🧠 Think and Discuss!
One day, Mustafa's lecture schedule at the university was as follows:
- 🕒 From 9:00 AM to 12:00 PM.
- 🕒 Then from 2:00 PM to 4:00 PM.
If the university's sports center 🏋️ is open daily from 10:00 AM to 1:00 PM, what is the suitable time interval for Mustafa to practice sports at the university's sports center on this day without it affecting his lecture attendance? Discuss.

✅ Solution
Here is a step-by-step solution to find the suitable time for Mustafa to go to the sports center:
- Identify Mustafa's Busy Times (Lectures):
- 🕒 Lecture 1: The interval [9:00 AM, 12:00 PM]
- 🕒 Lecture 2: The interval [2:00 PM, 4:00 PM]
- Identify the Sports Center's Open Hours:
🏋️ The center is open during the interval [10:00 AM, 1:00 PM].
- Find the Overlap:
We need to find a time when the sports center is open AND Mustafa is free.
Let's compare the sports center's open hours with Mustafa's lecture times.
- 🕒 The first lecture (9:00 AM - 12:00 PM) overlaps with the center's hours (10:00 AM - 1:00 PM). The specific time of conflict is from 10:00 AM to 12:00 PM. During this period, Mustafa is in a lecture, so he cannot go to the gym.
- 🕒 The second lecture (2:00 PM - 4:00 PM) does not overlap with the center's hours, as the center closes at 1:00 PM.
- Determine the Available Time Slot:
- 🏋️ The center is open from 10:00 AM to 1:00 PM.
- 🕒 Mustafa is busy from 10:00 AM until 12:00 PM.
- This leaves the period when the center is still open after his lecture ends.
- Conclusion:
✅ The only time Mustafa can go to the sports center is after his first lecture ends at 12:00 PM and before the center closes at 1:00 PM.
Therefore, the suitable time interval for Mustafa is from 12:00 PM to 1:00 PM.
📖 Learn!
📚 Intervals
Between any two real numbers, there is an infinite number of real numbers.
The set of real numbers and its subsets cannot be expressed by the listing method because it is impossible to list all their elements. Therefore, another method is used to express them, which is "intervals".
❓ Example (1)
Problem 1
Write in interval notation, then represent the set A on the number line:
\(A = \{x : x \in \mathbb{R}, 1 \le x \le 3\}\)
Does \(\sqrt{2} \in A\)?
Problem 2
Write in interval notation, then represent the set B on the number line:
\(B = \{x : x \in \mathbb{R}, -2 < x < 2\}\)
Does \(\sqrt[3]{-9} \in B\)?
Solution
Solution ①
\(A = [1, 3]\)
![Number line for the closed interval [1, 3] by Math Egypt - Mr. Ayman Hassan (خط الأعداد للفترة المغلقة [١، ٣] من ماث إيجيبت - مستر أيمن حسن)](https://i.postimg.cc/Gt71Kyvb/1.png)
Using a calculator:
\(\sqrt{2} \approx 1.4\)
\(\because 1 \le 1.4 \le 3\)
\(\therefore \sqrt{2} \in [1, 3]\)
Solution ②
\(B = \mathopen{]} -2, 2 \mathclose{[}\)
![Number line for the open interval ]-2, 2[ by Math Egypt - Mr. Ayman Hassan (خط الأعداد للفترة المفتوحة ]-٢، ٢[ من ماث إيجيبت - مستر أيمن حسن)](https://i.postimg.cc/90FhypRx/2.png)
Using a calculator:
\(\sqrt[3]{-9} \approx -2.08\)
\(\because \sqrt[3]{-9} < -2\)
\(\therefore \sqrt[3]{-9} \notin \mathopen{]} -2, 2 \mathclose{[}\)
♾️ Unbounded Intervals
- The set of real numbers extends on the number line in the positive direction to infinity \((\infty)\) and extends in the negative direction to negative infinity \((-\infty)\).
- The symbols \(\infty\) and \(-\infty\) are used to express unbounded sets as follows:
If \(a \in \mathbb{R}\), then:
The set of real numbers greater than or equal to \(a\)
\([a, \infty[\) = \(\{x : x \in \mathbb{R}, x \ge a\}\)

