11
$\begingroup$

Substituting $x=1$ into $px^{p-1}$, why do so many students get $p^{p-1}$? I saw this four or fives times in my office hours this week as students worked on the same problem. Not a single student spontaneously got this step right. Some write $p(1)^{p-1}$ as an intermediate step. I doubt that any of them would have had trouble evaluating $3x^2$ at $x=1$.

I tried to get them to articulate why they would think this, and none of them could. It could be that they don't understand that exponentiation has higher priority than multiplication, but I doubt they would have evaluated $3x^2$ to be 9.

It seems like the symbolic exponent makes their brains turn off. Many students couldn't evaluate $1^{p-1}$ by itself, even after I asked them what $1^{87}$ was, and they knew it was 1.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ At a glance, this looks a bit like $(px)^{p-1}$; in such a case, subbing $x=1$ would indeed give $p^{p-1}$. I expect they were operating procedurally and letting the one just be folded into the expression; perhaps they possess an incorrect piece of reasoning around (as Sue VH says) "the $1$ goes away." One question I'd have is: How did you try "to get them to articulate why they would think this"? Your post on why students get this wrong is very interesting; I'd also be interested in what teachers can do/say to understand better the students' processes in arriving at such an error. $\endgroup$ – Benjamin Dickman Sep 14 '14 at 1:47
  • 2
    $\begingroup$ I want to respond specifically to the part about 1^87. You say "what's 1^87?" They say "1." You say "what's 1^(p-1)?" and they can't do it. When I use this approach, I include many other steps in between. What is 1^87? 1. What is 1^777? 1. What is 1^cow? 1. What is 1^Frank? 1. What is 1^ Johnny? 1. What is 1^x? 1. What is 1^y? 1. What is 1^box? 1. What is 1^[actually draw a picture of a box]? 1. What is 1^[actually draw a picture of a box with a smiley face in it]? 1. What is 1^[actually draw a picture of a box with p-1 in it]? 1. What is 1^(p-1)? Hopefully they say 1. $\endgroup$ – Frank Newman May 15 '15 at 0:46
11
$\begingroup$

I think this is what goes on in their head:

$3x^2$ is very familiar to them, and it is clearly $3 \times (x^2)$. It helps that $3$ is a number and $x$ is a variable.

$px^{p-1}$ is not so familiar, and unless they are paying complete attention (using their System 2), their lazy System 1 automatically associates it with the nearest familiar expression, namely those of the form $31^2$ (in $px^{p-1}$, all are variables, and in $31^2$, all are numbers $-$ this probably makes them associated). But $31^2$ is, of course, $(31)^2$, and not $3 \times 1^2$.

If you ask them about it, however, they are forced to pay attention, and System 2 makes sure they think it through, so the mistake cannot be replicated. And as System 1 works automatically and subconsciously, they do not even understand what they were thinking before.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ What would happen if there was a multiplication sign between the p and the x? $\endgroup$ – Bernard Masse Sep 13 '14 at 3:51
  • 1
    $\begingroup$ @BernardMasse $p \times x^{p-1}$? I'm quite sure that now they will correctly reduce it to $p$ when $x = 1$, even if they process it subconsciously. In fact, I've seen many students write $(1.5)^2$ instead of $1.5^2$, probably because the decimal point in between makes $1.5$ look like it has two parts, though it's a single number, and $.$ is not an operator. But it looks close enough to an operator to make them uncomfortable without parentheses. $\endgroup$ – M. Vinay Sep 13 '14 at 5:08
11
$\begingroup$

It may not be helpful to think of it as their brains turning off. Instead, perhaps their brains have never turned on!They may never have thought about what the operations there mean, and have been working on a stimulus-response basis where they see certain symbols and respond in certain ways. A combination of symbols that doesn't fit neatly into one of the categories they are familiar with will be responded to a fairly random way.

My instinct is that their expected response to an expression with a pronumeral in it is to give a new expression which still has a pronumeral in it, matching with their experience of expanding and/or simplifying. The answer being a plain number like 1 is not something they expect to happen and so it doesn't even occur to them as an option.

Also, their experience with pronumerals in the exponent is probably very limited, and I would hazard a guess that in all of those experiences, there was still a pronumeral in the exponent after simplification. So the pull to keeping the pronumeral in the exponent is probably strong.

They are probably not consciously seeing a pronumeral as a representation of a possible number. They see numbers and pronumerals as completely different entities with different rules of reckoning. All they have in their heads are rules for manipulating things. They don't realise that a good strategy to assess their understanding is to try it out with various numbers to see if it makes sense for each case, because they don't see it as a description of something that works for multiple different numbers -- it's just symbols.

To go further on this line, in one of your comments to another answer you say

... $p1^{p-1} = p^{p-1}$. It seems like there must be some reason that nearly all my students think this same thing.

It is probably not true that the students think of these as the same thing at all. They probably don't even think of $3x^2$ and $3xx$, or $3 \times 1^2$ and $3$ as the same thing. All they see is manipulations that are or are not allowed and have expectations for what might or might not happen as a result.

In light of this discussion, you may need to really get them to talk about what they think expressions mean. Trying the expressions out with various different values of both x and p might help them to see that the expression is a description of how those numbers are combining to make an answer. I mention using different x's to highlight that 1 is a special case. Talking about how even if you don't know p, you can be sure the answer is 1 and how this would not be the case for most other x's would be helpful where the answer is expressable as a number, as opposed to not being able to do any manipulation.

Improve this answer
$\endgroup$
5
$\begingroup$

I don't think the proper question is why they do this. A better question is: How can we get them to keep their brains turned on while doing math?

I try to ask a student in this situation a series of questions: What is 1^87? You say that's 1? Why? (Good reason given.) Now what is 1^n? [If that is too hard, make some questions that get at their trouble. Maybe just ask 1^15, 1^20, possibly until they roll their eyes and say it's always 1. Oh? What does n represent here? ... What wrong answer did they give? Can you figure out where it comes from? That will help you decide what questions to ask them.] If that's not too hard, now you get to ask about 1^(n-1). Is n easier for them than p? That would be interesting.

I think what I'm suggesting would be called Socratic dialogue. Maybe we need some youtube videos of this sort of thing...

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ What you're describing is essentially what I did. I just lack insight into the error of thinking that $p1^{p-1}=p^{p-1}$. It seems like there must be some reason that nearly all my students think this same thing, which to me seems inexplicable. $\endgroup$ – Ben Crowell Sep 13 '14 at 1:05
  • 2
    $\begingroup$ I say they're just not thinking. The 1 goes away, as it so often does. ;^) $\endgroup$ – Sue VanHattum♦ Sep 13 '14 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.