3

So I got this question in my exams.
Find the HCF of $0.5$ and $2.5$.
My friends are saying that answer would be $0.5$.
I wrote $1$ because I couldn't see any factor bigger than it common to both.
Moreover, I have one more point that domain of $\gcd$ is positive integers.
So there can be two things, whether, the question is wrong, and even if we extend $\gcd(a,b)$ to rational numbers, answer would be $1$, because $1$ is bigger than $0.5$.
So please clear my doubts.

Aditya Agarwal
  • 4,595
  • I agree with you. The formal definition of gcd does specify the domain as integers. Your reasoning is fine. I would stick to it – Shailesh Sep 26 '15 at 04:32
  • 3
    Does the definition of the gcd apply to non-integer rationals? – Yes Sep 26 '15 at 04:32
  • That is what I am saying. @GudsonChou, I got this question in my exam, so let us assume it does. But still $1$ will be greater than $0.5$. – Aditya Agarwal Sep 26 '15 at 04:36
  • I prefer to say this is off-topic, since it certainly not a question about math. What can anyone do, in mathematical way, if the concept itself is not well-defined and can't be improved (since your teacher is not here and can't tell what HCF means to him/her)? – Asydot Sep 26 '15 at 04:45
  • Isn't MSE a site to get mathematical concepts cleared? Where would I go then? I think this is a topic worth discussing. And again, I clear, that I only want to know the logic and answer behind the "title" question. @Asydot. – Aditya Agarwal Sep 26 '15 at 04:55
  • 1
    One consistent definition they could be using is "The $\gcd$ is the smallest number writable as a sum (with integer coefficients) of the arguments" as in Bézout's identity - i.e. using $a|b$ to mean there is an integer so that $na=b$. This isn't the most unreasonable definition (since we get to keep Euler's algorithm), but it doesn't go without saying. – Milo Brandt Sep 26 '15 at 04:56
  • Having a HCF of two numbers that is greater than the minimum of the two would seem problematic to my mind would be my reason for supporting the answer of .5 here. – JB King Sep 26 '15 at 05:03
  • Why? @JBKing. Reasoning or logic would help? – Aditya Agarwal Sep 26 '15 at 05:10
  • @AdityaAgarwal: The point is that, with the usual definition, the question is nonsense; but if your professor has given a reasonable definition, then of course the situation is different. So you may try to see if your professor indeed gave another definition of gcd; if not, then the question is nonsense. :) – Yes Sep 26 '15 at 05:17
  • Because then there would be the question of how well could you really eliminate the numbers greater than a value? For example, could be there numbers greater than a million that divide 2 or 3? Could you exhaustively demonstrate that for each value without running out of time in your life? While this may seem ridiculous, it is an extension of where this could go. – JB King Sep 26 '15 at 05:17
  • In the post above, it is explained in an elementary way how to extend the concept from the integers to the rationals in a way that you lack generality but preserve non-triviality. Since it seems your interest is not of thorough and general discussion, I recommend you see the above post. Keep in mind that this is not the "right" (meaning: generalized) way of talking about $\text{gcd}$. For that, see my answer and/or https://en.wikipedia.org/wiki/Greatest_common_divisor#The_gcd_in_commutative_rings – Aloizio Macedo Sep 26 '15 at 05:22
  • 1
    @AloizioMacedo: In the possible duplicate you pointed to, I gave a historically-based answer that would indeed make the gcd equal to $0.5$. That said, in current common usage, the notion is restricted to pairs of natural numbers. – André Nicolas Sep 26 '15 at 05:29
  • Please see my answer. My confusions are cleared. Thank you all for contributing. (I got to that before that duplicate mark). – Aditya Agarwal Sep 26 '15 at 05:38
  • 1
    I think the question has a severe flaw in itself, and I'm ashamed myself that I didn't ask it in my first comment:

    What is your teacher's definition of GCD?

