Adding by a Form of 0 (Part 4)

In my previous post, I wrote out a proof (that an even number is an odd number plus 1) that included the following counterintuitive steps:

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

A common reaction that I get from students, who are taking their first steps in learning how to write mathematical proofs, is that they don’t think they could produce steps like these on their own without a lot of coaching and prompting. They understand that the steps are correct, and they eventually understand why the steps were necessary for this particular proof (for example, the conversion from 2k-1 to [2k - 1 -1]+1 was necessary to show that 2k-1 is odd).

Not all students initially struggle with this concept, but some do. I’ve found that the following illustration is psychologically reassuring to students struggling with this concept. I tell them that while they may not be comfortable with adding and subtracting the same number (net effect of adding by 0), they should be comfortable with multiplying and dividing by the same number because they do this every time that they add or subtract fractions with different denominators. For example:

\displaystyle \frac{2}{3} + \frac{4}{5} = \displaystyle \frac{2}{3} \times 1 + \frac{4}{5} \times 1

= \displaystyle \frac{2}{3} \frac{5}{5} + \frac{4}{5} \times \frac{3}{3}

= \displaystyle \frac{10}{15} + \frac{12}{15}

= \displaystyle \frac{22}{15}

In the same way, we’re permitted to change 2k-1 to 2k-1 + 0 to 2k -1 - 1 + 1.

Hopefully, connecting this proof technique to this familiar operation from 5th or 6th grade mathematics — here in Texas, it appears in the 5th grade Texas Essential Knowledge and Skills under (3)(H) and (3)(K) — makes adding by a form of 0 in a proof somewhat less foreign to my students.

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