Introduction - Part B

introduction - part b

Expand +

As the learners guess input numbers, the educator generates the output number. After each guess Fran has one of the learners “graph the point.”

introduction - part b

5th & 6th Grade Math – Multiple Representations of Numeric Patterning
Fran Dickinson, San Carlos Charter Learning Center, San Carlos School District, San Carlos, California

Next Up:   Introduction - Part C
Previous:  Introduction - Part A


STUDENT: I think it’s x times 3 minus 3, so x 3 minus 3, yeah, because, because, then.

STUDENT: Yeah. It works! Yeah!

STUDENT: No, because

STUDENT: Times 3 minus 3..

STUDENT: minus 3 would be negative 3…

STUDENT: Oh, yeah, so it is impossible!

STUDENT: Because, because, 0 times 3

STUDENT: Yeah, because anything times 0

STUDENT: and then minus 3, you’re going negative, and we’re only…

STUDENT: Yeah, yeah. I wonder why he said 0? He said that 0 could be a possibility.

STUDENT: I kind of understand where Griffin was coming from, though.

STUDENT: Yeah, like if he thought, that…

STUDENT: Yeah, it’s kind of fun.

OBSERVING TEACHER: Are you agreeing?

STUDENT: Then it’s minus 3.

FRAN DICKINSON: So tell me. What were you hearing at your table? Kristin , what did you hear at your table?

STUDENT: Um, well, we thought that it could work, because your parameter was greater than or equal to 0

FRAN DICKINSON: All right. Christina, what did you hear at your table?

STUDENT: Well, we thought it wouldn’t work, because, well, we think it would be negative 3. Instead of 0. Like, we didn’t think that 0 would work.

FRAN DICKINSON: What would be negative 3? Can you be more specific?

STUDENT: The, the…


STUDENT: y… the...

STUDENT: the x value…

STUDENT: it would be the y. The, um, output. Because, like, 0 wouldn’t be positive.

FRAN DICKINSON: 0, so, let me get some clarification now. So Griffin’s guess was, his guess was a y value. He guessed 0. So maybe I should write that on the board here. So. And my question to you was, is this possible? So I asked a very vague question there. What do you think, Robby?

STUDENT: It’s kind of, because, I kind of noticed this, it’s because the y, in every one, is bigger than the x, and your parameters say that x has to be greater than or equal to 0, so if the x is smaller than the y, it can’t be smaller than 0, so it wouldn’t work.

FRAN DICKINSON: Can someone put in different words what they hear Robby saying? Kylie.

STUDENT: Um, with all the numbers we’ve done so far, the y is bigger than x. so if the y is 0, then x would have to be lower than 0. But the parameters say that x has to be equal to or greater than 0. So it wouldn’t work.

FRAN DICKINSON: I see that I have some comments around the room. Eric, would you like to make a comment?

STUDENT: I think it works. I think, can I say what the x would be?


STUDENT: I think it would be 1, because, um. The other ones work that x times 3 minus 3, equals, um, y? and 1, so 1 times 3 is equal to 3, minus 3 is zero. So I thought it would work.

FRAN DICKINSON: I heard some “Ahhhhs.” Did I hear that somewhere? I see some agreement around the room as well. Teo, what are you agreeing with?

STUDENT: I’m agreeing, because we talked about the same thing at our table.Because I was thinking negative at first but then I saw it as a y value, not the x value. So…


STUDENT: And I thought, I don’t know. And then Griffin told me what he was thinking. About what he said.

FRAN DICKINSON: Very good! Can I see agreement in the room? Or disagreement. This point here, is 1, 0.

It was interesting to hear all of the different opinions of how to state the rule. I think this illustrates where we were as a group as far as our familiarity with algebraic expression goes. Great conversation between Eric and Sam about 3 groups of x versus x groups of 3.