stupid math question

[Balain]

Just studying for my exam and came across a question and I think I have been looking at math too long. LOL

Define a relation R on the set of positive integers by: for all a,b in the set of positive integers aRb if and only if gcd(a,b) > 1.

We have to show if the relation is reflexive, Symmetric or transitive.

So for the reflexive part if aRa then gcd(a,a) >1

since gcd(a,a) = a. Do we say it is reflexive or it is reflexive when a is not 1 or it is not reflexive because when a = 1 gcd(a,a) = 1 so gcd(a,a) > 1

[Simon of Aragon]

To be honest I was never any good with math, so I went into medicine.

Taking the standard topology on R: Let A~B <-->A\B is open
Reflexivity: A\A = is open.

Not symmetric: Let A = (0,1), B={2} then A\B = A is open. B\A= B is closed.
Not transitive: Let A = (0,1) and B=(2,3), C= (0,1/2) Then A\B=(0,1) is open, B\C=(2,3) is open. But A\C = [1/2,1) is not open.

I hope I got that right...

Now can you tell me how to resequence my DNA; GCTAACATGGATTACATG :D

[Jay R]

Just studying for my exam and came across a question and I think I have been looking at math too long. LOL

Define a relation R on the set of positive integers by: for all a,b in the set of positive integers aRb if and only if gcd(a,b) > 1.

We have to show if the relation is reflexive, Symmetric or transitive.

So for the reflexive part if aRa then gcd(a,a) >1

since gcd(a,a) = a. Do we say it is reflexive or it is reflexive when a is not 1 or it is not reflexive because when a = 1 gcd(a,a) = 1 so gcd(a,a) > 1

It only takes one counterexample. The relation is not reflexive.

It's clearly symmetric, since gcd(a,b) = gcd(b,a).

It isn't transitive. 4R6 and 6R9, but not 4R9


Now can you tell me how to resequence my DNA; GCTAACATGGATTACATG :D

Well, you see, when a mommy and a daddy love each other very, very much, ....

[kurokotetsu]

Just studying for my exam and came across a question and I think I have been looking at math too long. LOL

Define a relation R on the set of positive integers by: for all a,b in the set of positive integers aRb if and only if gcd(a,b) > 1.

We have to show if the relation is reflexive, Symmetric or transitive.

So for the reflexive part if aRa then gcd(a,a) >1

since gcd(a,a) = a. Do we say it is reflexive or it is reflexive when a is not 1 or it is not reflexive because when a = 1 gcd(a,a) = 1 so gcd(a,a) > 1
You said it yourself. 1 is not related to 1. So the relations isn't reflexivve for all elements of the positive integers. A counter exampleis enough.

Also get 2,6,9. 2R6 as gcd=2. 6R9 as gdc=3 but 2R9 isn't in the relation a gdc=1. So it is not transitive.

But the gcd doesn't change by order, so it is symmetric.

R isn't an equivalence relation, it is just symmetric.


To be honest I was never any good with math, so I went into medicine.

Taking the standard topology on R: Let A~B <-->A\B is open
Reflexivity: A\A = is open.

Not symmetric: Let A = (0,1), B={2} then A\B = A is open. B\A= B is closed.
Not transitive: Let A = (0,1) and B=(2,3), C= (0,1/2) Then A\B=(0,1) is open, B\C=(2,3) is open. But A\C = [1/2,1) is not open.

I hope I got that right...You are defining another relation, not the one Balian is using. You are defining a relation between open sets, as if A and B are members of the topology they are one, so you counterexample in the symmetric property is wrong, B is not member of a topology. Also, you are taking intervals, and Balian is working on the integers.

THe counter example of transitivity though is correct and shows why that isn't an equivalence relation (the symmetric difference is, if I remember correctly).