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# Tag Info

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I do not believe that this is a concern that would surface, or be worth surfacing, in the courses named by this question's title. In my reading, the question is analogous to worrying about whether you can ask about e.g. the thousands place of $7521$: To do so assumes that the base ten representation of $7521$ is unique, or, in particular, that "thousands ...

12

I'm going to rewrite this answer to clarify what I think the issue is. I think the OP is imagining a different definition of the ring $k[x]$ than most answerers are. Here are two reasonable definitions: $k[x]$ is the ring of formal expressions of the form $\sum_{j=0}^{\infty} p_j x^j$ with $p_j \in k$ and we require that $p_j$ is $0$ for $j$ sufficiently ...

1

Soon after introducing polynomials, students learn to add and subtract polynomials. Notice that if $f(x)$ and $g(x)$ are polynomials with degrees $n$ and $m$ respectively, and if $f(x)-g(x)=0$, then $n=m$. It follows that the degree of a polynomial is well-defined. So the proof that degree is well-defined is not difficult at all. On the other hand, students ...

3

In general, I agree with @Henry Towsner on the fact that the proofs should not always be presented in an elementary course. However, I have to disagree on the implicit "well-definition property" of any definition. Such a definition would require that some sort of uniqueness property has been proved, which cannot - or should not - be done prior to the ...

6

This: "On the other hand, without the proof the definition of the degree of polynomials is not even logically established." is not quite right. What is needed to establish the definition is the fact that the degree is well-defined, but this fact is stated (implicitly) by stating the definition. The expectation that all facts be proven is out of place in a ...

3

I'm kind of restating what other answers have said, but I wanted to practice expressing it in the clearest, most concise way I could think of. (Coincidentally, this came up with my partner tonight, so I got a test-run with it, and got an entirely satisfying result). Consider only natural numbers (i.e, positive integers). It seems like the number of ...

1

The problem here is defining primes in their own right rather than defining them in terms of factorization. Start with a number like 30. Writing 30 = 2 x 15 tells us something new; writing 30 = 1 × 30 doesn't. Writing 30 = 2 x 3 x 5 tells us something new again, whereas writing 30 = -2 x -15 doesn't. Once we get to 30 = 2 x 3 x 5, we can't break down any ...

0

Primes are called primes because all other integers above 1 are (multiplicatively) "built out of them." You can't build anything else (multiplicatively) out of 1s. No matter how many 1s you multiply together.

2

I would start by showing them, on paper, what they already know - that in the context of multiplication the number 1 is useless. It is the identity function. It simply reflects the original number. It is a mirror. "And just like your reflection in a mirror is not a real person, neither is 1 a real number when multiplying (not to be confused with a Real ...

1

\begin{align} & \begin{array}{cccccccccccccccccccc} & & & & & & & & & 840 \\[12pt] = {} & & & & & 28 & & & & \times & & & & & 30 \\[12pt] = {} & & & & 4 & \times & 7 & & & \times & & & & 5 & \times & 6 \\...

1

When I was at school, after being taught about integer division, I was told: "A natural number is prime if it has exactly 4 integer divisors" Then 2 is prime, as it can be divided by -2, -1, 1, and 2. Then one can be divided by -1 and 1, and those are only 2 divisors. Hence not prime. Zero can be divided by anything but itself, yielding zero, hence ...

4

This one is really very simple. First, tell them what a prime number is: A prime number has exactly two different factors. (If they don't know what factors are, and they ask about primes, the correct answer is "well, first you have to know about factors...") With that definition, it is very easy to figure out 0 and 1. Is 1 a prime? No, because it only ...

1

(If your context does not include negative numbers, turn all the negatives below positive. This almost won't change the discussion.) Everything divides zero, so zero can't be prime. $0 \cdot 7 = 0$ means $0$ and $7$ divide $0$. $0 \cdot -8 = 0$ means minus eight also divides zero. Can we see that everything divides zero, so zero is very far from being ...

3

We don't need the full FTA upfront if we limit our discussion for the moment to obvious examples of what's necessary for a factorization to be unique. The FTA provides analogous sufficiency conditions they'll probably guess on their own, even if they don't know how it's proven. You can say, $1$ isn't considered prime because then there wouldn't be unique ...

7

How should one talk about the question of 1 or 0 being prime ... with primary or middle school children? Depending on what you did before you will have an easy or a hard task: If the children were told: A prime number is a natural number which cannot be divided by other numbers than by 1 and by itself. ... you will have problems explaining why 1 is not ...

4

A good way to lead to the uniqueness of prime factorization and the convention that $1$ is not a prime is to build factor trees (that's common in elementary school these days in fourth grade, sometimes third grade). 24 24 24 8 3 6 4 2 12 2 4 3 2 2 2 3 ...

4

If you build each number n using n square blocks in rectangular configurations, there are multiple configurations for each composite number. (4 is 4 by 1 or 2 by 2.) The primes are the ones that can only be built as a 1 by n rectangle. It seems clear that 0 would be neither prime nor composite, when looked at this way. The easiest way to understand why we ...

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