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I'm not a math expert either, but rise over run sounds like a perfectly good definition of slope to me, at least for the purposes of an introduction to high sch

I'm not a math expert either, but rise over run sounds like a perfectly good definition of slope to me, at least for the purposes of an introduction to high school algebra. It is certainly how I was taught about slope, and I went on to graduate from MIT, so whatever nuance this definition allegedly didn't catch did me no harm. Once you get to calculus, things get a little more complicated, as the slope at a given point is defined using derivatives, but I'm pretty sure that ends up being the same as the rise over the run of a tangent line. In any case, anyone who would make such a lame criticism, should just STFU. If that's the best criticism they can come up with, they surely can't have much of value to say. Also, when teaching something like algebra, it's more important to make the material approachable and comprehensible, rather than define everything to a level of rigor that would make Russell and Whitehead happy.

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pflats 14y ago
>It is certainly how I was taught about slope, and I went on to graduate from MIT, so whatever nuance this definition allegedly didn't catch did me no harm. In that case, this debate really isn't about you. You were probably an exceptional student, who saw the connections between mathematical concepts easily, regardless of instruction. You probably found yourself predicting the next concept a teacher would introduce, because it just "makes sense." Not all students are that way. Most are not. The two people involved here are fighting over two different ideas. Sal is being pedantic, but is right, slope is defined as ∆x/∆y. What the other guy was saying is that slope represents rate of change, which is a much more important concept to early algebra, and the underpinning of why you actually care about slope in physics and calculus. You probably made the connection effortlessly. I assure you, many students do not. I teach high school mathematics to both honors and special needs students, and it's important to keep in mind that the instruction is very different between the two populations.
- deleted 14y ago
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- rm999 14y ago
  "slope is defined as ∆x/∆y" Do you mean ∆y/∆x?
  - pflats 14y ago
    Yup. Caught the typo when I wrote the post, even, but I apparently retyped it the same way.
  - pflats 14y ago
    Yup. Caught the typo when I wrote the post, even, but I apparently retyped it the same way.
- nessus42 14y ago
  I'm still not sure I understand the criticism. I'm sure that Kahn must eventually get to rate of change in his algebra course. The criticism is then supposed to be that Kahn didn't motivate his students on why they should care about slope soon enough? If so, that's a completely different criticism, however, from the criticism that Khan is putatively making an alarmingly dense stream of gross factual errors. I think that we can all agree that Kahn is not the best possible teacher that exists in the world for each given subject. Is that a decent argument against what he has done? Hardly! That would be letting the perfect be the enemy of the good. Considering that so many people learn from the Kahn Academy these days, an argument can certainly be made that Kahn's lectures should all eventually be replaced with lectures by the actual best teacher in the world for that given topic. For all we know, this is already in the works.
- saraid216 14y ago
  > slope represents rate of change You may as well say it represents a tangent. (Pun not entirely intended.)
  - hobin 14y ago

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pflats 14y ago

>It is certainly how I was taught about slope, and I went on to graduate from MIT, so whatever nuance this definition allegedly didn't catch did me no harm. In that case, this debate really isn't about you. You were probably an exceptional student, who saw the connections between mathematical concepts easily, regardless of instruction. You probably found yourself predicting the next concept a teacher would introduce, because it just "makes sense." Not all students are that way. Most are not. The two people involved here are fighting over two different ideas. Sal is being pedantic, but is right, slope is defined as ∆x/∆y. What the other guy was saying is that slope represents rate of change, which is a much more important concept to early algebra, and the underpinning of why you actually care about slope in physics and calculus. You probably made the connection effortlessly. I assure you, many students do not. I teach high school mathematics to both honors and special needs students, and it's important to keep in mind that the instruction is very different between the two populations.

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rm999 14y ago
"slope is defined as ∆x/∆y" Do you mean ∆y/∆x?
- pflats 14y ago
  Yup. Caught the typo when I wrote the post, even, but I apparently retyped it the same way.
- pflats 14y ago
  Yup. Caught the typo when I wrote the post, even, but I apparently retyped it the same way.
nessus42 14y ago
I'm still not sure I understand the criticism. I'm sure that Kahn must eventually get to rate of change in his algebra course. The criticism is then supposed to be that Kahn didn't motivate his students on why they should care about slope soon enough? If so, that's a completely different criticism, however, from the criticism that Khan is putatively making an alarmingly dense stream of gross factual errors. I think that we can all agree that Kahn is not the best possible teacher that exists in the world for each given subject. Is that a decent argument against what he has done? Hardly! That would be letting the perfect be the enemy of the good. Considering that so many people learn from the Kahn Academy these days, an argument can certainly be made that Kahn's lectures should all eventually be replaced with lectures by the actual best teacher in the world for that given topic. For all we know, this is already in the works.
saraid216 14y ago
> slope represents rate of change You may as well say it represents a tangent. (Pun not entirely intended.)
- hobin 14y ago