tc to clarify.

Totally_clueless said...

Hi Cecil & Dave,

For (3), I was thinking more on the lines of a geometric description of the method on the lines of where the tangent to the curve is perpendicular to the line from the origin. As Cecil says, solving the general equation may turn out to be quite difficult.

Rephrasing the question: For a given curve f(x,y)=0, show that at the point closest to the origin, the tangent is perpendicular to the line from the origin to the point.

This can easily be shown using calculus and minimizing the distance from the origin.



Totally_clueless said...

On a related note, a problem that stumped me (only once) in high school:

Find the derivative of the function

y=x^3-3x^2+2x-1 at the point (2,3)



mathmom said...

tc, is the "point" that that's a trick question?

Eric Jablow said...


That's why the sticklers among us distinguish between the graph of a function and the function itself.


It's interesting that for the ordinary real algebraic case you can turn f(x)≥0 into an equation by adding a variable: f(x) -t^2 = 0. However, you can't turn f(x) > 0 into an equation by adding a variable. If f is continuous, roughly speaking the set where f(x) ≥0 is the closure of the set where f(x)>0. If you could have a F such that F(x,t) = 0 implied that f(x) > 0, then you could do this: choose a point where f is zero, and find a sequence of points tending to it where f>0. Now plug them into F: F(x1, t1), F(x2, t2), .... The t values should tend to a limit t, and F(x,t) = 0, though that isn't a proof.

I was thinking: the best lesson in classical mathematics you could base on Diophantine equations is the problem of finding all primitive Pythagorean triples using two different methods.

Number theory let a^2 + b^2 = c^2, a and c odd, b even. Then, a^2 = (c-b) (c+b), and relative primality shows that a=mn, c-b= m^2, c+b = n^2.

Geometry: let a^2 + b^2 = c^2. Then let x = a/c, y=b/c. x^2 + y^2 = 1. x and y are rational, so the line from (x,y) to (1,0) has rational slope. The converse is true too. A line through (1,0) with rational slope meets the circle in two points: (1,0) and a point with rational coefficients. Just combine x^2+y^2=1 with y=m(x-1) with m rational. The resulting quadratic in x has rational coefficients and has a root at x=1. By results on the sum and the products of roots, both roots are rational. y is then rational too.

Totally_Clueless said...

Hi Mathmom,

As with most things (particularly in math), it is obvious once you know it.


Dave Marain said...

I enjoyed your 'derivative at a point' problem! Eric's 'point' about textbooks or teachers using technically correct language is actually a critical message.

Also, tc, I've taught the 'normal' solution for finding the point closest to the graph of f(x,y)=0 for years and it's one of my favorites. Thanks for reminding me. One has to be careful about this method however. If the point is (0,0), for example, then the line joining a point (x,y) on the graph to the origin has slope y/x. I always cautioned students that a 'possible' exceptional solution to our problem occurs when x=0. This must be handled separately from solving y/x = -dx/dy. Further, all the 'solutions' of this equation along with x=0 need to then be tested to see which one(s) are in fact closest to the origin. This is still less cumbersome than the traditional solution and much more meaningful to students who 'see' it geometrically. You're motivating me to post a calculus investigation of this nature, requiring both methods and solving it in terms of a parameter. I may just do that!

I've never seen the parametric approach to solving f(x)≥ 0 -- that is 'way cool'! I am familiar with your 2 approaches to solving for primitive triples. Students rarely get to see any of this except perhaps as an enrichment in honors geometry or in a math club, but this is the kind of 'richer' experience mathmom has been suggesting for talented students, instead of simply accelerating them through calculus by 10th grade. If we could only get that message across!

Now, Eric, several thoughts ran through my brain as I was reading another of your exceptionally clear and profound comments last night.
(1) If I ever compile some of my favorite posts into a book, I would have to entitle it: Professor Jablow's Insights. I would only get credit for the foreword! Seriously, the breadth and depth of your contributions would make for one heck of a resource for teachers and highly motivated students.
(2) As much as I've enjoyed interviewing Prof. Steen and Alec Klein, I think my readers would relish my interviewing you! I've wanted to do this with a professional mathematician like yourself and, in my mind, you would be perfect, even though you might not be actively engaged in research at this stage of your life. Your story may serve as inspiration for some student or teacher somewhere... Are you interested? I won't be offended if you aren't. I would use a chat format with AIM or something like that which would make it easier for me to copy and paste your replies. My interview with Alec is taking me forever to transcribe. Again, Eric, I fully understand if you would not be comfortable with this idea.