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What this question attempts to ask is, perhaps, the rigorous definition of factor itself. I browsed the web for a definition of factor, and everywhere it was defined loosely (That common sense definition of factors).

But I have 3 questions -

  1. Can we extend the concept of factors to the realm of real numbers? like the one my original question seems to ask?
  2. Factors must necessarily divide a number perfectly (i.e., integrally) but the question is - must that factor also be an integer?
  3. if we agree that $\pi$ is a factor of $3\pi$ then that means every number has infinite number of factors like $\pi /2$, etc. Should we, then, redefine prime numbers?
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    $\begingroup$ As to the question of if we should redefine prime numbers, absolutely not. All nonzero real numbers are units in $(\Bbb R,+,\times)$. Units are neither prime nor composite. Look up the definition for a prime ideal of a ring which is one of the more arbitrary definitions we use for "primality" $\endgroup$ – JMoravitz Mar 11 '17 at 7:37
  • $\begingroup$ Sorry, I didn't get what you said. Well, I am not that proficient in advanced mathematics and could not get your comment. Can you please elaborate it in simple terms, if possible? $\endgroup$ – Sarthak123 Mar 11 '17 at 7:39
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    $\begingroup$ I'll try... We have a definition for primality in arbitrary rings (sets with "addition" and "multiplication" satisfying some properties) which involve special subsets called ideals. Ideals are special in that if you take any element in the ideal and any element in the ring (in or out of that ideal), their product will remain in the ideal. For example how even integers times any other integer will again be an even integer. With the ring of real numbers however, there is no interesting ideal, just $\{0\}$ and all of $\Bbb R$ itself. $\endgroup$ – JMoravitz Mar 11 '17 at 7:45
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    $\begingroup$ Our definition of prime ideals requires that the ideal be a proper ideal (not the entire ring) satisfying the nice property that whenever $ab\in P$ then $a\in P$ or $b\in P$ (reminiscent of for integers if $p\mid ab$ then $p\mid a$ or $p\mid b$), but since there aren't any proper ideals of $\Bbb R$, it becomes a moot point. $\endgroup$ – JMoravitz Mar 11 '17 at 7:50
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    $\begingroup$ It very well might be as elemental as you think in the specific case of the integers (there are a number of equivalent definitions for prime integers) and that same intuitive definition can usually be applied to other more abstract settings, but for the case of the real numbers you either get "everything is a factor of everything else" or "everything is a multiple of everything else" and a lot of the importance we put on being a factor or a multiple is lost or doesn't make sense (e.g. your original question). $\endgroup$ – JMoravitz Mar 11 '17 at 8:27
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One reason for your confusion is probably that the term "factor" is used for two different, but related concepts.

The first concept is quite simple: If you write down a specific product, each of the things you multiply is a factor of the product. For this meaning of "factor" the answer to your question is easy: $3\pi$ clearly is a product, and the factors of that product are $3$ and $\pi$. Note that according to this definitions, $3$ is not a factor of $6\pi$; the factors of that product are $6$ and $\pi$; on the other hand it is a factor of $2\cdot 3\cdot\pi$. In other words, $3$ is not a factor of the real number $3\pi$, but of the expression $3\pi$, that is, of the expression "three times pi".

The second meaning of "factor" is as synonym to "divisor" (but then, "divisor" also has two separate meanings; only one of them is synonymous with this meaning of "factor"). This second meaning is related to the fact that if you write down all products of integers whose value is a given integer $n$, then you'll find that some integers will occur as factor (1st meaning) in the product, while other integers will not. For example, take $n=12$. Then you can write $n$ as product of integers in several ways: $$12 = 1\cdot 12 = 2\cdot 6 = 3\cdot 4$$ As you can see, only the numbers $1$, $2$, $3$, $4$, $6$ and $12$ occur as factors (1st meaning) in any of those products. Note that this is also called divisor, because those are exactly those numbers by which you can divide $12$ and get an integer result.

Now we call an integer $k$ a factor (2nd meaning) of $n$ if you can write a product with value $n$ where $k$ occurs as factor (1st meaning), or equivalently, if you can divide $n$ by $k$ and obtain an integer result.

Note that unlike the first meaning of factor, this second meaning refers to the number, not to the expression. For example, $4$ is a factor (2nd meaning) of $2\cdot 6$, because $2\cdot 6=12=3\cdot 4$. However $4$ is not a factor (1st meaning) of $2\cdot 6$ because there's no $4$ in the expression $2\cdot 6$.

