r/matheducation • u/One_Signature_9415 • 4d ago
Extend the zero exponent
Have you ever needed to show why negative exponents are positive in the denominator? I recall (when I was in school) just getting a definition and running with it. My students are giving me puzzled looks in this part of the unit, and I'm considering showing them the reasoning. If you have explained the logic (I assume it's an extension of the zero exponent) was it effective?
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u/yeolcoatl 4d ago
You have to be a little careful, because there is a right way to do this and a wrong way to do this, and they look nearly identical.
The key idea is that this is an extension of a mathematical definition, which means we are changing the definition. Changing a definition is a choice that mathematicians (including math students) can make. But that doesn't mean that a change in a definition is arbitrary. There are arguments and reasons why a definition should change.
So we start with our original definition: multiplication by a natural number is repeated multiplication, so 2^4 = 2*2*2*2 (4 times) and a^m = a* a * ... * a (m times).
From here, we can take two paths. The pattern path that droopy_headquarters suggested is more concrete and better for younger children, the more formal properties path does more work for higher mathematics.
In the pattern path, we notice that every time we increase the exponent by one, we multiply by the base. so 2^4=2^3 * 2, a^{m+1} = a^m * a, and this lets us go backwards.... 2^3=2^4 / 2, a^{m} = a^{m+1} / a ... and from here we can construct a new meaning for zero exponents and negative exponents. 2^0 = 2^1 / 2 = 1, and a^0 = a^{m+1} / a = 1 (when a≠0). Then 2^(-1) = 2^0 /2 = 1/2, a^(-1) = a^0 / a = 1/a (when a≠0), and so on.
Two important things to note here: 1) we have changed the definition of exponent. It no longer means repeated multiplication in the same way. 2^(-3) isn't 2 * 2 *... *2 (-3 times)... instead we have done something else. We have used a property of exponents (that going down by 1 exponent is dividing by the base) to decide what "2 * 2 *... *2 (-3 times)" should mean.
The second thing to notice is that this doesn't work when a=0. 0^0 is a whole thing on it's own, where this idea of choosing a definition really stands out. There are good arguments to be made that 0^0 should be 0, that it should be 1, that it should be undefined, or that it should be indeterminate. In Calculus, it's indeterminate. In Computer science, it's 1. These are choices (and again, those choices have reasons).
The second pathway is the properties pathway. In this pathway, we notice that exponents have a particular property: 2^3 * 2^ 4 = (2* 2 * 2) * (2 * 2 * 2 * 2). On the right we multiply 3 times, then 4 times for a total of 3+4=7 times. So 2^3 * 2^4 = 2^(3+4)... similar reasoning generates the full property, a^m * a^n = a^(m+n). We then take this new property to be the new definition of exponent.
Using this new definition, we can choose values for integer exponents.
2^3 * 2^0 = 2^(3+0) = 2^3, so 2^0 should be 1, and similarly,
2^3 * 2^(-3) = 2^(3-3) = 2^0 = 1, so 2^(-3) should be 1/(2^3)... etc.
You can make similar arguments for a^0 and a^(-m), but again, you have to be careful about a=0.
0^3 * 0^ 0 = 0^3 = 0. So 0^0 could be 1, but because 0^3 is already you would also get the same result if you chose any number to be the value of 0^0, which is why 0^0 is indeterminate in some systems and 1 in others.
Later in math, the definition of exponent gets changes again, based on Taylor series, but this will work for now.
I said be careful because there's a wrong way to do this. The wrong way to do this is to completely lose the idea that the definition changes, and teach these extensions of a^0 and a^(negative) as inevitable. There are three problems with this.
1) In reality, there are simply very good arguments for these new definitions. Treating a definition as inevitable makes it harder to talk about situations where definitions are more clearly chosen (like 0^0).
2) Choosing and manipulating definitions, and looking at the consequences of those choices is the work that mathematicians actually do. Mathematicians don't calculate things, they explore systems or rules and prove theorems in those systems to see what a system can and can't do.
3) Most importantly, If students don't realize that their thinking is supposed to change and get stuck on "exponent means repeated multiplication" then they can't fit these new ideas into their old framework and have trouble remembering the new rules. Students who struggle with exponents later are the ones who never realized that their thinking had to change.
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u/theadamabrams 4d ago
it's an extension of the zero exponent
Why is a0 considered to be 1? Whatever pattern/logic you used for that will probably extend into negative exponents, yes.
There are two main methods I know of:
- Sequence: look at, say, 24 and 23 and 22 and 21. Each time you divide by 2. Continuing after 21 = 2, we must get 20 = 2/2 = 1 and then 2-1 = 1/2 and then 2-1 = (½)/2 = 1/4 and then 2-3 = (¼)/2 = 1/8 and so on.