The set of real numbers less than or equal to \(a\)
\(]-\infty, a]\) = \(\{x : x \in \mathbb{R}, x \le a\}\)
![Number line for the unbounded interval (-infinity, a] by Math Egypt - Mr. Ayman Hassan (خط الأعداد للفترة غير المحدودة (-∞، أ] من ماث إيجيبت - مستر أيمن حسن)](https://i.postimg.cc/RFCST92Q/8.png)
The set of real numbers greater than \(a\)
\(]a, \infty[\) = \(\{x : x \in \mathbb{R}, x > a\}\)

The set of real numbers less than \(a\)
\(]-\infty, a[\) = \(\{x : x \in \mathbb{R}, x < a\}\)

Note that:
The interval \(]-\infty, \infty[\) represents the set of real numbers \(\mathbb{R}\), and \(\infty \notin \mathbb{R}\) and \(-\infty \notin \mathbb{R}\).
📝 Example (2)
Solution
Solution ①
\([0, \infty[\)

Solution ②
\( ]-\infty, 0] \)
![Number line for the set of non-positive real numbers (-infinity, 0] by Math Egypt - Mr. Ayman Hassan (خط الأعداد لمجموعة الأعداد الحقيقية غير الموجبة (-∞، ٠] من ماث إيجيبت - مستر أيمن حسن)](https://i.postimg.cc/Fz5xqrX8/2.png)
Solution ③
\([-2, \infty[\)

Solution ④
\( ]-\infty, 1[\)

📘 Notes on Real Number Sets
Note that:
-
•
\( \mathbb{R} = \mathbb{R}^+ \cup \{0\} \cup \mathbb{R}^- \)Where:
\(\mathbb{R}^+\) is the set of positive real numbers.
\(\mathbb{R}^-\) is the set of negative real numbers. -
•
The set of non-negative real numbers = \( \mathbb{R}^+ \cup \{0\} \)
-
•
The set of non-positive real numbers = \( \mathbb{R}^- \cup \{0\} \)
📐 Operations on Intervals
Using the number line, find each of the following:
- 1. \( [-3, 5] \cup [2, 7[ \)
- 2. \( \mathopen{]}1, 5] \cap \mathopen{]}-\infty, 3] \)
- 3. \( \mathopen{]}-\infty, 6] - \mathopen{]}4, \infty[ \)
- 4. \( \mathopen{]}-\infty, 3[' \)
Solution
Solution (1)

Solution (2)

Solution (3)

Solution (4)

⭐ Example (4): Choose the Correct Answer
1. \(4 \in \text{...........}\)
Correct Answer: (c). The number 4 is greater than 2 and less than 5.
2. \(\sqrt[3]{-8} \text{...........} [-8, -2[\)
Correct Answer: (b). Because \(\sqrt[3]{-8} = -2\), and the interval \( [-8, -2[ \) is open at -2, which means -2 is not included in the interval.
3. \(\{1, 6\} \text{...........} ]1, 6]\)
Correct Answer: (d). For a set to be a subset, all its elements must belong to the other set. Here, \(1 \notin ]1, 6]\) because the interval is open at 1.
4. If \(x \in [-5, \infty[\), then ...........
Correct Answer: (b). The interval \( [-5, \infty[ \) includes all real numbers greater than or equal to -5.
5. The sum of the real numbers in the interval \( [-3, 3[ \) is ...........
Correct Answer: (b). For every number \(x\) in the interval \( (0, 3) \), its additive inverse \(-x\) is also in the interval \( [-3, 0) \). These pairs sum to zero. The number 0 is also in the interval. The only number remaining that does not have its additive inverse within the interval is -3. Therefore, the conceptual sum is -3.
📝 Example (5): Operations on Intervals
If \(X = [-3, 3]\) and \(Y = [-1, 5[\), find using the number line:
Solution

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