melanoman: (Default)
This morning I spent the better part of an hour writing a review on Walt Hickey's recent article about controversial math facts and then promptly deleted my work by trusting the browser with too much text and not saving partial work. I'm going to rewrite the part of the article I actually care about and leave the rest of the review in the bitbucket.

I'm only going to look at two of the answers, #3 and #9. (slide 14 and slide 49 for those who don't want to read the whole presentation).

Let's take #9 first. This is a proof of why the harmonic summation (let's call that H) diverges to positive infinity. Hickey takes a similar summation (let's call it A) and expresses it in two forms. One form shows that H is obviously greater than A. The other form shows that A is obviously infinite. The proof concludes that if H is greater than something infinite, it must also be infinite. Great explanation, solid proof; I have no complaints here.

But then there is #3. In this section, Hickey claims that there are the same number of positive even numbers as counting numbers. He sets up a 1-to-1 correspondence between the to sets and concludes that they are both countably infinite and therefore it anyone who says that one set has more elements than the other is wrong.

So close, but fail. The fundamental misunderstanding here comes from the reflexive property of numbers which states
Every number is equal to itself
This isn't exactly the most controversial property ever. The problem is when you try to apply that rule to infinity, it doesn't work because infinity is not a number.

In problem 9, the key step of the proof was that H > A. Since H was infinity and A was infinity, that means that in this case infinity > infinity, not infinity = infinity. This is fine, since infinity is not a number and therefore doesn't have to follow the reflexive property.

Applying the same logic to problem 3, we see that it makes no sense to say that there is the same number of elements in the sets because both cardinalities (set sizes) are infinite. The moral of the story is that infinity doesn't play nice with concepts like equals, addition, multiplication and the like, so be careful about treating it like a number, or before you know it you'll be claiming that infinity minus infinity is zero or infinity divided by infinity is one, which turn out to be variants of the same mistake Hickey made.
melanoman: (Default)

Solving Inequalities with One Absolute Value (II)


These steps explain how to solve an inequality with one absolute value and one variable when there are variable terms both inside and outside the absolute value sign.


Step 1: Isolate the absolute value


Exactly like Part I of this lesson.


3 - | x – 3 | > - x           è - | x – 3 | > - x – 3               è | x – 3 | < x + 3

2 | y2 – 4 | > 6y è | y2 – 4 | > 3y


Step 2: Is the expression in the absolute value negative or nonnegative?


This can be a difficult step for complicated expressions, but for the simple polynomials we are usually working with, there is an easy way.  Solve the related equation where the expression is equal to zero.  This will tell you where the graph crosses zero.  Mark the variable in this related equation so you don’t confuse it with the original problem.


x0 – 3 = 0         è x0 = 3

y02 – 4 = 0        è y0 = -2 or 2


Divide the domain into sections using the roots above and mark each section as positive or negative by plugging in a test value.


x – 3    è negative where x < 3, nonnegative where x >= 3

y2 – 4   è nonnegative where y < -2, negative where -2 < y < 2, nonnegative where y > 2


Step 3: Split the problem into sections


Write a separate inequality for the negative section and the positive sections.  Mark the variables so you don’t mix the two up.  To make the positive inequality, just drop the absolute value sign.  To make the negative inequality, multiply the contents of the absolute value sign by -1.  If step 2 showed that your expression was always positive or always negative, you only need to make one inequality.


xp - 3 < xp + 3

3 – xn < xn + 3

yp2 – 4 > 3yp

4 – yn2 > 3yn


Step 4: Solve using basic algebra


xp - 3 < xp + 3              è -3 < 3                     è true (everything)

3 – xn < xn + 3              è 0 < 2xn                    è xn > 0


yp2 – 4 > 3yp                 è yp2 – 3yp – 4 > 0     è (yp – 4) (yp + 1) > 0  è yp > 4 or yp < -1

4 – yn2 > 3yn                 è - yn2 – 3yn + 4 > 0   è (1 – yn) (yn + 4) > 0  è -4 < yn < 1


Step 5: Take the intersection for each section


Now we combine the domain notes from step 2 with the partial solutions from step 4 by taking the intersection for each section.  If you got an always-true result in step 4, take the entire corresponding section.  If you got an always-false result, take nothing for that section.  At this point we can drop our special markings for the sections.  Be extra careful when the section has an OR in it --- you may want to sketch a graph in two colors.


xp >=3 AND true         è x >= 3

xn < 3 AND xn > 0       è 0 < x < 3


(yp <= -2 OR yp >= 2) AND (yp > 4 OR yp < -1)         è y <= -2 OR y > 4

(-2 < y < 2) AND (-4 < yn < 1)                                    è -2 < y < 1


Step 6: Take the union of the two sections


0 < x < 3 OR x>=3                              è x > 0

y <= -2 OR y > 4 OR -2 < y < 1          è y < 1 or y > 4


melanoman: (Default)

Solving Inequalities with One Absolute Value (I)


Step 1: Isolate the Absolute value


Treat the absolute value signs like a set of parenthesis that you can’t break, or like the whole thing was one variable, then solve for that on the left side.  Remember to flip the inequality if you multiply by a negative number.


1-2|x-3| > -5  è -2|x-3| > -6 è |x-3| < 3


If the right side is a constant, solve using the steps below.  If the right side includes a variable, this problem requires special handling (learn how in part II of this lesson)


Step 2: Is the right side negative or zero?


If the right side is or could be negative, this problem needs special treatment.  If the right side is a negative constant, then the answer will either be “no solution” or “all real numbers” depending on whether the equality is “greater than” or “less than”


|x| < -1             è no solution

|x| <= -1           è no solution

|x| > -1             è all real numbers

|x| >= -1           è all real numbers


The same goes for a right side of zero, UNLESS the inequality is “less than or equal.” In this very special case the contents of the absolute value equal zero.


|x| <= 0            è x = 0


Step 3: Get rid of the absolute value marker


Now we split the equation into two parts.  We will compare the left side without the absolute value markers to the right side and it’s opposite.  If the inequality uses “less than” then we bound the contents of the absolute value marker between the right side of the equation and it’s opposite.  If the inequality uses “greater than” then we split the equation into the union of two inequalities.


|x+1| < 3          è -3 < x +1 < 3

|x| <= 5            è -5 <= x <= 5

|x-3| > 7           è x -3 > 7 OR x -3 < -7

|x| >= 13          è x >= 13 OR x <= -13


Step 4: Simplify (if needed)

Use basic algebra to simplify as needed.


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