- Algebra: 2n · 2m should be 2n+m because if you literally write out, for example 23 · 24 you get (2·2·2)·(2·2·2·2) which is 3+4=7 twos multiplied together. In order to keep the bold fact true, 2n · 20 must be 2n+0 = 2n, and if "2n · Thing = 2n" then that "Thing" must be 1. So 20 = 1. Similarly 21 · 2-1 = 20, meaning 2 · (2-1) = 1, and therefore 2-1 must be 1/2.
The algebra version can also lead you to x1/2 = √x pretty easily, and if you eventually get to logarithms it's the analogue of log₂(ab) = log₂(a) + log₂(b).
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u/Just__Liberty 4d ago
Directly by the rules of adding exponents and the zero exponent, x^(-a) * x^a = x^(-a+a) = x^0 = 1, so x^(-a) = 1/x^a.
Do your students know logarithms? Using the various properties: log(1/x^a) = log(1) - log(x^a) = 0 -a*log(x) = log(x^-a).
Additionally, plot an example like y = 2^a with a as the x-axis... which gives you visually your 'extension of the zero exponent).
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u/yamomwasthebomb 4d ago
I’d argue nearly every topic should be taught so that students understand (if not discover themselves) the reasons why mathematics looks the way it does.
For this topic, I’d recommend making a table with columns for expressions, value, and meaning. The first row should start with 5^3 which is 125 since it represents 5x5x5. The next rows should be 5^2, 5^1, and 5^0. Before memorizing that 5^0 is 1, students should be asked to predict what it is from the numerical pattern… as we move down the table, we are dividing by 5. Since 5^1 is 5, then 5^0 seems like it should be 5/5 or 1. Continuing this logic, 5^-1 would be 1 divided by 5, or 1/5, and so forth. Repeat for different bases like 3, 10, -2, and 1/2.
Another way to do this would be with a number line (as close to scale as possible). Since 5^1 (5), we take a jump that’s 5 times bigger to 25 and then 125. Similarly, we can divide to jump backwards from our landing spot of the 3rd jump (125) to the landing spot of the second (25), then from the second to the first. But that begs the question: where did our first jump begin? Intuitively it feels like 0, but then we’d be multiplying by 0… and also 5/5 is 1, so that must’ve been our true starting place. From there, our first jump backwards would be 5 times smaller for 1/5, and so forth. Again, repeat for other bases including negatives and fractions.
If they have some experience with exponential functions, I would also build connections there. In the same way the function is never below the x-axis, our values aren’t negative. Our jumps forward get astronomically large, where our jumps backwards get microscopically small. In the same way our starting place on the number line was 1, our y-intercept is 1.
I’d also recommend trying the same logic as the above for 0^n to help them see that 0^0 is undefined since we’d be dividing by 0. Therefore it’s actually NOT true that “everything to the 0 power is 1.”
Hope this helps.
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u/UnderstandingPursuit Physics BS, PhD 4d ago
Perhaps start with
- 23 25 = [8] [32] = [256] = 28 = 23+5
Then
- [ 25 ] / [ 23 ] = [32] / [8] = [4] = 22 = 25-3
The general form is
- xm xn = xm+n
which applies for any real {m, n}. If m>0 and n<0, this only works if n<0 is dividing.
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u/AndyC1111 4d ago
I’ve had the privilege of talking to some pretty impressive mathematicians about this. Their response is normally a reference to a proof (25 * 2-5 = 20, so 20 = 1).
They’ve never said anything that I’ve felt would really help children feel good about it. I pretty actively try to avoid “because I said so” when teaching children, so I spent decades thinking about this. This is what I came up with…
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u/OutrageousPair2300 4d ago
Fractions, division, and negative exponents are all interconnected. Why they're taught as entirely separate things baffles me.
Numbers on the top of the fraction line are multipliers. Numbers on the bottom are divisors. When you multiply by the same number more times, it increases the exponent. When you divide by the same number more times, it decreases the exponent.
Dividing by a number (say) five more times is the same as multiplying by that number five fewer times, which is why moving a number with an exponent from the bottom to the top (or vice versa) changes the exponent from a positive to a negative (or the other way around.)
The exponents just keep track of how many times you're multiplying or dividing by a number.
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u/KiwasiGames 4d ago
I normally teach this right after I’ve taught them the rules for dividing indices.
Students know that a^5 / a^3 = a^2. They can prove this pretty well by expanding. They know they can work out this out by subtracting the indices.
So then I swap it around, and give them a^3 / a^5 . If we solve it by expanding, we get 1 / a^2 . If we solve it by subtracting indices we get a^-2 . This is normally enough for students to click that a^-2 = 1/a^2 .
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u/droopy_headquarters 4d ago
A lot of students get it when you show them the pattern. Write out 2^3, 2^2, 2^1, 2^0 on the board and ask what happens each time you go down a step. They'll see you're dividing by 2 each time. Then keep going into the negatives and it clicks for most of them