User Name Remember Me? Password

 Linear Algebra Linear Algebra Math Forum

 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
Newbie

Joined: Jun 2018
From: Brasil

Posts: 20
Thanks: 0

photo

phto
Attached Images site ingles.jpg (63.8 KB, 9 views) 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
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 311
Thanks: 109

Math Focus: Number Theory, Algebraic Geometry
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
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 311
Thanks: 109

Math Focus: Number Theory, Algebraic Geometry
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 Tags algebraic, structures Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Roberto 37 Linear Algebra 7 September 16th, 2018 06:05 PM Roberto 37 Algebra 1 September 16th, 2018 05:21 AM fifa14 Computer Science 4 September 19th, 2013 04:11 PM minidu Computer Science 0 July 1st, 2011 12:15 PM ElMarsh Abstract Algebra 0 November 4th, 2009 04:23 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      