# Adding by a Form of 0: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on adding by a form of 0 (analogous to multiplying by a form of 1).

Part 1: Introduction.

Part 2: The Product and Quotient Rules from calculus.

Part 3: A formal mathematical proof from discrete mathematics regarding equality of sets.

Part 4: Further thoughts on adding by a form of 0 in the above proof.

# Adding by a Form of 0 (Part 3)

As part of my discrete mathematics class, I introduce my freshmen/sophomore students to various proof techniques, including proofs about sets. Here is one of the examples that I use that involves adding and subtracting a number twice in the same proof.

Theorem. Let $A$ be the set of even integers, and define

$B = \{ n: n = m+1 for some odd integer m\}$

Then $A = B$.

Proof (with annotations). Before starting the proof, I should say that I expect my students to use the formal definitions of even and odd:

• An integer $n$ is even if $n = 2k$ for some integer $k$.
• An integer $n$ is odd if $n = 2k+1$ for some integer $k$.

To prove that $A = B$, we must show that $A \subseteq B$ and $B \subseteq A$. The first of these tends to trickiest for students.

Part 1. Let $n \in A$. By definition of even, that means that there is an integer $k$ so that $n = 2k$.

To show that $n \in B$, we must show that $n = m + 1$ for some odd integer $m$. To this end, notice that $n = (n-1) + 1$. Thus, we must show that $n - 1$ is an odd integer, or that $n -1$ can be written in the form $2k+1$. To do this, we add and subtract 1 a second time:

$n = 2k$

$= (2k - 1) + 1$

$= ([2k - 1 - 1] + 1) + 1$

$= ([2k-2] + 1) + 1$

$= (2[k-1] + 1) + 1$.

By the closure axioms, $k-1$ is an integer. Therefore, $2[k-1] + 1$ is an odd number by definition of odd, and hence $n \in B$.

The above part of the proof can be a bit much to swallow for students first learning about proofs. For completeness, let me also include Part 2 (which, in my experience, most students can produce without difficulty).

Part 2. Let $n \in B$, so that $n = m + 1$ for some odd integer $m$. By definition of odd, there is an integer $k$ so that $m = 2k+1$. Therefore, $n = (2k+1) + 1 = 2k+2 = 2(k+1)$. By the closure axioms, $k +1$ is an integer. Therefore, $n$ is even by definition of even, and so we conclude that $n \in A$.

$\square$

For what it’s worth, this is the review problems for which I recorded myself talking through the solution for the benefit of my students.

In my opinion, the biggest conceptual barriers in this proof are these steps from Part 1:

$2k = (2k - 1) + 1 = ([2k - 1 - 1] + 1) + 1$.

These steps are undeniably awkward. Back in high school algebra, students would get points taken off for making the expression more complicated instead of simplifying the answer. But this is the kind of jump that I need to train my students to do so that they can master this technique and be successful in their future math classes.

TED-Ed made a very good video describing the Infinite Hotel Paradox, a thought experiment to describe how injective (one-to-one) functions can be used to examine countably infinite sets.

# Mathematicians Explain Sports to Each Other

Just when I think I’ve heard every math pun under the sun, Math With Bad Drawings comes up with a new one. This week’s edition is called “Mathematicians Explain Sports to Each Other.” The one on basketball is my favorite.

To see the others (on tennis, soccer, golf, baseball, hockey, marathon, tee ball, cricket, and skiing), click here.

# Another poorly written word problem (Part 6)

Textbooks have included the occasional awful problem ever since Pebbles Flintstone and Bamm-Bamm Rubble chiseled their homework on slate tablets while attending Bedrock Elementary. But even with the understanding that there have been children have been doing awful homework problems since the dawn of time (and long before the advent of the Common Core), this one is a doozy.

There’s no sense having a debate about standards for elementary mathematics if textbook publishers can’t construct sentences that can be understood by students (or their parents).

This one makes my blood boil. According to its advocates, the whole point of the Common Core standards was to increase the rigor in secondary mathematics. However, this one is SIMPLY WRONG.

The textbook does correctly note that the proper definition of a function is a set of ordered pairs. The “correct” answer, according to the textbook, is answer G — the plotted points do not match the ordered pairs.

However, answer H is also wrong. The textbook would have students believe that order is important when listing the elements of a set. However, order is not important — the domain of $\{-3, 1, -1, 3\}$ is the same as $\{-3, -1, 1, 3\} or$latex \{3, -3, -1, 1\}\$. This is standard mathematical notation — in an ordered pair (or ordered $n-$tuple), the order is important. For a set, the order is not important.

Specifying that the domain is $\{-3,-1,1,3\}$ and the range is $\{2,5,8,11\}$ does not uniquely determine the function. In fact, there are 24 different functions that have this domain and range (where we distinguish between the range of a function and its codomain).

In other words, in trying to be clever about properly defining a function and showing different representations of a function, the textbook promotes a misconception about sets… which makes me wonder if the textbook’s attempt at trying to be ultra-careful about the definition of a function is really worth it.