Determine whether the relation R on the set of all real numbers is reflexive, symmetric or transitive? where (x,y) is an element of R Explain why and why not by proving and giving examples.

(a) x=1
(b) xy ≥ 0
(c) x-y is a rational number
(d)x+y = 0

I need help urgently!!!! pls reply asap!!! thanks so much!!

(a) To determine if the relation R is reflexive, we need to check if (x, x) belongs to R for every x in the set of all real numbers.

In this case, x = 1, so we need to check if (1, 1) belongs to R.

Since there is no given rule or condition for the relationship between x and y, we cannot determine if (1, 1) satisfies the relation or not. Therefore, we cannot conclude if the relation R is reflexive or not based on this information alone.

(b) To determine if the relation R is symmetric, we need to check if (x, y) belongs to R whenever (y, x) belongs to R for every x and y in the set of all real numbers.

In this case, the given inequality is xy ≥ 0.

Consider the example x = 1 and y = -2:

(1, -2) does not satisfy the given inequality (1 * -2 = -2 < 0).

But, (y, x) = (-2, 1) satisfies the given inequality (-2 * 1 = -2 < 0).

Since (1, -2) belongs to R but (y, x) = (-2, 1) does not belong to R, the relation R is not symmetric.

(c) To determine if the relation R is transitive, we need to check if for every (x, y) and (y, z) belonging to R, (x, z) also belongs to R for every x, y, and z in the set of all real numbers.

In this case, the given condition is x - y is a rational number.

Consider the example x = 2, y = 1, and z = 0:

(x, y) = (2, 1) satisfies the given condition (2 - 1 = 1 is a rational number).

(y, z) = (1, 0) also satisfies the given condition (1 - 0 = 1 is a rational number).

But, (x, z) = (2, 0) does not satisfy the given condition (2 - 0 = 2 is not a rational number).

Since (x, y) = (2, 1) and (y, z) = (1, 0) belong to R, but (x, z) = (2, 0) does not belong to R, the relation R is not transitive.

(d) To determine if the relation R is reflexive, symmetric, or transitive based on the condition x + y = 0, we need to follow the same steps as above.

Consider the example x = 2 and y = -2:

(x, y) = (2, -2) satisfies the given condition (2 + (-2) = 0).

But, (y, x) = (-2, 2) does not satisfy the given condition (-2 + 2 = 0).

Since (x, y) = (2, -2) satisfies the given condition, but (y, x) = (-2, 2) does not, the relation R is not reflexive, symmetric, or transitive based on this condition.

To summarize:
- (a) The relation R cannot be determined as reflexive based on the given information.
- (b) The relation R is not symmetric.
- (c) The relation R is not transitive.
- (d) The relation R is not reflexive, symmetric, or transitive.

Sure! I'll be happy to help you. Let's go through each relation (a), (b), (c), and (d) and determine whether they are reflexive, symmetric, or transitive.

(a) x = 1:
To check reflexivity, we need to see if (1, 1) is in the relation R. Since (1, 1) satisfies the condition x = 1, it is reflexive.

Reflexive: Yes

(b) xy ≥ 0:
To check reflexivity, we need to see if (x, x) is in the relation R for all real numbers x. Since multiplication of real numbers is commutative, (x, x) will always satisfy the condition xy ≥ 0, as any number multiplied by itself is non-negative. Hence, it is reflexive.

Reflexive: Yes

To check symmetry, we need to see if for every (x, y) in R, (y, x) is also in R. In this case, if (x, y) is in R, then xy ≥ 0. We also know that multiplication of real numbers is commutative, i.e., xy = yx. Therefore, (y, x) will also satisfy xy ≥ 0. Hence, it is symmetric.

Symmetric: Yes

To check transitivity, we need to see if for every (x, y) and (y, z) in R, (x, z) is also in R. In this case, if (x, y) is in R, then xy ≥ 0, and if (y, z) is in R, yz ≥ 0. However, we cannot guarantee that the product of xy and yz will be non-negative, and hence, we cannot establish the transitive property. Therefore, it is not transitive.

Transitive: No

(c) x - y is a rational number:
To check reflexivity, we need to see if (x, x) is in the relation R for all real numbers x. Since x - x is always zero, which is a rational number, (x, x) satisfies the condition x - y is a rational number. Hence, it is reflexive.

Reflexive: Yes

To check symmetry, we need to see if for every (x, y) in R, (y, x) is also in R. In this case, if (x, y) is in R, then x - y is a rational number. However, we cannot guarantee that y - x will also be a rational number, as the difference between two numbers is not necessarily the same as the difference in the reverse order. Hence, it is not symmetric.

Symmetric: No

To check transitivity, we need to see if for every (x, y) and (y, z) in R, (x, z) is also in R. In this case, if (x, y) is in R, then x - y is a rational number, and if (y, z) is in R, then y - z is a rational number. The difference of two rational numbers can be either rational or irrational, so we cannot establish the transitive property. Therefore, it is not transitive.

Transitive: No

(d) x + y = 0:
To check reflexivity, we need to see if (x, x) is in the relation R for all real numbers x. Since x + x is equal to 2x, and this is not equal to zero unless x is zero, (x, x) does not satisfy the condition x + y = 0. Hence, it is not reflexive.

Reflexive: No

To check symmetry, we need to see if for every (x, y) in R, (y, x) is also in R. In this case, if (x, y) is in R, then x + y = 0. We can rewrite this equation as y + x = 0, which satisfies the condition. Therefore, it is symmetric.

Symmetric: Yes

To check transitivity, we need to see if for every (x, y) and (y, z) in R, (x, z) is also in R. In this case, if (x, y) is in R, then x + y = 0, and if (y, z) is in R, then y + z = 0. Adding these two equations together, we get (x + y) + (y + z) = 0, which simplifies to x + z = 0. Hence, it satisfies the transitive property.

Transitive: Yes

I hope this helps! Let me know if you have any further questions.