[If you absolutely can't wait for a challenge problem, go to the bottom, but it might be worth reading through this first...]
Absolute value equations and inequalities, in particular, are notoriously difficult for most students. If you are a math educator, is this a topic you relish?
So what method of solution usually works best for the student? What method of presentation is most effective for the instructor? If the equations are straightforward such as |x+3| = 7, most students seem comfortable with expressing the equation as a disjunction: x+3 = 7 or
x+3 = -7. Some instructors, in preparing students for the technical definition (using cases), require the student to express this as x+3 = 7 or -x-3 = 7. Rarely have I observed instructors introduce the full-blown piecewise definition using cases early on in algebra:
x+3 = 7 if x>=-3
OR
-x-3= 7 if x<-3. This is generally believed to be too sophisticated for an introductory treatment. Motivating the technical definition of |x| usually comes later on in Algebra 2. However, I have always been a bit uncomfortable teaching the traditional algorithm for absolute value inequality problems such as |x+3| "<" 7 which leads to the conjunction x+3 ">" -7 AND x+3 "<" 7 or, in combined form, -7 "<" x+3 "<" 7. Math instructors devise creative mnemonics to help students recall the procedure. This all begins when prealgebra students are exposed to the verbal description of the piecewise definition of the absolute value function:
The absolute value of a positive number is that number and the absolute value of a negative number is its opposite. The absolute value of zero is zero.
This is immediately followed by a number of numerical examples, guided practice with some more complicated variations involving mixed operations and an assignment. The student sees this as another example of something they're supposed to learn in math without much meaning attached. Most catch on to the idea by repetition and errors are generally caused by weaknesses with signed numbers or order of operations.
However, some educators prefer the 'distance' interpretation of absolute value to make this notion more meaningful. Technically one has to distinguish between the real number, x, and the graph of x on a number line, but this distinction is often sacrificed for clarity:
The absolute value of x is its distance from zero.
Thus, both 8 and -8 have the same absolute value because they are the same distance from zero on a number line. Since distance cannot be negative, we have a powerful visual model that students can use to make sense of this idea, although it seems limited when solving more complicated equations or inequalities later on.
Stay tuned for more on this topic (including a 2-dimensional graphing approach using functions), but, since some readers are disappointed if there is no challenge problem, here's one for you. It's not that difficult, a version of this has recently been tested on the SAT and it can be approached in a variety of ways (formal algorithm, guess-test, etc.), but I will challenge you to solve it using the distance model! Don't hesitate to take strong exception to my comments above!
For how many positive integer values of x is
|x-2007| > x?
Tuesday, April 24, 2007
Absolute Zero Part I: Do Absolute Values Leave Some Out in the Cold?
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3 comments:
Dave, for |x + 3| = 7 I do a number of things.
1st, geometric approach. This is a 1 dimensional circle, center at -3, radius 7. Easy to plot.
2nd, algebraic. number line under the inequality, -3 circled, vertical line descending from -3. On the left side, -(x+3) = 7, on the right x+3 = 7. Each solution must be checked to see that it is consistent with the "side" it is on.
You read once, I think, how I extend this to abs. val. ineq.s, right?
The analysis I would suggest to a student would be:
Problem: How many positive integers x satisfy |x − 2007| > x.
Case 1: x − 2007 ≥ 0.
Then, the inequality reduces to x − 2007 ≥ x, which is obviously impossible.
Case 2: x − 2007 < 0.
This implies that 0 < x < 2007. Then, the inequality becomes
−(x − 2007) > x and 0 < x < 2007
2007 − x > x and x < 2007.
2007 > 2x and 2007 > x > 0.
Since x is positive, the first inequality subsumes the second:
2007 > 2x > 0.
Divide by 2:
1003½ > x > 0.
Since x is an integer, the solution set is the integers 1, 2, …1003, and the solution is 1003.
Now, let's look at a more interesting item for your students to look at considering your last few assignments. One thing mathematicians have worried about is metrics. A metric on a set S is a function d:S×S→ R such that
d(x, y) = 0 if and only if x = y,
d(x, y) = d(y, x), and
d(x, y) + d(y, z) ≥ d(x, z). [Triangle inequality]
You can discuss the metric topology on a metric space; open sets have a base given by the open balls:
B(x, ε) = {y: d(x, y) < ε}
Then, you can discuss metrics with equivalent topologies. Compare the Euclidean metric on the plane to the L^{1} or taxicab metric,
d((x,y), (z, w)) = |x−z| + |y−w|.
Then, ask your students what sort of interesting metrics you could have on the integers with two additional property: d(a, b) = d(a+c, b+c) [translation invariance], and d(0, ab) = d(0, a)d(0, b) [scaling].
It turns out that essentially you have two types of metrics: the Euclidean metric d(a, b) = |a−b| is one. the other is the family of "p-adic metrics", where p is a prime:
d(a, b) = 1 if p does not divide a−b.
d(a, b) = 1/p^{n} if p^{n} divides a−b, but p^{n+1} does not.
d(a, a) = 0. Essentially, every power of p divides 0, so the limit of 1/p^{n} as n goes to infinity is zero.
For the computer buffs among your students, consider |x|_{2} = d(0, x) where x is any number, and d is the 2-adic metric. The distance reflects the number of zero-bits in the rightmost part of the expansion of x. Multiply x by two, and you get another zero, and the distance is halved.Do this enough times (16 times if it's a 16-bit integer), and the zeros take up the entire machine word, so that the number on the computer is now 0.
The idea of 'large numbers' having 'small distances' might be alien to your students, but the fact that this is important in mathematics might pursuade them that something is going on.
You can extend this from Z to Q easily:
|r|_{p} = p^{−n} if r = p^{n}a/b, and both a and b are not divisible by p.
Now, the completion of Q with the Euclidean metric is the real numbers R. What is the completion of the rationals under the p-adic metric? You get numbers of the form
1 + 3p + 2p^2 + 5p^3 + … + 10 p^121 + …
That should shock your students.
jonathan--
I really like the center, radius model for |x - a|= r. It's so much easier for students to make sense of, retain and apply in many situations. Your algebraic approach is also excellent since it is the basis for solving polynomial inequalities as well ('sign charts') that is needed in calculus. I do seem to recall how you did this for inequalities, but I'll have to look that up. The algebraic approach is generally required for more complicated abs values equations and inequalities, particularly those involving more than one expression with abs values. I may post a challenge of this type later.
eric--
your solution of the challenge problem by cases was dead on and extremely thorough. The rest of your post brought back memories of my general topology course from the ancient past! Metrics, pseudo-metrics, p-adic,... Just a tad beyond most of my students!
Ok, here's an outline of a 'distance' solution for the challenge:
Rewrite the inequality as
|x-2007|>|x-0| since x is positive.
This can then be interpreted as x is CLOSER to 0 than to 2007! The solution follows easily. Most students, IMO, who solve this problem, would 'plug in' x = 1, 2, 3, etc., and come to the same conclusion. Not that there's anything wrong with that...
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