My hypothesis is this: You need four more 2s (than are in the numerator) to make 10. But you need 8 more .5s (than are in the numerator) to make 5. Now, it should really be 9 more .5s. But be that as it may, the (hypothesized) algorithm is a valid one, right?

2/5 is greater than 3/12 because you need 1.5 more 2s to make 5, and you need 3 more 3’s to make 12.

Essentially, the (hypothesized) strategy is to compare the scale factor from the numerator to the denominator (minus 1) in each fraction.

Or think of a circle in 10 wedges. Two of them are shaded. You need to shade 4 more sets of 2 to fill the circle. But with the circle in 5 wedges, half of a wedge is shaded. You need to shade in 9 more half wedges (kid says 8) to fill the circle.

Now, an argument could be made that the 8 represents a more meaningful error. Maybe every time there is a fractional numerator, the kid counts the fractionally shaded piece as a whole one. But the fact that the kid then understands (or seems to, anyway) that 4 whole fifths can be expressed as 8 half-fifths suggests that this error should be easy to overcome.

This is similar to my most cogent guess, with a slightly different argument for the .5->5. You need four more 2’s to get from 2 to 10. You need four more 1’s to get from 1 to 5, and since .5 is half of 1 you’ll need 8 .5’s to get from .5 to 5.

Just wondering, Avery: have you spoken with the student about this?

The student is working in base 8.

Woah. Nice catch. Could the student have used a calculator set to base 8? It’s too perfect of a fit to not check out.

Then shouldn’t they say 0.5 goes into 5 10 times? After all, there’s no such thing as “8” in octal. (Perhaps they multiplied by 2 three times and already knew that was equivalent to multiplying by eight, or added 0.5 to itself repeatedly?)

## 6 replies on ““0.5 goes into five 8 times…””

My hypothesis is this: You need four more 2s (than are in the numerator) to make 10. But you need 8 more .5s (than are in the numerator) to make 5. Now, it should really be 9 more .5s. But be that as it may, the (hypothesized) algorithm is a valid one, right?

2/5 is greater than 3/12 because you need 1.5 more 2s to make 5, and you need 3 more 3’s to make 12.

Essentially, the (hypothesized) strategy is to compare the scale factor from the numerator to the denominator (minus 1) in each fraction.

Or think of a circle in 10 wedges. Two of them are shaded. You need to shade 4 more sets of 2 to fill the circle. But with the circle in 5 wedges, half of a wedge is shaded. You need to shade in 9 more half wedges (kid says 8) to fill the circle.

Now, an argument could be made that the 8 represents a more meaningful error. Maybe every time there is a fractional numerator, the kid counts the fractionally shaded piece as a whole one. But the fact that the kid then understands (or seems to, anyway) that 4 whole fifths can be expressed as 8 half-fifths suggests that this error should be easy to overcome.

This is similar to my most cogent guess, with a slightly different argument for the .5->5. You need four more 2’s to get from 2 to 10. You need four more 1’s to get from 1 to 5, and since .5 is half of 1 you’ll need 8 .5’s to get from .5 to 5.

Just wondering, Avery: have you spoken with the student about this?

The student is working in base 8.

Woah. Nice catch. Could the student have used a calculator set to base 8? It’s too perfect of a fit to not check out.

Then shouldn’t they say 0.5 goes into 5 10 times? After all, there’s no such thing as “8” in octal. (Perhaps they multiplied by 2 three times and already knew that was equivalent to multiplying by eight, or added 0.5 to itself repeatedly?)