Question

please any one solve the questions in brief. ​

Answer 1

Step-by-step explanation:

Note that i will use $\sim$ for the relation symbol instead of the cursive S.

a)

Reflexive: $a\sim a \iff a - a = 0$ is divisible by 5. As zero is divisible by any integer then the relation is reflexive.

Symmetric: We need that $a \sim b \iff b \sim a$. If $a \sim b$ then $a - b = c \in \mathbb{Z}$, where c is divisible by 5. Then $b\sim a = b-a = -c$. As c is divisble by 5, then -c must also be divislbe by 5. Hence, the relation is symmetric.

Transitive: We need that $a \sim b \sim c \implies a \sim c$. By modular arithmetic we have that $a - b \equiv 0 \mod 5$ and $b-c \equiv 0 \mod 5$, as $a\sim b$ and $b \sim c$. Adding the equations yield $a - b + b - c \equiv 0 \mod 5$, which simplfies to

$a - c \equiv 0\mod 5$. Hence a - c is divisible by 5 and the relation is therefore transitive.

b)

Reflexive: Trivially, all lines lie in the same plane as it self. It is reflexive.

Symmetric: If a line L lies in the same plane as M, then M must lie in the same plane as L. It is symmetric.

Transitive: If two lines lie in a plane, $L \sim M$, and we also have $M \sim P$, then

L, M and P are all in the same plane, or we can create a third plane that contains both L and P. Hence, it is transitive.

c)

Reflexive: Trivially, any number is a divisior of it self.

Symmetric: If $a > b$ and $a \sim b$, then a divides b. However, as $a > b$, then b cannot divide a (this would yield a fraction/decimal). Not symmetric.

Transitive: If $a \sim b \sim c$, then $a \sim b \implies$ $a = bp$, for some number p. As $b \sim c$

then $b = cq$ for soome q. Then by substitution we have $a = cqp$ . Hence c is a divisor of a, and we have that $a\sim c$. It is transitive.