Reading Proofs for Validation and Comprehension: an Expert-Novice Eye-Movement Study

Abstract

Does reading a mathematical proof for validation engender different behaviors from reading it for comprehension? Experts and novices each read two mathematical proofs under different sets of instructions: they were asked to understand one proof, and to assess the validity of the other. Their eye movements were recorded while they read and were analyzed to investigate possible differences in attention allocation, in cognitive demand and in the mathematical reading process. We found negligible differences in reading behaviors under the two sets of instructions, and we discuss the implications of this for theoretical development, research methodology and pedagogical practice.

Appendix

Proof P _n

Theorem: For n, k ∈ ℕ, the equation (n − k) !  = n^k − 1 has exactly three solutions:

(n₁, k₁) = (2, 1),  (n₂, k₂) = (3, 1) and (n₃, k₃) = (5, 2).

Proof: Let n ∈ ℕ and assume that there exists a solution with n > 2 and n even.

Because n is even, (n − 1)! is also even.

But n^k − 1 is odd for all k ∈ ℕ, so (n − k) !  ≠ n^k − 1 for all even n > 2.

Now assume that there exists a solution with n > 5 and n odd.

We know that (n − 1) ∣ (n − 2)!, so it follows that (n − 1)² ∣ (n^k − 1).

By factorization of (n^k − 1) we have n^k − 1 = (n − 1)(1 + n + n² + … + n^k − 1).

On the one hand the second factor is divisible by (n − 1).

On the other hand by dividing this factor by (n − 1) we get the remainder k.

Thus we have shown that (n − 1) ∣ k, but this contradicts n^n − 1 − 1 > (n − 1)!.

Proof P _a

Theorem: ∫x⁻¹dx = ln(x) + c.

Proof: We know that $\int {x}^k dx=\frac{x^{k+1}}{k+1}+c$ for k ≠  − 1.

Rearranging the constant of integration gives

$\int {x}^k dx=\frac{x^{k+1}-1}{k+1}+{c}^{\prime }$ for k ≠  − 1.

Set $y=\frac{x^{k+1}-1}{k+1}$, and take the limit as k →  − 1 as follows.

Let m = k + 1, and rearrange $y=\frac{x^{k+1}-1}{k+1}$ to give

x^m = 1 + ym or $x={\left(1+ ym\right)}^{\frac{1}{m}}$.

Set $n=\frac{1}{m}$. Then $x={\left(1+ ym\right)}^{\frac{1}{m}}=\left(1+\frac{1}{m}\right)\to {e}^y$ as n →  ∞ .

by properties of e.

As n → ∞, we have m → 0, so k →  − 1.

In other words, x → e^y as k →  − 1, so y → ln(x) as k →  − 1.

So $\int {x}^k dx=\frac{x^{k+1}-1}{k+1}+{c}^{\prime }=y+{c}^{\prime}\to \ln (x)+{c}^{\prime }$ as k →  − 1.

So ∫x⁻¹dx = ln(x) + c^′.