     – Aloizio Macedo Sep 26 '15 at 05:39
  • I don't think it is necessary now, all the confusions are cleared. And still, my teacher was wrong. We had no reason for his answer's support. He just had a book, which used the traditional method of finding the GCD, his only support. – Aditya Agarwal Sep 26 '15 at 05:41
  • 1
    @AdityaAgarwal This can serve as a reference for other people eventually. Your confusions may have been cleared, but the question has its own existence now. Leaving a signifcant hole such as an ambiguous definition (or rather, the inexistence of one) in the question can be very harmful to potential confused people, as it was to you.

    EDIT: For example: you linked a plethora of links, each of which provided only an algorithmical way to solve, without a proper definition. This may have confused you.

     – Aloizio Macedo Sep 26 '15 at 05:44
  • Ok, I understand. So what things I have to clarify? – Aditya Agarwal Sep 26 '15 at 05:45
  • 2
    The definition of GCD you are using. – Aloizio Macedo Sep 26 '15 at 05:46
  • If you want to know my professor's logic. All I can provide is "And he has no logic, he just says that "$1$ is a universal factor, it is for all. So here the answer is $0.5$". – Aditya Agarwal Sep 26 '15 at 05:47
  • If you want, I can delete the convo in between, and set the question simply as finding the gcd of $(0.5,1.5)$ – Aditya Agarwal Sep 26 '15 at 05:47
  • 2
    The duplicate is half of the story. There are two reasonable things one might mean when the $\gcd(1/2, 5/2)$ comes up in a problem. The linked duplicate is one of those two (and in that meaning, $1$ does not divide either number evenly). In the other, both $1$ and $1/2$ (along with every nonzero rational number) are correct, equivalent answers to $\gcd(1/2, 5/2)$. –  Sep 26 '15 at 09:48

3 Answers3

5

This question has the assumption that a factor of a non-integer rational number can be specified. In particular, it assumes that factors of 0.5 and 2.5 can be specified.

I believe that this shows a fundamental misunderstanding on the part of the creator of the question.

The algorithm stated is this:

$HCF(a/b, c/d) =\frac{\gcd(a, c)}{\operatorname{lcm}(a, c)} =\frac{\gcd(a, c)}{ac/\gcd(a, c)} =\frac{\gcd^2(a, c)}{ac} $.

This does not even use the denominators!

To use Pauli's famous phrase, this is not only not right, it is not even wrong.

Martin Sleziak
  • 53,687
marty cohen
  • 107,799
3

First and foremost, I would ask you to edit your question in order to only present itself as a question. Other details are not only irrelevant but also distracting.

Now, two ways of defining $d=\text{mdc}(a,b)$ are:

  1. $d$ satisfies:

    • $d | a$, $d| b$
    • $ e|a, e|b \implies e | d $

or

  1. $d$ is a generator of the ideal generated by $a,b$.

My algebra may be a bit rusty... so these definitions may not be equivalent when dealing with some spaces (for instance, non-PID's), but they are for our discussion here.

There is a problem when we try to apply these definitions for $\mathbb{Q}$. If you check, given any two (non-trivial) numbers, any number will be a $\text{gcd}$ of both. The discussion is quite trivial in $\mathbb{Q}$.

Whereas, in $\mathbb{Z}$, generators of ideals are defined up to a sign (since units are only $1$ and $-1$), hence there is no problem in defining $\text{gcd}$ well enough.

Aloizio Macedo
  • 34,292
2

I got my answer on my own! After a lot of thinking (elementary terms only), I deduced that the answer is indeed $0.5$. Because,

first and the foremost, we look at Euclid's Division Lemma, which states that any positive integer $a$ can be expressed as $bq+r$ where $b,q$ and $r$ are integers!

Now even if we extend this to rational numbers, $q$ will indeed remain integer. (think it over, you will get why, because if it would not have been the case then gcd would have been non-sensical).

Coming back to my question, $\gcd(\frac12,\frac32)$, if we take $1$ to be the common factor, $1$ is not a common factor because on dividing $\frac12$ by $1$, we get a fraction, not an integer. Same with $\frac32$. And the largest number satisfying this rule (quotient is an integer) is indeed $0.5$.
Thus, $$\gcd(\frac12,\frac32)=\frac12$$ P.S.: Sorry, if any one already got there.

Aditya Agarwal
  • 4,595