Note however, that the equivalence stated above is true only in the integers. If you try to extend that to the real numbers, you get vastly different results: For any real number $a$ and every non-zero real number $b$, you can find a real number $c$ such that $a=bc$. So the first definition of factor (2nd meaning) applied to the real numbers will give you that every non-zero number is a factor of every number. On the other hand, if you generalize the second definition of factor (2nd meaning) to the real numbers, you indeed get that of $3\pi$ only $\pi$ is a factor (because $3\pi/\pi = 3\in\mathbb Z$, but $3\pi/3 = \pi\notin\mathbb Z$).

However neither option is really satisfying, therefore one generally doesn't use the notion of factor (2nd meaning) for real numbers at all. Note that if you do need the second relation, you'd normally say "$3\pi$ is an integer multiple of $\pi$; the term factor is not commonly used here.

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  • $\begingroup$ Thanks, this clears a lot of confusion. $\endgroup$ – Sarthak123 Mar 12 '17 at 7:47
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Here is some rather elementary information which might be helpful.

The answer to your title-question is no, the statement is false.

Reasoning: Whenever you have a valid representation of a number as product of other numbers, then each of these numbers is called a factor.

Let's assume we are working with real numbers. A representation \begin{align*} 3\cdot \pi \end{align*} is valid and we can conclude $3$ is a factor of $3\pi$ as well as $\pi$.

A few remarks to your questions:

  • Ad (1) Can we extend the concept of factors to the realm of real numbers?

The answer is: Yes, we can. Whenever we have a multiplication $\cdot$, as we do have when working with reals, we can also talk about factors which are the constituents, the building blocks to do this multiplication.

  • Ad (2) Factors must necessarily divide a number perfectly (i.e., integrally) but the question is - must that factor also be an integer?

The formulation of this question is somewhat problematic. But first this answer: If there is a valid representation of real numbers given as product of factors, the factors need not be integers.

Example 1: The product $3\pi$ consists of two factors $3$ and $\pi$. One of them is an integer, the other is not an integer. No problem at all.

Example 2: The product $\pi \cdot \pi$ consists of two factors $\pi$ and $\pi$. None of them is an integer. No problem at all.

Note: The formulation divide a number perfectly (i.e., integrally) is problematic, since perfect is not a technical term in this context and integrally is not necessary in this context (see e.g. $\pi \cdot \pi$).

  • Ad (3) if we agree that $\pi$ is a factor of $3\pi$ then that means every number has infinite number of factors like $\pi /2$, etc. Should we, then, redefine prime numbers?

This is really a clever question and it touches algebraic structures and number theory. It addresses questions like

  • Which kind of numbers do we want to work with? Are these natural numbers, integers, reals or complex numbers or $\ldots$?

  • What kind of multiplication is useful for these numbers?

  • If we have specified a certain kind of multiplication, what are the consequences. What are the factors, what is divisibility, which numbers of them should be called prime.

The appropriate definition of a number being prime is part of algebra and will be defined there, when studying ring theory.

Final note: Sometimes you might work with and stick at integers without using any other numbers. In this specific context whenver you consider a product \begin{align*} a\cdot b \end{align*} then $a$ and $b$ are integers and nothing else. In such cases a representation $3\cdot \pi$ is not valid and not under consideration, simply because $\pi$ is not an integer. The question is $\pi$ a factor? is in this context meaningless.

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The definition of a factor for integers is:

Suppose $a$ and $b$ are integers. Then $a$ is a factor of $b$ if and only if there is an integer $x$ such that $a x = b$.

I think there are two reasonable ways to define a factor for real numbers. The first way is:

Suppose $a$ and $b$ are real numbers. Then $a$ is a factor of $b$ if and only if there is a real number $x$ such that $a x = b$.

Unfortunately, this definition isn't very interesting, because it's equivalent to: "The number $a$ is a factor of $b$ if and only if $a \ne 0$ or $b = 0$."

The second way is:

Suppose $a$ and $b$ are real numbers. Then $a$ is a factor of $b$ if and only if there is an integer $x$ such that $a x = b$.

Under this definition, your original statement is true, and factors don't have to be integers. However, I'm not aware of this definition being standard or commonly used.

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  • $\begingroup$ > I think there are two reasonable ways to define a factor for real numbers. That's my question. Which of the 2 definitions is (more) appropriate and formally accepted? $\endgroup$ – Sarthak123 Mar 11 '17 at 8:58
  • $\begingroup$ @Sarthak123 Depends on the context. Both definitions are useful, at different times. $\endgroup$ – Tanner Swett Mar 11 '17 at 9:05
  • $\begingroup$ I agree. And this is what made me to ask the question because there seems not to be one clear-cut definition of factors of real number. A bigger problem is that what should be, now, the answer of my original question? True or False? or perhaps 'it depends' (sounds more reasonable)? $\endgroup$ – Sarthak123 Mar 11 '17 at 9:10

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