discrete math. sets problem help?

Question

Let

S = {(x, y) ∈ R × R | 2x + 3y < 8},

T = {(x, y) ∈ R | y^2 + y > 6x^2}.


(a) Exists a ∈ R so that (a, −10) ∈ ST?
Why?

(b) Exists b ∈ R so that (−10, b) ∈ TS^−1
? Why?

Answer 1

Given Information:

S = {(x, y) ∈ ℝ × ℝ | 2x + 3y < 8}

T = {(x, y) ∈ ℝ | y^2 + y > 6x^2}

Part (a): Does there exist a ∈ ℝ so that (a, -10) ∈ S∩T?

For (a, -10) to be in S∩T, it must be in both S and T.

Checking if (a, -10) ∈ S:

We need: 2a + 3(-10) < 8

2a - 30 < 8

2a < 38

a < 19

Checking if (a, -10) ∈ T:

We need: (-10)^2 + (-10) > 6a^2

100 - 10 > 6a^2

90 > 6a^2

15 > a^2

-√15 < a < √15

Approximately: -3.87 < a < 3.87

Answer: Yes, such an a exists. We need both conditions satisfied: a < 19 AND -√15 < a < √15. The intersection of these is -√15 < a < √15, which is non-empty. For example, a = 0 works.

Part (b): Does there exist b ∈ ℝ so that (-10, b) ∈ T\S?

For (-10, b) to be in T\S, it must be in T but NOT in S.

Checking if (-10, b) ∈ T:

We need: b^2 + b > 6(-10)^2

b^2 + b > 600

b^2 + b - 600 > 0

Using the quadratic formula: b = (-1 ± √(1 + 2400))/2 = (-1 ± 49)/2

So b < -25 or b > 24

Checking if (-10, b) ∉ S:

We need: 2(-10) + 3b ≥ 8 (the negation of being in S)

-20 + 3b ≥ 8

3b ≥ 28

b ≥ 28/3 ≈ 9.33

Answer: Yes, such a b exists. We need (b < -25 or b > 24) AND b ≥ 28/3. Since 28/3 ≈ 9.33 < 24, the condition simplifies to b > 24. For example, b = 25 works.