Math, asked by sarbanibiswas04538, 8 months ago

please any one solve the questions in brief. ​

Attachments:

Answers

Answered by nooblygeek
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.

Similar questions