Lots of good, interesting advice. Bowen Kerins said:

Another way to deal with speed demons is to give several problems in a row that are related and have the same answer. Speed demons may not even notice this happening, and the result is they're more likely to "look around" a little more before and after working a problem.I liked that. And it made me think about an interesting thing that happened to me today.

I'm teaching beginning algebra at a community college. My first two mastery tests are on pre-algebraic topics. There is one fraction story problem. On version 3 of the test, it goes like this...

I have a lot of books at my house, especially after all the math books I bought during my sabbatical. Right now 3/7ths of my books are math-related; 3/10ths of those are kids’ books. What fraction of my books are kids’ math books?I had marked my student's subtraction problem wrong, and was explaining to her why it would be multiplication. As I finished up, she pointed to her answer, which was the same as mine.

I said I hate marking people wrong when they have the right answer, but that this was a fluke. Subtraction doesn't solve this sort of problem. On the spot, I made up another problem, 1/2 of 1/3. Guess what.

Both answers are the same again.

Optional homework: When does this happen? ;^)

(I give my students lots of optional homework. Most of it is: "Read this cool book and write a review.")

I liked his post, too.

ReplyDeleteMy homework is now for a time period also. Much fairer among different students, and to all students to know what the expectations are. (Hour per workshop, but this is college.)

I like the idea of homework being for a purpose: if it's a skill, do it until you have the skill. Demonstrate the skill.

I do give choice home workshops, and have found that - on the whole - students make amazing choices.

( let b be a rational number in (0, 1/2] and )

ReplyDeletelet a = b/(1-b). then

a - b = b/(1-b) - b

= [b - b(1-b)]/(1-b)

= b^2/(1-b)

= b*[b/(1-b)]

=ba = ab.

i got a last-second fill-in gig

at big state u starting yesterday

(hired the day before). so i'm

teaching again after 5 quarters

in exile. also madeline has set

up the net in her home again

(alas). so maybe i'll be back

to posting in some quasi-regular

way. doesn't seem likely.

http://vlorbik.wordpress.com

(Owen! It's so good to see you here! Will email.)

ReplyDeleteBut, how'd you discover a? I couldn't think with your fractions being single letters, so I gave b the form n/d, and worked through your steps that way.

And I'm not seeing why b has to be less than 1/2.

i've assumed we want 1 > a > b > 0

ReplyDeleteso that a & b 'll be "common fractions"

(as i think i've heard said for suchlike

rational numbers a & b) and so the

difference a-b 'll be positive.

then you just set up ab = a-b

and fire up the algebra:

b = a- ab = a(1-b)

and out pops a = b(/1-b).

then you just shove it

into the old "theorem/proof"

obfuscation machine.

lose the parenthetical assumptions

for the general case of course

(or rather, replace with b\=1).

ot

much more interestingly (as i hope):

ReplyDeleteletting b = n/d in our context gives

a = [n/d]/[1-n/d]

a = n/(d-n).

(b, a) = (n/d, n/(d-n) )

is a more interesting formula

i think... particularly in light

of the *matching numerators*...

and students might be less

intimidated (subtractions

are easier than divisions).

one is reminded here of the theory

of "partial fractions" from (typically)

calc ii...

the character-string "(/"

ReplyDeletedoes not appear

in elementary algebra

and should be replaced

with "/(":

"out pops a = b/(1-b)".