Guide to written work

Math 95 – Froemke – Spring 2011 – Guidelines for submissions

WRITING PROJECTS:

In each question I am looking for three things:

1) When you make a claim, do you justify the claim correctly?

2) Does your mathematics work, and does it work to answer the question?

3) Do you answer the question being asked?

Not every question will need all three of those – a question asking you to find a definition for ‘congruence’ and then explain that definition, for instance, might not need you to provide any mathematics that would ‘work’.

I grade the projects by starting everyone with a score of ten points, and then subtract for the following:

• A missing question, or one that fails in one or more of the three ways above, gets minus two.

• A question that speaks to the above three ways, but is missing some details (or contains minor inaccuracies) receives a minus one, and some indication of where you need to give more details or provide more work. Some questions might get more than one point taken off in this way.

• Occasionally, I’ll take off one-half point for a very small inaccuracy.

Written projects can then be fixed and resubmitted for an increased score, up to a perfect score of 10 points. Multiple resubmissions are fine. The deadline for submissions is the day of the final exam. Resubmissions must include the original project (so I can see what needs fixing) and your new work, clearly indicated as to which fixes are which. You don’t necessarily need to rewrite each problem on which you miss points – just give your fixes.

I reserve the right to reject any project that I find illegible or too confusing to read. For some of us, using a word processor is the only way to make our writing legible.

Writing Example

Question: Solve 2x – 3 = 7 and explain each step in detail

· First, I add three to both sides, to get rid of the -3 on the left (the side with the variable):

o 2x – 3 + 3 = 7 + 3

· Then I simplify each side:

o 2x = 10

· Then I divide both sides by 2:

o 2x/2 = 10/2

· Then I simplify each side:

o X = 5

· And there it is: 2x – 3 = 7 solved for x.

· Then I check, by plugging in x = 5 to the original equation:

o 2(5) – 3

o =10 – 3

o =7

· Which is what I have on the right side, so x = 5 is a solution to 2x – 3 = 7.

(NOTE: you do not have to do it in exactly this way, of course, but this is about the level of detail I’m looking for.)