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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.

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    $\begingroup$ 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. $\endgroup$ Sep 14, 2014 at 1:47
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    $\begingroup$ 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. $\endgroup$ May 15, 2015 at 0:46

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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.

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    $\begingroup$ What would happen if there was a multiplication sign between the p and the x? $\endgroup$
    – MasB
    Sep 13, 2014 at 3:51
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    $\begingroup$ @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, 2014 at 5:08
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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.

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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...

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    $\begingroup$ 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. $\endgroup$
    – user507
    Sep 13, 2014 at 1:05
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    $\begingroup$ I say they're just not thinking. The 1 goes away, as it so often does. ;^) $\endgroup$
    – Sue VanHattum
    Sep 13, 2014 at 1:43
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  1. It's a moderately complicated expression. With x, p, and a -1....and then putting a 1 into it. It's not complicated for you (or for strong students), but for neophytes and weak students, verily, it is.

  2. I think there's some weakness on order of operations. Even if the kids mostly get it right, they don't have it down gnat's ass. So in a tough situation, they flub.

  3. Some of them are not being enough "step by step". Yes, you showed that some had an intermediate step. But many didn't. And even for those, they'd be more likely to get it, by using multiple intermediate steps.


Finally, I think this question falls into a class that we often see on here. The "why don't kids get something that is easy for me". It shows a little bit of a lack of empathy or at least of savviness. People are different. Your students are (likely) both less experienced than you and less innately strong at math.

However, to give it credit, it looks for some solution, some pedagogical trick to solve the student's gap. But this is not the right way to think about things (lock and key fix). Instead, you need to break the problem down. AND then drill EXTREMELY SIMILAR problems over and over.

Yes, occasionally there will be a lock and key aha. But even then the aha will differ by person...and often will only make sense after significant struggle (like "getting your kip" in gymnastics).


As a practical solution, what I recommend is just to ask them what (1)^(5) is. Then ask them (1)^87 is. then (1)^n is. The (1)^(n-1) is. Then (1)^(p-1) is. Even that last one is a simpler expression than the problem statement. So finally ask that (maybe even covering it up). And then having walked them through this and eventually step by step coaxed them to doing the final problem right, don't expect that things will click and no problems in the future. In general, weaker students will need lots of SUBSEQUENT drill to do this right...not just the one coaching session.

Finally, I highly encourage you to make the kids show step by step manipulations. Eventually they may not need them. But for neophytes or intrinsically weak students, they help a lot. There's a nice little essay by Feynman where he talks about being able to quickly "get" solutions of x in little one degree, one unknown problems...and that he did it "by arithmetic", rather than by combining terms, adding to each side of the equation, dividing by coefficient, etc. Well...that's fine for Dick. But for your average/weak neophyte, those step by step processes (and writing them) in pre-algebra (or algebra) are a Godsend.

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