High school mathematics teachers’ perspectives on the purposes of mathematical proof in school mathematics

Abstract

Proof serves many purposes in mathematics. In this qualitative study of 17 high school mathematics teachers, we found that these teachers perceived that two of the most important purposes for proof in school mathematics were (a) to enhance students’ mathematical understanding and (b) to develop generalized thinking skills that were transferable to other fields of endeavor. We found teachers were divided on the characteristics (or features) of proofs that would serve these purposes. Teachers with less experience tended to believe that proofs in the high school should adhere to strict standards of language and reasoning while teachers with more experience tended to believe that proofs based on concrete or visual features were well suited for high school mathematics. This study has implications for teacher preparation because it appears that there is a wide variation in how teachers think about proof. It seems likely that students would experience proof very differently merely because they were seated in different classrooms.

Appendix

Odd Squares I

Claim::

If a ² is odd, then a is odd as well.

Proof::

We will prove the claim by proving the contrapositive: If a is even, then a ² is even.

Let a = 2n, then a ² = 4n ², and a ² = 2(2a ²). So, a ² is even.

Therefore if a is even, then a ² is even and if a ² is odd, then a is odd.

Odd Squares II

Claim::

If a ² is odd, then a is odd as well.

Proof::

We will prove the claim by proving the contrapositive: If a is even, then a ² is even.

Even numbers end in 0, 2, 4, 6, or 8.

The square of any number ending in 0 ends in 0.

The square of any number ending in 2 ends in 4.

The square of any number ending in 4 ends in 6.

The square of any number ending in 6 ends in 6.

The square of any number ending in 8 ends in 4.

These squares are then all even. Therefore, the square of an even number is even.

So if an odd number is a square, then its square root is also odd.

Interior Angles I

Claim::

The sum of the degree measures of the interior angles of a triangle is 180.

Proof:Footnote

Proof adapted from Klutch et al. (1991), p. 154.

:

Statements	Reasons
1. Through B, draw $\overleftrightarrow{ DE}\left\Vert \overline{ AC}\right..$	1. Through a point not on a given line, there exists one and only one line parallel to the given line.
2. ∠1 ≅ ∠4, ∠3 ≅ ∠5	2. If two parallel lines are cut by a transversal, then each pair of the alternate interior angles is congruent.
3. ∠4 and ∠ABE are supplementary.	3. Supplement axiom
4. m∠4 + m∠ABE = 180	4. Definition of supplementary angles
5. m∠ABE = m∠2 + m∠5	5. Angle addition axiom
6. m∠4 + m∠2 + m∠5 = 180	6. Substitution axiom
7. m∠1 + m∠4, m∠3 =m∠5	7. Definition of congruent angles
8. m∠1 + m∠2 + m∠3 = 180	8. Substitution axiom

Interior Angles II

Claim::

The sum of the interior angles of a triangle is 180˚.

Proof:Footnote

Proof adapted from http://www.math.csusb.edu/courses/m129/tri180.html. Retrieved in the spring of 2006 but at the time of this writing, this link was dead.

:

Cut a triangle from a piece of paper. Arrange it so that its longest side is the base.

Fold the triangle on a line parallel to the base so that the top vertex lies on the base as shown below.

Now fold the triangle so that the right vertex touches the top vertex (in its folded down position). The two parts of the right side of the triangle should line up, and the right part of the bottom side should be folded along itself as shown below.

Fold in the left vertex in the same way.

You should see the three interior angles of the triangle meeting at a point, forming an 180˚ angle with no overlapping.

Quadratic Formula

Claim::

The solution to ax ² + bx + c = 0 is $x=\frac{-b\pm \sqrt{b^2-4 ac}}{2a}.$

Proof::

Solve: ax ² + bx + c = 0

If: ax ² + bx + c = 0

Then: ${x}^2+\frac{b}{a}x+\frac{c}{a}=0$

By completing the square, we get:

$$ {\left(x+\frac{b}{2a}\right)}^2-{\left(\frac{b}{2a}\right)}^2+\frac{c}{a}=0 $$

Solving this for x, we get the following sequence of steps:

$$ \begin{array}{l}{\left(x+\frac{b}{2a}\right)}^2={\left(\frac{b}{2a}\right)}^2-\frac{c}{a}\hfill \\ {}x+\frac{b}{2a}=\pm \sqrt{{\left(\frac{b}{2a}\right)}^2-\frac{c}{a}}\hfill \\ {}x=-\frac{b}{2a}\pm \sqrt{{\left(\frac{b}{2a}\right)}^2-\frac{c}{a}}\hfill \\ {}x=-\frac{b}{2a}\pm \sqrt{\frac{b^2}{4{a}^2}-\frac{c}{a}}\hfill \\ {}x=-\frac{b}{2a}\pm \sqrt{\frac{b^2}{4{a}^2}-\frac{4 ac}{4{a}^2}}\hfill \\ {}x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4 ac}}{2a}\hfill \end{array} $$

And so the solution to the quadratic equation ax ² + bx + c = 0 is:

$$ x=\frac{-b\pm \sqrt{b^2-4 ac}}{2a} $$

Completing the Square

Claim::

${x}^2+ ax={\left(x+\frac{a}{2}\right)}^2-{\left(\frac{a}{2}\right)}^2$

Proof:Footnote

Proof adapted from Nelsen (1993), p. 19.

:

Chocolate Bar

You have a rectangular chocolate bar marked into m × n squares, and you wish to break up the bar into its constituent squares. At each step, you may pick up one piece and break it along any of its marked vertical or horizontal lines.

Claim::

Every method finishes in the same number of steps

Proof:Footnote

Problem adapted from Winkler (2004), p. 82.

:

Each time you break the bar, you end up with one more piece than you had before. So it will take mn – 1 breaks (no matter how you proceed) to go from one large piece to mn little pieces.