I have no records to see how old these particular students were; but the material is high-school level (pre-calculus), and I’ve taught it in community college (as preparation for college-level classes). My approach to it, however, is perhaps deeper than many textbooks at this level, because I think more is needed in order to fully understand than is commonly taught. (Some of the other posts on this topic explicitly explore underlying ideas like function composition.)

]]>As I explained, the theorem does not say that any lower bound will produce alternating signs, but only that any number that produces alternating signs is a lower bound. The implication is only in one direction.

And in fact, if you continue further, you will find that -6 produces alternating signs, so we know that is a lower bound — and in fact it is! There is no zero lower than -6.

So the test works just a described (though not as many students initially expect).

]]>

`-4 | 1 5 -2 -24`

|__-4_-4__24

1 1 -6 0

These do not alternate in sign. Yet I know that -4 is a lower bound. So I thought maybe I need to be below -4 so I tried -5 Here I get

`-5 | 1 5 -2 -24`

|___-5_0__10

1 0 -2 -14

The quotient alternates but when the remainder is included it doesn’t alternate. So am I doing something wrong?? It doesn’t seem like the lower bound test is working as described.

]]>Since the point of this post is largely to show the variety of forms that are called “expanded”, it’s entirely reasonable to add another! I suppose we can call it “nested expanded form”.

Of course, then we have to think about when it might be useful. The ones we discussed are primarily intended to help young students understand the meaning of place value and standard notation, and this probably is not very useful as a notation at that level (as it is less than obvious what it means, especially for students who have not yet learned about parentheses); but it is very useful in practice, both for evaluating polynomials efficiently, and for base conversion.

The latter is where I have indeed seen this form used heavily. Your conversion of time units is an example of this. As another example, we can convert 3D1C in hexadecimal to decimal by expanding it as \(10(10(10(3)+D)+1)+C\) in hexadecimal, or, in decimal, \(16(16(16(3)+13)+1)+12=15,644\).

This form is also equivalent to synthetic division, used to evaluate polynomials (synthetic substitution); and we can similarly use synthetic division to efficiently evaluate numbers in non-decimal bases, too.

Ultimately, I don’t tend to think of this as a **notation**, but it is an excellent way to **apply **the concepts.

4 x 100,000 + 7 x 10,000 + 4 x 1,000 + 1 x 100 + 3 x 10 + 6 x 1

for [ 474,136 ], just write it as

10*(10*(10*(10*(10* ( 4 )+7)+4)+1)+3)+6

This way all the digits remain grouped together, and it fully showcases exponents without actually having to use the power operator (or big chunks of 100,000,…….). So many textbooks keep saying this nested form is only used for polynomials, when in fact, it’s just as good, if not better, way to present numbers in any base as a composite of coefficients and powers of the base.

And all you need is 5 multiplies and 5 adds spanning a total of 11 numeric operands, instead of

4 x 10^5 + 7 x 10^4 + 4 x 10^3 + 1 x 10^2 + 3 x 10 + 6

which is 5 multiplies, 5 adds, 4 exponentiations, and 15 numeric operands.

To find out the largest exponent in the nested form, just count # of copies of “10 x” on the left ( and also for each digit ). A similar approach can also be used for so many other scenarios :

Time to seconds conversion becomes

17 : 43 : 29

60 * (60 * ( 17 ) + 43 ) + 29

————————————

6 3, 8 0 9