Assessment of proof
In STACK the basic assumption is that a student's answer will be a mathematical expression, e.g. a polynomial or an equation. The facilities for assessing a student's free-form proof is limited.
A discussion of Practical Online Assessment of Mathematical Proof was given by Sangwin, C.J. and Bickerton, R. (2023), International Journal of Mathematical Education in Science and Technology}, 53(10) doi:10.1080/0020739X.2021.1896813. This included more structured questions making use of
- faded worked examples and,
- reading comprehension questions.
Colleagues assessing proof might also consider semi-automatic marking.
From STACK 4.3 we support the following CSS styles to enable consistent display of mathematical proof, and arguments in general. General style was also introduced with the HELM materials.
This class is a general high level container.
Note, the proof container is minimal, and subtle and does not intrude too much but contains a proof as a distinct entity.
This class allows the teacher to highlight sub-components of a proof. It is intended to be a nested sub-proof block of a main proof.
Greater typeset structure can be gained by marking up the proof in more detail, if so desired.
A typical proof consists of individual lines. Each line is numbered, so we can refer to them later. The main "meat" of the line is the "proof step" and there is an optional comment. The proof comment is designed for a comment, to allow an input to ask a question or to act as the second column in a two column proof.
Alternatively, you can use columns, which are fixed at a width of 48%. The design assumes two columns and line numbers.
proof-column-2 has a different visual style for emphasis.
It is possible to use
proof-column-2 style in the first column, but this looks odd and is not recommended. The
proof-column-2 style is intended for only the second column. The
proof-column style is intended to be used in both columns.
Here is a proof by induction.
This class enables ordered lists to have connected numbered statements.
- Let A be an n by n matrix and let R be the reduced row echelon form of A.
- We will show first that det(A) is not zero if and only if det(R) is not zero.
- Then Er ... E1 A = [[input:ans1]]]
- Taking determinants of both sides, we obtain: det(Er) ... det(E_1) det(A) = det(R).
- The determinants of all the elementary matrices are all [[input:ansdropdown]]]
- We conclude that det(A) is not zero if and only if det(R) is not zero.
- Now suppose that A is invertible.
- Then R = [[input:ans4]]
- Thus, det(R) = [[input:ans3]] and hence det(A) is not zero.
- Conversely, if det(A) is not zero then det(R) is not zero, so R cannot have a zero row.
- Thus R = I and A is invertible as required.