Question

aRb (a less or equal to (-b))
and it consist of all integers.

is it symmetric,reflexive, and transitive.

oh man I hate relations And I bet my answers are all wrong ,but I just wanted to check :(

Answer 1

so u have
\[\large (\forall a,b\in\mathbb{Z})(aRb\Leftrightarrow a\leq-b) \]
right?

Answer 2

yup

Answer 3

do u know the definitions?

Answer 4

what definitions ?

I just want the answers , since I just had a mid term exam and this one was the last question.

Answer 5

ok

Answer 6

do u think that
\[\large a\leq-a \]
for every integer a?

Answer 7

it's not reflexive for what I know.

anyways I said that they're all true since he didn't give me the quantifiers for the values of x and y , it just said that they're integers , after the exam I thought of it and I think we should work on every integer not just give a counter example of some integer

Answer 8

well, in this the quantifiers were implicit (they usually are)

Answer 9

Our prof only gives one example of each topic and I remember him using a counter example to solve most of the examples.

anyways I think that they're all false since we're talking about all integers.

Answer 10

counterexamples are used to show something is not true when the quantifier is \(\forall\)

Answer 11

:(

anyways are they all false or true ?

Answer 12

in this case the relation IS symmetric

Answer 13

if aRb then
\[\large a\leq -b \]
multiplying by (-1) we get
\[\large b\leq -a \]
which mean bRa.

So R is symmetric

Answer 14

Even though I said it's symmetric , I think I won't get any marks since he wants a full explanations or he'll simply consider it false and give a 0 for it.

Answer 15

so it's only symmetric ?

Answer 16

it is not transitive because
\[\large 1R(-2)\quad\text{because}\quad1\leq2 \]
and
\[\large (-2)R0\quad\text{because}\quad-2\leq0 \]
but 1R0 is false because
\[\large 1\not\leq0 \]

Answer 17

Thank you :)

Answer 18

u r welcome