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 September 16th, 2018, 03:59 AM #1 Newbie   Joined: Jun 2018 From: Brasil Posts: 20 Thanks: 0 algebraic structures A binary relation is a set formed by pairs taken from the Cartesian product between two sets, according to a "rule" that varies from relation to relation. In particular, let us consider, in the set of positive integers, the binary relation * defined by a * b = c, where c is the greatest divisor common between a and b. Class V for true sentences and F for false sentences: () * is commutative. () * is associative. () 1 is the neutral element. () a * a = a, for all a. () For each a, there exists b such that a * b = 1. Now, check the alternative that shows the sequence CORRECT: * a) F-V-F-F-F. * b) V-F-V-V-V. * c) F-V-F-F-V. * d) V-V-V-V-F. Can someone help please?
 September 16th, 2018, 05:35 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry It's not your fault, but this question is awful. The definition it gives of a binary relation is very imprecise and is more complicated than is needed: a binary relation on a set $X$ is a subset $R$ of the cartesian product $X \times X$. That's it. While most binary relations we're interested in might be given by some "rule" (knowing that $(a,b) \in R$ if and only if $a$ and $b$ satisfy some nicely formulated property), this is certainly not necessary. Furthermore, the example it gives is of a binary operation, and not a binary relation. Note that it doesn't really make sense to talk about commutativity or associativity of a binary relation (the closest analogue to commutativity is symmetry, which says $(a,b) \in R$ if and only if $(b,a) \in R$, and there isn't an analogue of associativity). A binary operation on $X$ is a function $f: X \times X \to X$, where we often write $a * b$ (or even just $ab$) as shorthand for $f(a,b)$. Here we have $X = \mathbb{Z}_{>0}$ and $a * b = \operatorname{gcd}(a,b)$. Now to get to the actual problem: 1) This is simply asking whether $\operatorname{gcd}(a,b) = \operatorname{gcd}(b,a)$ for all positive integers $a$ and $b$. It should be clear that this is true. 2) This is asking whether $\operatorname{gcd}(a, \operatorname{gcd}(b,c)) = \operatorname{gcd}(\operatorname{gcd}(a,b), c)$ for all positive integers $a, b$ and $c$. This isn't hard to show directly from the definition of $\operatorname{gcd}$; alternatively you could use the fundamental theorem of arithmetic and work with prime decompositions of $a,b$ and $c$ to get it. 3) Given that the first two parts are true, we are locked into option d) which means this is also true. However, this is a little odd: when we say $e$ is a "neutral" or "identity" element, we usually mean that it leaves other elements unchanged: $a * e = e * a = a$ always. But here $1$ kills off all other elements: $a * 1 = \operatorname{gcd}(a,1) = 1$ for all positive integers $a$. I guess here the word "neutral" is being used in the sense of "neutralizing" or "neutering". (That, or the question is wrong.) 4) This says $\operatorname{gcd}(a,a) = a$. This one should also be clear to you. 5) This says for every positive integer, there is a positive integer coprime to it. This is true (hint: see my discussion for part 3)). So really we have V-V-V/F(depending on what "neutral" actually means here)-V-V, which is option e): none of the above. Thanks from Roberto 37 Last edited by cjem; September 16th, 2018 at 05:42 AM.
 September 16th, 2018, 06:04 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 Can a photograph of the original question be posted, so that we can check for typing or translation errors?
 September 16th, 2018, 01:08 PM #4 Newbie   Joined: Jun 2018 From: Brasil Posts: 20 Thanks: 0 Sorry, it's in Portuguese.
September 16th, 2018, 01:11 PM   #5
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 September 16th, 2018, 01:35 PM #6 Newbie   Joined: Jun 2018 From: Brasil Posts: 20 Thanks: 0 V - V - V - V - F is it corret? do you think?
September 16th, 2018, 02:05 PM   #7
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Quote:
 Originally Posted by Roberto 37 V - V - V - V - F is it corret? do you think?
As I said, the answer should be V - V - F - V - V (or V - V - V - V - V if "neutral"/"neutro" is being meant in a weird way). All of the given options a), b), c) and d) are incorrect.

 September 16th, 2018, 04:42 PM #8 Newbie   Joined: Jun 2018 From: Brasil Posts: 20 Thanks: 0 but is for options below, only 1. * a) F-V-F-F-F. * b) V-F-V-V-V. * c) F-V-F-F-V. * d) V-V-V-V-F. I think letter d. Last edited by Roberto 37; September 16th, 2018 at 04:46 PM.
September 17th, 2018, 08:17 AM   #9
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Quote:
 Originally Posted by Roberto 37 but is for options below, only 1. * a) F-V-F-F-F. * b) V-F-V-V-V. * c) F-V-F-F-V. * d) V-V-V-V-F. I think letter d.
None of the options are correct.

 September 17th, 2018, 01:22 PM #10 Newbie   Joined: Jun 2018 From: Brasil Posts: 20 Thanks: 0 but I need to check 1

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